Chapter 9 Some Applications of Trigonometry (Additional Questions)

The angle formed by the line of sight with the horizontal when the object viewed is below the horizontal level is defined as the Angle of Depression.

---

Therefore, the correct option is (B) Angle of Depression.

The line joining the observer's eye to the object being viewed (in this case, the top of a tree) is known as the line of sight.

---

Therefore, the correct option is (C) Line of sight.

The angle of elevation is the angle formed between the horizontal level through the observer's eye and the line of sight to an object above the horizontal level.

---

In this case, the boy is the observer, the bird is the object, and the angle of elevation is formed by the horizontal line from the boy's eye and the line joining the boy's eye to the bird (the line of sight).

---

Therefore, the correct option is (B) The horizontal level and the line of sight.

The angle of depression is defined as the angle formed between the horizontal level through the observer's eye and the line of sight when the object being viewed is below the horizontal level.

---

In this scenario, the person on top of the building is the observer, and the car is the object below the horizontal level. The angle of depression is formed by the horizontal line extending from the person's eye and the line connecting the person's eye to the car (the line of sight).

---

Therefore, the correct option is (C) The horizontal level through the person's eye and the line of sight.

Let the height of the pole be $h$ and the length of the shadow be $s$. The angle of elevation of the sun be $\theta$.

We are given:

Height of the pole, $h = 10$ m

Length of the shadow, $s = 10$ m

---

The pole, its shadow, and the line of sight from the tip of the shadow to the top of the pole form a right-angled triangle.

In this right-angled triangle, the height of the pole is the side opposite to the angle of elevation, and the length of the shadow is the side adjacent to the angle of elevation.

We can use the tangent trigonometric ratio:

$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} $

$ \tan(\theta) = \frac{h}{s} $

---

Substitute the given values:

$ \tan(\theta) = \frac{10 \text{ m}}{10 \text{ m}} $

$ \tan(\theta) = 1 $

---

To find the angle $\theta$, we need to find the angle whose tangent is 1.

We know that $ \tan(45^\circ) = 1 $.

Therefore,

$ \theta = 45^\circ $

---

The angle of elevation of the sun is $45^\circ$.

---

Therefore, the correct option is (B) $45^\circ$.

Given:

Length of the ladder = 13 m

Distance of the foot of the ladder from the wall = 5 m

---

To Find:

The angle of elevation of the top of the ladder.

---

Solution:

Let the ladder be represented by $L$. The length of the ladder is the hypotenuse of a right-angled triangle formed by the wall, the ground, and the ladder.

Let the height the ladder reaches on the wall be $h$. Let the distance of the foot of the ladder from the wall be $d$. Let the angle of elevation be $\theta$.

We have $L = 13$ m and $d = 5$ m.

The angle of elevation $\theta$ is formed at the foot of the ladder, between the ground (horizontal) and the ladder (line of sight to the top).

In the right-angled triangle, the distance from the wall ($d$) is the side adjacent to the angle of elevation $\theta$, and the length of the ladder ($L$) is the hypotenuse.

We can use the cosine trigonometric ratio:

$ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} $

$ \cos(\theta) = \frac{d}{L} $

---

Substitute the given values:

$ \cos(\theta) = \frac{5}{13} $

---

To find the angle $\theta$, we take the inverse cosine (arccosine) of $\frac{5}{13}$.

$ \theta = \cos^{-1}\left(\frac{5}{13}\right) $

---

Using a calculator, we find the approximate value:

$ \theta \approx \cos^{-1}(0.3846) $

$ \theta \approx 67.38^\circ $

---

Comparing this value with the given options, we find that option (C) is the closest approximation.

---

Therefore, the correct option is (C) $67.38^\circ$ (approx.).

Given:

Angle of elevation of the top of the tower from a point on the ground, $\theta = 30^\circ$.

Distance of the point from the foot of the tower, $d = 30$ m.

---

To Find:

The height of the tower, $h$.

---

Solution:

Let the tower be represented by a vertical line segment. The point on the ground, the foot of the tower, and the top of the tower form a right-angled triangle. The height of the tower is the side opposite the angle of elevation, and the distance from the foot of the tower is the side adjacent to the angle of elevation.

We can use the tangent trigonometric ratio, which relates the angle of elevation to the opposite and adjacent sides:

$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} $

$ \tan(\theta) = \frac{h}{d} $

---

Substitute the given values:

$ \tan(30^\circ) = \frac{h}{30} $

---

We know the value of $\tan(30^\circ)$ is $\frac{1}{\sqrt{3}}$.

$ \frac{1}{\sqrt{3}} = \frac{h}{30} $

---

Now, we can solve for $h$ by multiplying both sides by 30:

$ h = 30 \times \frac{1}{\sqrt{3}} $

$ h = \frac{30}{\sqrt{3}} $

---

The height of the tower is $\frac{30}{\sqrt{3}}$ m.

---

Therefore, the correct option is (B) $\frac{30}{\sqrt{3}}$ m.

Given:

Height of the building, $H = 50$ m.

Angle of depression of a car from the top of the building, $\delta = 45^\circ$.

---

To Find:

The distance of the car from the building, $D$.

---

Solution:

Let the top of the building be $T$, the foot of the building be $F$, and the position of the car on the road be $C$. The building is represented by the vertical line segment $TF$. The height of the building is $TF = 50$ m.

The angle of depression from the top of the building ($T$) to the car ($C$) is the angle between the horizontal line through $T$ and the line of sight $TC$. Let the horizontal line through $T$ be $TH$, parallel to the ground $FC$. The angle of depression is $\angle HTC = 45^\circ$.

Since the horizontal line $TH$ is parallel to the ground $FC$, the angle of depression $\angle HTC$ is equal to the angle of elevation $\angle TCF$ (alternate interior angles).

So, the angle of elevation from the car to the top of the building is $\angle TCF = 45^\circ$.

Now consider the right-angled triangle $\triangle TFC$, where $\angle TFC = 90^\circ$ (assuming the building is vertical and the road is horizontal).

In $\triangle TFC$, we have:

The side opposite to the angle $\angle TCF$ is the height of the building $TF = H = 50$ m.

The side adjacent to the angle $\angle TCF$ is the distance of the car from the building $FC = D$.

We can use the tangent trigonometric ratio:

$ \tan(\angle TCF) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

$ \tan(45^\circ) = \frac{TF}{FC} $

$ \tan(45^\circ) = \frac{50}{D} $

---

We know that the value of $\tan(45^\circ)$ is 1.

$ 1 = \frac{50}{D} $

---

Multiply both sides by $D$ to solve for $D$:

$ D = 50 $

The distance of the car from the building is 50 m.

---

Therefore, the correct option is (A) 50 m.

Given:

Angle of elevation of the top of the tower from a point on the ground, $\theta = 60^\circ$.

Height of the tower, $h = 75$ m.

---

To Find:

The distance of the point from the foot of the tower, $d$.

---

Solution:

Let the tower be represented by a vertical line segment. The point on the ground, the foot of the tower, and the top of the tower form a right-angled triangle. The height of the tower is the side opposite the angle of elevation, and the distance from the foot of the tower is the side adjacent to the angle of elevation.

We can use the tangent trigonometric ratio:

$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} $

$ \tan(60^\circ) = \frac{h}{d} $

---

Substitute the given value of $h$ and the value of $\tan(60^\circ)$, which is $\sqrt{3}$:

$ \sqrt{3} = \frac{75}{d} $

---

Now, we need to solve for $d$. Multiply both sides by $d$ and then divide by $\sqrt{3}$:

$ d \sqrt{3} = 75 $

$ d = \frac{75}{\sqrt{3}} $

---

The distance of the point from the foot of the tower is $\frac{75}{\sqrt{3}}$ m.

---

Therefore, the correct option is (C) $\frac{75}{\sqrt{3}}$ m.

Given:

Angle of elevation of the top of a tower from point P on the ground, $\theta = 30^\circ$.

Height of the tower, $H = h$.

---

To Find:

The distance of point P from the foot of the tower, $D$.

---

Solution:

Let the tower be represented by a vertical line segment, and let its foot be at point F. Let P be the point on the ground. The tower, the ground, and the line of sight from P to the top of the tower form a right-angled triangle, say $\triangle PFT$, where T is the top of the tower.

In the right-angled triangle $\triangle PFT$, the height of the tower $FT$ is the side opposite to the angle of elevation $\angle FPT = \theta = 30^\circ$. The distance of P from the foot of the tower $PF$ is the side adjacent to the angle of elevation.

We can use the tangent trigonometric ratio, which relates the angle of elevation to the opposite and adjacent sides:

$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} $

---

Substitute the given height $H=h$ and the known value of $\tan(30^\circ)$, which is $\frac{1}{\sqrt{3}}$:

$ \frac{1}{\sqrt{3}} = \frac{h}{D} $

---

Now, we need to solve the equation for $D$. We can cross-multiply or take the reciprocal of both sides:

$ D \times 1 = h \times \sqrt{3} $

$ D = h\sqrt{3} $

---

The distance of point P from the foot of the tower is $h\sqrt{3}$.

---

Therefore, the correct option is (B) $h\sqrt{3}$.

Let's analyze the Assertion (A) and the Reason (R) separately.

---

Assertion (A): If an observer is looking at an object above the horizontal, the angle of elevation is positive.

The angle of elevation is defined for an object that is above the observer's horizontal level. This angle is measured upwards from the horizontal line of sight. By convention, angles measured upwards from a reference line are considered positive. Therefore, the Assertion (A) is True.

---

Reason (R): Angles of elevation are measured upwards from the horizontal line.

This is the standard definition of the angle of elevation. It is precisely the angle formed between the horizontal line through the observer's eye and the line of sight to the object, when the object is above the horizontal. The measurement is explicitly defined as being upwards from the horizontal. Therefore, the Reason (R) is True.

---

Now, let's consider if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that the angle of elevation is positive when the object is above the horizontal. Reason (R) states how the angle of elevation is measured – upwards from the horizontal. Because the angle is measured upwards from the horizontal line, when the object is above this level, the resulting angle is positive. Thus, the way the angle of elevation is defined and measured (as stated in R) directly leads to it being a positive angle when the object is above the horizontal (as stated in A).

Therefore, Reason (R) is the correct explanation for Assertion (A).

---

Based on the analysis:

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains why Assertion (A) is true.

---

Therefore, the correct option is (A) Both A and R are true and R is the correct explanation of A.

Let's analyze the Assertion (A) and the Reason (R) separately.

---

Assertion (A): The angle of elevation of an object from an observer is always equal to the angle of depression of the observer from the object.

Consider an observer at point A looking up at an object at point B. The angle of elevation is the angle between the horizontal line through A and the line of sight AB. Now, consider the object at point B looking down at the observer at point A. The angle of depression is the angle between the horizontal line through B and the line of sight BA. Geometrically, these two angles are equal.

Therefore, Assertion (A) is True.

---

Reason (R): The horizontal lines at the observer's level and the object's level are parallel, and the line of sight acts as a transversal, forming equal alternate interior angles.

Let the horizontal line through the observer (A) be $H_A$ and the horizontal line through the object (B) be $H_B$. Since both are horizontal lines in the same plane, $H_A$ is parallel to $H_B$. The line of sight AB connects the observer and the object and intersects both horizontal lines. Thus, the line of sight AB acts as a transversal. The angle of elevation at A and the angle of depression at B are indeed alternate interior angles formed by the parallel lines $H_A$ and $H_B$ and the transversal AB.

According to the properties of parallel lines intersected by a transversal, alternate interior angles are equal.

Therefore, Reason (R) is True.

---

Now, let's evaluate if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states the equality of the angle of elevation and the angle of depression in this context. Reason (R) provides the geometrical justification for this equality by identifying the relevant parallel lines, the transversal, and the property of alternate interior angles. The fact that these angles are equal directly follows from the geometric configuration described in Reason (R).

Thus, Reason (R) correctly explains why Assertion (A) is true.

---

Based on the analysis:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

---

Therefore, the correct option is (A) Both A and R are true and R is the correct explanation of A.

Let's recall the definitions of the trigonometric ratios in a right-angled triangle with respect to an acute angle $\theta$:

$ \sin(\theta) = \frac{\text{Opposite side}}{\text{Hypotenuse}} $

$ \cos(\theta) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} $

$ \tan(\theta) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

$ \text{sec}(\theta) = \frac{\text{Hypotenuse}}{\text{Adjacent side}} $

---

Now let's match each scenario from Column A to the relevant ratio in Column B:

(i) Given angle and adjacent side, find opposite side:

We need a ratio that involves the angle, the adjacent side (given), and the opposite side (to be found). The tangent ratio relates the opposite and adjacent sides.

$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} $

Thus, if $\theta$ and the adjacent side are known, we can find the opposite side using the tangent. So, (i) matches with (c) Tangent.

---

(ii) Given angle and opposite side, find hypotenuse:

We need a ratio that involves the angle, the opposite side (given), and the hypotenuse (to be found). The sine ratio relates the opposite side and the hypotenuse.

$ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} $

Thus, if $\theta$ and the opposite side are known, we can find the hypotenuse using the sine. So, (ii) matches with (b) Sine.

---

(iii) Given angle and adjacent side, find hypotenuse:

We need a ratio that involves the angle, the adjacent side (given), and the hypotenuse (to be found). The cosine ratio relates the adjacent side and the hypotenuse.

$ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} $

Thus, if $\theta$ and the adjacent side are known, we can find the hypotenuse using the cosine. So, (iii) matches with (a) Cosine.

---

(iv) Given angle and opposite side, find adjacent side:

We need a ratio that involves the angle, the opposite side (given), and the adjacent side (to be found). The tangent ratio relates the opposite and adjacent sides.

$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} $

Thus, if $\theta$ and the opposite side are known, we can rearrange the tangent formula to find the adjacent side:

$ \text{Adjacent} = \frac{\text{Opposite}}{\tan(\theta)} $

So, (iv) matches with (c) Tangent.

---

The matches are:

(i) - (c)

(ii) - (b)

(iii) - (a)

(iv) - (c)

---

Comparing these matches with the given options:

(A) (i)-(c), (ii)-(b), (iii)-(a), (iv)-(c) - Matches our findings.

(B) (i)-(c), (ii)-(a), (iii)-(b), (iv)-(c) - Incorrect.

(C) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(a) - Incorrect.

(D) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b) - Incorrect.

---

Therefore, the correct option is (A) (i)-(c), (ii)-(b), (iii)-(a), (iv)-(c).

Let's examine each statement:

---

(A) The angle of elevation is always measured from the observer's eye level.

The angle of elevation is defined relative to the horizontal line passing through the observer's eye. This is the standard convention used in trigonometry for angle of elevation problems. This statement is True.

---

(B) The angle of depression is always measured from a horizontal line above the observer's eye level.

When determining the angle of depression, the observer is typically at a higher point looking downwards. The horizontal reference line is drawn at the level of the observer's eye. The angle of depression is the angle between this horizontal line and the line of sight downwards to the object. This statement is True.

---

(C) If you walk towards the base of a tower, the angle of elevation to the top increases.

Let $h$ be the height of the tower and $d$ be the horizontal distance from the observer to the base of the tower. The angle of elevation $\theta$ is related by $ \tan(\theta) = \frac{h}{d} $. If you walk towards the base, $d$ decreases. Since $h$ is constant, $\frac{h}{d}$ increases as $d$ decreases. For acute angles, the tangent function is increasing. Therefore, if $ \tan(\theta) $ increases, the angle $ \theta $ also increases. This statement is True.

---

(D) If you walk away from the base of a tower, the angle of depression from the top to your position increases.

Let the observer be at the top of the tower (height $h$) and you are at a distance $d$ from the base on the ground. The angle of depression $\delta$ from the top of the tower to your position on the ground is equal to the angle of elevation $\theta$ from your position on the ground to the top of the tower (alternate interior angles). The relationship is $ \tan(\delta) = \tan(\theta) = \frac{h}{d} $. If you walk away from the base, the distance $d$ increases. Since $h$ is constant, $\frac{h}{d}$ decreases as $d$ increases. For acute angles, the tangent function is increasing, so if $ \tan(\delta) $ decreases, the angle $ \delta $ also decreases. This means the angle of depression decreases as you walk away from the base. The statement says the angle of depression increases, which is False.

---

Based on the analysis, the false statement is (D).

---

Therefore, the correct option is (D) If you walk away from the base of a tower, the angle of depression from the top to your position increases.

Given:

Angle made by the broken part of the pole with the ground, $\theta = 30^\circ$.

Distance of the foot of the pole from the point where the top touches the ground, $d = 8$ m.

---

To Find:

The height of the pole before it was broken, $H$.

---

Solution:

Let the vertical pole be $AF$, where $A$ is the top and $F$ is the foot. Let the pole be broken at a point $B$, such that $BF$ is the unbroken lower part and $AB$ is the broken upper part.

After breaking, the top $A$ touches the ground at a point $G$, such that $FG = 8$ m. The broken part $AB$ now becomes $BG$. Thus, $AB = BG$.

The point $B$ is at a height $BF$ above the ground. The triangle $\triangle BFG$ is a right-angled triangle, with the right angle at $F$ (assuming the pole was vertical and the ground is horizontal).

In $\triangle BFG$:

The angle $\angle BGF = 30^\circ$ (given).

The side $FG = 8$ m (adjacent to the angle $30^\circ$).

The side $BF$ is the height of the unbroken part (opposite to the angle $30^\circ$). Let $BF = y$.

The side $BG$ is the broken part of the pole (hypotenuse). Let $BG = x$.

---

The original height of the pole before it was broken is the sum of the unbroken part and the broken part:

$ H = BF + AB $

Since $AB = BG$, we have:

$ H = y + x $

---

Using trigonometric ratios in the right-angled triangle $\triangle BFG$ with angle $30^\circ$ and adjacent side 8 m:

To find the opposite side ($y$):

$ \tan(30^\circ) = \frac{BF}{FG} = \frac{y}{8} $

$ \frac{1}{\sqrt{3}} = \frac{y}{8} $

$ y = \frac{8}{\sqrt{3}} $

---

To find the hypotenuse ($x$):

$ \cos(30^\circ) = \frac{FG}{BG} = \frac{8}{x} $

$ \frac{\sqrt{3}}{2} = \frac{8}{x} $

$ x = \frac{8}{\frac{\sqrt{3}}{2}} = 8 \times \frac{2}{\sqrt{3}} = \frac{16}{\sqrt{3}} $

---

Now, calculate the total height $H = y + x$:

$ H = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} $

$ H = \frac{8 + 16}{\sqrt{3}} $

$ H = \frac{24}{\sqrt{3}} $

---

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$:

$ H = \frac{24}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} $

$ H = \frac{24\sqrt{3}}{3} $

$ H = 8\sqrt{3} $

---

The height of the pole before it was broken is $8\sqrt{3}$ m.

---

Comparing this with the given options:

(A) $8\sqrt{3}$ m

(B) $\frac{8}{\sqrt{3}}$ m

(C) $8 + \frac{8}{\sqrt{3}}$ m

(D) $8 + \frac{16}{\sqrt{3}}$ m

Our calculated height matches option (A).

---

Therefore, the correct option is (A) $8\sqrt{3}$ m.

Given:

Height of the observer = 1.5 m.

Distance of the observer from the chimney = 28.5 m.

Angle of elevation of the top of the chimney from the observer's eyes, $\theta = 45^\circ$.

---

To Find:

The height of the chimney.

---

Solution:

Let the height of the observer be $h_o = 1.5$ m. Let the distance of the observer from the chimney be $d = 28.5$ m.

Consider a right-angled triangle formed by the horizontal line from the observer's eyes to the chimney, the vertical line representing the upper part of the chimney above the observer's eye level, and the line of sight from the observer's eyes to the top of the chimney.

Let $x$ be the height of the chimney above the observer's eye level.

In the right-angled triangle, the horizontal distance is the adjacent side ($d = 28.5$ m), the height above the observer's eye level is the opposite side ($x$), and the angle of elevation is $\theta = 45^\circ$.

We use the tangent trigonometric ratio:

$ \tan(\theta) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

$ \tan(45^\circ) = \frac{x}{d} $

---

Substitute the given values:

$ \tan(45^\circ) = \frac{x}{28.5} $

---

We know that $ \tan(45^\circ) = 1 $.

$ 1 = \frac{x}{28.5} $

Multiply both sides by 28.5:

$ x = 28.5 $ m

---

This value $x$ is the height of the chimney above the observer's eye level. To find the total height of the chimney, we need to add the observer's height:

Total height of chimney, $H = x + h_o$

$ H = 28.5 \text{ m} + 1.5 \text{ m} $

$ H = 30.0 \text{ m} $

---

The height of the chimney is 30 m.

---

Therefore, the correct option is (A) 30 m.

Given:

Initial angle of elevation from point P to the top of the tower T, $\angle TPA = 30^\circ$.

Distance moved towards the tower, PQ = 20 m.

New angle of elevation from point Q to the top of the tower T, $\angle TQA = 60^\circ$.

---

To Find:

The height of the tower, TA = $h$.

---

Solution:

Let the height of the tower be TA = $h$ and the distance from the initial point P to the foot of the tower A be PA = $x$. The point Q is between P and A, such that PQ = 20 m. The distance from Q to A is QA = PA - PQ = $x - 20$.

Consider the right-angled triangle $\triangle TPA$. The angle of elevation is $30^\circ$.

$ \tan(30^\circ) = \frac{TA}{PA} $

$ \frac{1}{\sqrt{3}} = \frac{h}{x} $

---

Now consider the right-angled triangle $\triangle TQA$. The angle of elevation is $60^\circ$. The distance QA is $x - 20$.

$ \tan(60^\circ) = \frac{TA}{QA} $

$ \sqrt{3} = \frac{h}{x - 20} $

---

Now we have a system of two equations with two variables, $h$ and $x$. We can substitute the value of $x$ from equation (i) into equation (ii):

$ \sqrt{3}((h\sqrt{3}) - 20) = h $

Distribute $\sqrt{3}$:

$ \sqrt{3} \times h\sqrt{3} - 20\sqrt{3} = h $

$ 3h - 20\sqrt{3} = h $

---

Now, solve for $h$. Subtract $h$ from both sides and add $20\sqrt{3}$ to both sides:

$ 3h - h = 20\sqrt{3} $

$ 2h = 20\sqrt{3} $

---

Divide both sides by 2:

$ h = \frac{20\sqrt{3}}{2} $

$ h = 10\sqrt{3} $ m

---

The height of the tower is $10\sqrt{3}$ m.

---

Therefore, the correct option is (B) $10 \sqrt{3}$ m.

Given:

Initial angle of elevation from point P to the top of the tower T, $\angle TPA = 30^\circ$.

Distance moved towards the tower, PQ = 20 m.

New angle of elevation from point Q to the top of the tower T, $\angle TQA = 60^\circ$.

---

To Find:

The height of the tower, TA = $h$.

---

Solution:

Let the height of the tower be TA = $h$. Let the initial distance from the point P on the ground to the foot of the tower A be PA = $x$. The point Q is located 20 m closer to the tower from P, so the distance from Q to A is QA = PA - PQ = $x - 20$.

Consider the right-angled triangle $\triangle TPA$. The angle of elevation from P is $30^\circ$.

Using the tangent ratio:

$ \tan(\angle TPA) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

$ \tan(30^\circ) = \frac{TA}{PA} $

$ \frac{1}{\sqrt{3}} = \frac{h}{x} $

---

Now consider the right-angled triangle $\triangle TQA$. The angle of elevation from Q is $60^\circ$. The distance from Q to A is $x - 20$.

Using the tangent ratio:

$ \tan(\angle TQA) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

$ \tan(60^\circ) = \frac{TA}{QA} $

$ \sqrt{3} = \frac{h}{x - 20} $

---

We now have a system of two equations with two variables ($h$ and $x$). Substitute the expression for $x$ from equation (i) into equation (ii):

$ \sqrt{3}((h\sqrt{3}) - 20) = h $

Distribute the $\sqrt{3}$ on the left side:

$ \sqrt{3} \times h\sqrt{3} - \sqrt{3} \times 20 = h $

$ (\sqrt{3} \times \sqrt{3})h - 20\sqrt{3} = h $

$ 3h - 20\sqrt{3} = h $

---

Now, rearrange the equation to solve for $h$. Subtract $h$ from both sides and add $20\sqrt{3}$ to both sides:

$ 3h - h = 20\sqrt{3} $

$ 2h = 20\sqrt{3} $

---

Divide both sides by 2:

$ h = \frac{20\sqrt{3}}{2} $

$ h = 10\sqrt{3} $ m

---

The height of the tower is $10\sqrt{3}$ m.

---

Comparing this result with the given options:

(A) 10 m

(B) $10 \sqrt{3}$ m

(C) 20 m

(D) $\frac{20}{\sqrt{3}}$ m

Our calculated height matches option (B).

---

Therefore, the correct option is (B) $10 \sqrt{3}$ m.

Given:

Height of the lighthouse, $h = 75$ m.

Angle of depression of a boat from the top of the lighthouse, $\delta = 30^\circ$.

---

To Find:

The distance of the boat from the foot of the lighthouse, $d$.

---

Solution:

Let the top of the lighthouse be T, the foot of the lighthouse be F, and the position of the boat be B. The lighthouse is represented by the vertical line segment TF. The height of the lighthouse is $TF = h = 75$ m.

The angle of depression from the top of the lighthouse (T) to the boat (B) is the angle between the horizontal line through T and the line of sight TB. Let the horizontal line through T be TH, parallel to the ground FB. The angle of depression is $\angle HTB = 30^\circ$.

Since the horizontal line TH is parallel to the ground FB, and the line of sight TB is a transversal, the angle of depression $\angle HTB$ is equal to the angle of elevation from the boat to the top of the lighthouse, $\angle TBF$ (alternate interior angles).

So, the angle of elevation from the boat to the top of the lighthouse is $\angle TBF = 30^\circ$.

Now consider the right-angled triangle $\triangle TFB$, where $\angle TFB = 90^\circ$ (assuming the lighthouse is vertical and the ground/sea level is horizontal).

In $\triangle TFB$, we have:

The side opposite to the angle $\angle TBF = 30^\circ$ is the height of the lighthouse $TF = h = 75$ m.

The side adjacent to the angle $\angle TBF = 30^\circ$ is the distance of the boat from the foot of the lighthouse $FB = d$.

We can use the tangent trigonometric ratio:

$ \tan(\angle TBF) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

---

Substitute the given values into equation (i):

$ \tan(30^\circ) = \frac{75}{d} $

---

We know that the value of $ \tan(30^\circ) $ is $ \frac{1}{\sqrt{3}} $.

$ \frac{1}{\sqrt{3}} = \frac{75}{d} $

---

To solve for $d$, we can cross-multiply:

$ 1 \times d = 75 \times \sqrt{3} $

$ d = 75\sqrt{3} $ m

---

The distance of the boat from the foot of the lighthouse is $75\sqrt{3}$ m.

---

Comparing this result with the given options:

(A) 75 m

(B) $75 \sqrt{3}$ m

(C) $\frac{75}{\sqrt{3}}$ m

(D) 150 m

Our calculated distance matches option (B).

---

Therefore, the correct option is (B) $75 \sqrt{3}$ m.

Given:

Let the height of the tower be $h$.

Let the two points on the ground in the same straight line with the base of the tower be P and Q.

Distance of point P from the base of the tower = $a$.

Distance of point Q from the base of the tower = $b$.

The angles of elevation from P and Q to the top of the tower are complementary.

---

To Find:

The height of the tower, $h$.

---

Solution:

Let the top of the tower be T and the foot of the tower be A. So, TA = $h$. Let the points P and Q be on the line passing through A. Assume P and Q are on the same side of the tower, or on opposite sides. The problem statement "in the same straight line with it" and the context of angles of elevation usually implies the points are on the same side. Let's assume P and Q are on the same side of the base A, and the distances are measured from A.

Let the angle of elevation from point P to the top of the tower be $\alpha$ and the angle of elevation from point Q to the top of the tower be $\beta$.

We are given that the angles are complementary, which means their sum is $90^\circ$.

From equation (i), we can write $\beta = 90^\circ - \alpha$ or $\alpha = 90^\circ - \beta$.

---

Consider the right-angled triangle $\triangle TAP$. The angle of elevation from P is $\alpha$, and the distance PA = $a$.

Using the tangent trigonometric ratio:

$ \tan(\alpha) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{TA}{PA} $

---

Consider the right-angled triangle $\triangle TAQ$. The angle of elevation from Q is $\beta$, and the distance QA = $b$.

Using the tangent trigonometric ratio:

$ \tan(\beta) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{TA}{QA} $

---

Now use the complementary angle relationship from equation (i). Substitute $\beta = 90^\circ - \alpha$ into equation (iii):

$ \tan(90^\circ - \alpha) = \frac{h}{b} $

We know the trigonometric identity $ \tan(90^\circ - \alpha) = \cot(\alpha) $.

$ \cot(\alpha) = \frac{h}{b} $

---

We also know that $ \cot(\alpha) = \frac{1}{\tan(\alpha)} $. So, from equation (ii), $ \cot(\alpha) = \frac{1}{\frac{h}{a}} = \frac{a}{h} $.

Equating the two expressions for $ \cot(\alpha) $:

$ \frac{a}{h} = \frac{h}{b} $

---

Cross-multiply the equation:

$ a \times b = h \times h $

$ ab = h^2 $

---

Take the square root of both sides to solve for $h$. Since height must be positive, we consider the positive square root:

$ h = \sqrt{ab} $

---

The height of the tower is $ \sqrt{ab} $.

---

Comparing this result with the given options:

(A) $ab$

(B) $\sqrt{ab}$

(C) $\frac{a+b}{2}$

(D) $a+b$

Our calculated height matches option (B).

---

Therefore, the correct option is (B) $\sqrt{ab}$.

Given:

Let the height of the building be $h_b$.

Let the height of the tower be $h_t = 50$ m.

Let the distance between the foot of the building and the foot of the tower be $d$.

Angle of elevation of the top of the building from the foot of the tower = $30^\circ$.

Angle of elevation of the top of the tower from the foot of the building = $60^\circ$.

---

To Find:

The height of the building, $h_b$.

---

Solution:

Let the foot of the building be A and the top of the building be B. Let the foot of the tower be C and the top of the tower be D. Assume A and C are points on the ground, and the building and tower are vertical. The distance between A and C is $d$. The height of the building is AB = $h_b$, and the height of the tower is CD = $h_t = 50$ m.

The angle of elevation of the top of the building (B) from the foot of the tower (C) is $\angle ACB = 30^\circ$. Consider the right-angled triangle $\triangle ABC$, with the right angle at A.

Using the tangent trigonometric ratio in $\triangle ABC$:

$ \tan(30^\circ) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{AB}{AC} $

---

The angle of elevation of the top of the tower (D) from the foot of the building (A) is $\angle CAD = 60^\circ$. Consider the right-angled triangle $\triangle ACD$, with the right angle at C.

Using the tangent trigonometric ratio in $\triangle ACD$:

$ \tan(60^\circ) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{CD}{AC} $

---

We are given $h_t = 50$ m. Substitute this value into equation (ii):

$ \tan(60^\circ) = \frac{50}{d} $

We know that $ \tan(60^\circ) = \sqrt{3} $.

$ \sqrt{3} = \frac{50}{d} $

---

Solve this equation for the distance $d$:

$ d = \frac{50}{\sqrt{3}} $

---

Now, substitute the value of $d$ from equation (iii) into equation (i):

$ \tan(30^\circ) = \frac{h_b}{d} $

$ \frac{1}{\sqrt{3}} = \frac{h_b}{\frac{50}{\sqrt{3}}} $

---

Simplify the right side of the equation:

$ \frac{1}{\sqrt{3}} = h_b \times \frac{\sqrt{3}}{50} $

---

Now, solve for $h_b$. Multiply both sides by 50 and divide by $\sqrt{3}$ (or multiply by $\frac{50}{\sqrt{3}}$):

$ \frac{1}{\sqrt{3}} \times \frac{50}{\sqrt{3}} = h_b $

$ h_b = \frac{50}{\sqrt{3} \times \sqrt{3}} $

$ h_b = \frac{50}{3} $

---

The height of the building is $\frac{50}{3}$ m.

---

Comparing this result with the given options:

(A) $\frac{50}{\sqrt{3}}$ m

(B) $50 \sqrt{3}$ m

(C) $\frac{50}{3}$ m

(D) $50/2 = 25$ m

Our calculated height matches option (C).

---

Therefore, the correct option is (C) $\frac{50}{3}$ m.

Given:

Angle made by the broken part of the tree with the ground, $\theta = 30^\circ$.

Distance from the foot of the tree to the point where the top touches the ground, $d = 8$ m.

---

To Find:

The height of the part of the tree that broke.

---

Solution:

Let the tree be represented by a vertical line. When it breaks, the unbroken lower part, the distance on the ground, and the broken upper part form a right-angled triangle.

Let the point where the tree broke be B, the foot of the tree be F, and the point where the top touches the ground be G.

The angle made by the broken part with the ground is $\angle BGF = 30^\circ$.

The distance from the foot of the tree to the point where the top touches the ground is $FG = 8$ m.

The height of the part that broke is the length of the segment BG. Let this length be $x$. In the right-angled triangle $\triangle BFG$ (right-angled at F), BG is the hypotenuse.

We have the angle $\angle BGF = 30^\circ$, the adjacent side $FG = 8$ m, and we want to find the hypotenuse $BG = x$.

The trigonometric ratio that relates the adjacent side and the hypotenuse is cosine:

$ \cos(\theta) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} $

$ \cos(30^\circ) = \frac{FG}{BG} $

---

We know that the value of $ \cos(30^\circ) $ is $ \frac{\sqrt{3}}{2} $.

Substitute this value into equation (i):

$ \frac{\sqrt{3}}{2} = \frac{8}{x} $

---

Now, we solve for $x$. Cross-multiply:

$ x \times \sqrt{3} = 8 \times 2 $

$ x\sqrt{3} = 16 $

---

Divide both sides by $\sqrt{3}$:

$ x = \frac{16}{\sqrt{3}} $ m

---

The height of the part that broke is $\frac{16}{\sqrt{3}}$ m.

---

Comparing this result with the given options:

(A) 8 m

(B) $8\sqrt{3}$ m

(C) 16 m

(D) $\frac{16}{\sqrt{3}}$ m

Our calculated height matches option (D).

---

Therefore, the correct option is (D) $\frac{16}{\sqrt{3}}$ m.

Given:

Let the height of the building be $h$.

Let A and B be two points on the ground in a straight line with the foot of the building.

Angle of elevation of the top of the building from point A = $30^\circ$.

Distance moved from A towards the building to point B = 10 m.

Angle of elevation of the top of the building from point B = $45^\circ$.

---

To Find:

The height of the building, $h$.

---

Solution:

Let the top of the building be T and the foot of the building be C. So, TC = $h$. The points A and B are on the ground such that A, B, and C are collinear, and B is between A and C.

The distance AB = 10 m.

Let the distance from point B to the foot of the building be BC = $x$.

Then, the distance from point A to the foot of the building is AC = AB + BC = $10 + x$.

---

Consider the right-angled triangle $\triangle TBC$. The angle of elevation from B is $\angle TBC = 45^\circ$. The opposite side is TC = $h$, and the adjacent side is BC = $x$.

Using the tangent trigonometric ratio:

$ \tan(\angle TBC) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

$ \tan(45^\circ) = \frac{TC}{BC} $

Since $ \tan(45^\circ) = 1 $:

$ 1 = \frac{h}{x} $

---

Consider the right-angled triangle $\triangle TAC$. The angle of elevation from A is $\angle TAC = 30^\circ$. The opposite side is TC = $h$, and the adjacent side is AC = $10 + x$.

Using the tangent trigonometric ratio:

$ \tan(\angle TAC) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

$ \tan(30^\circ) = \frac{TC}{AC} $

Substitute the value of $ \tan(30^\circ) = \frac{1}{\sqrt{3}} $ into equation (iii):

$ \frac{1}{\sqrt{3}} = \frac{h}{10 + x} $

---

Now we have a system of equations. Substitute $x = h$ from equation (ii) into this equation:

$ \frac{1}{\sqrt{3}} = \frac{h}{10 + h} $

---

Cross-multiply to solve for $h$:

$ 1 \times (10 + h) = h \times \sqrt{3} $

$ 10 + h = h\sqrt{3} $

---

Rearrange the equation to isolate the terms involving $h$ on one side:

$ 10 = h\sqrt{3} - h $

$ 10 = h(\sqrt{3} - 1) $

---

Divide both sides by $ (\sqrt{3} - 1) $ to find $h$:

$ h = \frac{10}{\sqrt{3} - 1} $

---

This form of the height matches option (C).

---

Comparing this result with the given options:

(A) $10(\sqrt{3}-1)$ m

(B) $10(\sqrt{3}+1)$ m

(C) $\frac{10}{\sqrt{3}-1}$ m

(D) $\frac{10}{\sqrt{3}+1}$ m

Our calculated height $h = \frac{10}{\sqrt{3} - 1}$ matches option (C).

---

Therefore, the correct option is (C) $\frac{10}{\sqrt{3}-1}$ m.

Given:

Height of the hill, $h = 100$ m.

Angle of depression from the top of the hill to the object, $\delta = 30^\circ$.

---

To Find:

The distance of the object from the base of the hill, $d$.

---

Solution:

Let the top of the hill be T, the base of the hill be B, and the position of the object on the ground be O. The height of the hill is the vertical distance TB = $h = 100$ m. The distance of the object from the base of the hill is the horizontal distance BO = $d$.

The angle of depression from the top of the hill (T) to the object (O) is the angle between the horizontal line through T and the line of sight TO. Let the horizontal line through T be TH, parallel to the ground BO. The angle of depression is $\angle HTO = 30^\circ$.

Since the horizontal line TH is parallel to the ground BO, and the line of sight TO is a transversal, the angle of depression $\angle HTO$ is equal to the angle of elevation from the object to the top of the hill, $\angle TOB$ (alternate interior angles).

So, the angle of elevation from the object to the top of the hill is $\angle TOB = 30^\circ$.

Now consider the right-angled triangle $\triangle TBO$, where $\angle TBO = 90^\circ$ (assuming the hill is vertical and the ground is horizontal).

In $\triangle TBO$, we have:

The side opposite to the angle $\angle TOB = 30^\circ$ is the height of the hill $TB = h = 100$ m.

The side adjacent to the angle $\angle TOB = 30^\circ$ is the distance of the object from the base of the hill $BO = d$.

We can use the tangent trigonometric ratio:

$ \tan(\angle TOB) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

---

Substitute the given values into equation (i):

$ \tan(30^\circ) = \frac{100}{d} $

We know that the value of $ \tan(30^\circ) $ is $ \frac{1}{\sqrt{3}} $.

$ \frac{1}{\sqrt{3}} = \frac{100}{d} $

---

To solve for $d$, we can cross-multiply:

$ 1 \times d = 100 \times \sqrt{3} $

$ d = 100\sqrt{3} $ m

---

The distance of the object from the base of the hill is $100\sqrt{3}$ m.

---

Comparing this result with the given options:

(A) 100 m

(B) $100 \sqrt{3}$ m

(C) $\frac{100}{\sqrt{3}}$ m

(D) 200 m

Our calculated distance matches option (B).

---

Therefore, the correct option is (B) $100 \sqrt{3}$ m.

Given:

Height of the hill (soldier's height above the base), $h = 100$ m.

Angle of depression from the top of the hill to the object, $\delta = 30^\circ$.

---

To Find:

The straight-line distance from the soldier's eye (top of the hill) to the object on the ground (line of sight).

---

Solution:

Let the top of the hill be T, the base of the hill be B, and the position of the object on the ground be O. The height of the hill is the vertical distance TB = $h = 100$ m. The line of sight is the segment TO.

The angle of depression from T to O is $\delta = 30^\circ$. Let the horizontal line through T be TH, parallel to the ground BO. The angle of depression is $\angle HTO = 30^\circ$.

Since the horizontal line TH is parallel to the ground BO, and the line of sight TO is a transversal, the angle of depression $\angle HTO$ is equal to the angle of elevation from the object to the top of the hill, $\angle TOB$ (alternate interior angles).

So, the angle of elevation from the object to the top of the hill is $\angle TOB = 30^\circ$.

Now consider the right-angled triangle $\triangle TBO$, where $\angle TBO = 90^\circ$.

In $\triangle TBO$, we have:

The side opposite to the angle $\angle TOB = 30^\circ$ is the height of the hill $TB = h = 100$ m.

The hypotenuse is the straight-line distance TO (line of sight). Let TO = $L$.

We can use the sine trigonometric ratio, which relates the opposite side and the hypotenuse:

$ \sin(\angle TOB) = \frac{\text{Opposite side}}{\text{Hypotenuse}} $

---

Substitute the given values into equation (i):

$ \sin(30^\circ) = \frac{100}{L} $

We know that the value of $ \sin(30^\circ) $ is $ \frac{1}{2} $.

$ \frac{1}{2} = \frac{100}{L} $

---

To solve for $L$, we can cross-multiply:

$ 1 \times L = 100 \times 2 $

$ L = 200 $ m

---

The straight-line distance from the soldier's eye to the object is 200 m.

---

Comparing this result with the given options:

(A) 100 m

(B) $100 \sqrt{3}$ m

(C) 200 m

(D) $\frac{100}{\sqrt{3}}$ m

Our calculated distance matches option (C).

---

Therefore, the correct option is (C) 200 m.

Given:

Let the height of the lighthouse be $h$.

Angles of depression from the top of the lighthouse to two ships are $30^\circ$ and $45^\circ$.

Distance between the two ships = 100 m.

The ships are in line with the lighthouse and on the same side.

---

To Find:

The correct equation relating the height $h$ and the given distance.

---

Solution:

Let T be the top of the lighthouse and F be its foot. The height of the lighthouse is TF = $h$. Let S1 and S2 be the positions of the two ships on the ground, in line with F. Since the angles of depression are $30^\circ$ and $45^\circ$, the ship with the larger angle of depression ($45^\circ$) is closer to the lighthouse, and the ship with the smaller angle of depression ($30^\circ$) is farther away.

Let S1 be the ship closer to the lighthouse and S2 be the ship farther away. Both S1 and S2 are on the same side of the foot of the lighthouse F.

The angle of depression from T to S1 is $45^\circ$. The angle of elevation from S1 to T is equal to the angle of depression (alternate interior angles).

The angle of depression from T to S2 is $30^\circ$. The angle of elevation from S2 to T is equal to the angle of depression (alternate interior angles).

---

Consider the right-angled triangle $\triangle TFS1$, where $\angle TFS1 = 90^\circ$.

Using the tangent ratio:

$ \tan(45^\circ) = \frac{TF}{FS1} = \frac{h}{FS1} $

$ 1 = \frac{h}{FS1} $

$ FS1 = h $

---

Alternatively, using the cotangent ratio:

$ \cot(45^\circ) = \frac{FS1}{TF} = \frac{FS1}{h} $

$ 1 = \frac{FS1}{h} $

---

Consider the right-angled triangle $\triangle TFS2$, where $\angle TFS2 = 90^\circ$.

Using the tangent ratio:

$ \tan(30^\circ) = \frac{TF}{FS2} = \frac{h}{FS2} $

$ \frac{1}{\sqrt{3}} = \frac{h}{FS2} $

$ FS2 = h\sqrt{3} $

---

Alternatively, using the cotangent ratio:

$ \cot(30^\circ) = \frac{FS2}{TF} = \frac{FS2}{h} $

$ \sqrt{3} = \frac{FS2}{h} $

---

The distance between the two ships is S1S2 = 100 m. Since S2 is farther from the lighthouse than S1, the distance between them is the difference between their distances from the foot of the lighthouse:

$ S1S2 = FS2 - FS1 $

---

Substitute the expressions for FS1 and FS2 from equations (i) and (ii) into equation (iii):

$ 100 = h \cot(30^\circ) - h \cot(45^\circ) $

---

Factor out $h$ from the terms on the right side:

$ 100 = h (\cot 30^\circ - \cot 45^\circ) $

---

This equation relates the height $h$ of the lighthouse to the distance between the ships and the cotangents of the angles of depression (or elevation). This matches option (A).

---

Let's verify the values: $ \cot(30^\circ) = \sqrt{3} $ and $ \cot(45^\circ) = 1 $. The equation becomes $ 100 = h(\sqrt{3} - 1) $, which is a valid equation for finding $h$.

---

Comparing our derived equation with the given options:

(A) $h (\cot 30^\circ - \cot 45^\circ) = 100$ - Matches.

(B) $h (\cot 45^\circ - \cot 30^\circ) = 100$ - Would lead to $h(1 - \sqrt{3}) = 100$, implying a negative height.

(C) $h (\tan 30^\circ - \tan 45^\circ) = 100$ - Would lead to $h(\frac{1}{\sqrt{3}} - 1) = 100$, implying a negative height.

(D) $h + 100 (\tan 30^\circ) = h (\tan 45^\circ)$ - Incorrect structure.

---

Therefore, the correct equation calculation approach is given by option (A).

---

The correct option is (A) $h (\cot 30^\circ - \cot 45^\circ) = 100$.

Given:

The shadow of a tower is $\sqrt{3}$ times its height.

---

To Find:

The angle of elevation of the sun.

---

Solution:

Let the height of the tower be $h$.

Let the length of the shadow be $s$.

We are given that the shadow of the tower is $\sqrt{3}$ times its height:

$ s = \sqrt{3} h $

---

Let the angle of elevation of the sun be $\theta$. The tower, its shadow, and the line of sight from the tip of the shadow to the top of the tower form a right-angled triangle.

In this right-angled triangle, the height of the tower is the side opposite to the angle of elevation $\theta$, and the length of the shadow is the side adjacent to the angle of elevation $\theta$.

We can use the tangent trigonometric ratio:

$ \tan(\theta) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

$ \tan(\theta) = \frac{h}{s} $

---

Substitute the given relationship $s = \sqrt{3}h$ into the equation:

$ \tan(\theta) = \frac{h}{\sqrt{3}h} $

---

Cancel out $h$ from the numerator and denominator (assuming $h \neq 0$, which is true for a tower):

$ \tan(\theta) = \frac{\cancel{h}}{\sqrt{3}\cancel{h}} $

$ \tan(\theta) = \frac{1}{\sqrt{3}} $

---

To find the angle $\theta$, we need to find the angle whose tangent is $\frac{1}{\sqrt{3}}$.

We know that $ \tan(30^\circ) = \frac{1}{\sqrt{3}} $.

Therefore,

$ \theta = 30^\circ $

---

The angle of elevation of the sun is $30^\circ$.

---

Comparing this result with the given options:

(A) $30^\circ$

(B) $45^\circ$

(C) $60^\circ$

(D) $90^\circ$

Our calculated angle matches option (A).

---

Therefore, the correct option is (A) $30^\circ$.

Given:

Initial angle of elevation of the top of the tree from the bank = $60^\circ$.

Distance moved away from the bank = 40 m.

New angle of elevation of the top of the tree = $30^\circ$.

---

To Find:

The height of the tree.

---

Solution:

Let the height of the tree be $h$. Let the width of the river be $x$.

Let T be the top of the tree, B be the base of the tree (on the opposite bank), and P be the initial position of the man on the bank. Let Q be the new position of the man after moving 40 m away from the bank.

The distance from the bank to the base of the tree is the width of the river, $BP = x$.

The distance from the new position Q to the base of the tree B is $BQ = BP + PQ = x + 40$.

Consider the right-angled triangle formed by the tree, the river width, and the line of sight from the initial position P ($\triangle TBP$). The angle of elevation from P is $\angle TPB = 60^\circ$.

Using the tangent trigonometric ratio:

$ \tan(60^\circ) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{TB}{BP} $

$ \tan(60^\circ) = \frac{h}{x} $

---

From equation (i), we can express $x$ in terms of $h$:

---

Now consider the right-angled triangle formed by the tree, the distance from the new position Q, and the line of sight from Q ($\triangle TBQ$). The angle of elevation from Q is $\angle TQB = 30^\circ$.

Using the tangent trigonometric ratio:

$ \tan(30^\circ) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{TB}{BQ} $

$ \tan(30^\circ) = \frac{h}{x + 40} $

---

Substitute the expression for $x$ from equation (ii) into equation (iii):

$ \frac{1}{\sqrt{3}} = \frac{h}{\frac{h}{\sqrt{3}} + 40} $

---

Simplify the denominator on the right side by finding a common denominator:

$ \frac{h}{\frac{h}{\sqrt{3}} + 40} = \frac{h}{\frac{h + 40\sqrt{3}}{\sqrt{3}}} = h \times \frac{\sqrt{3}}{h + 40\sqrt{3}} = \frac{h\sqrt{3}}{h + 40\sqrt{3}} $

So, the equation becomes:

$ \frac{1}{\sqrt{3}} = \frac{h\sqrt{3}}{h + 40\sqrt{3}} $

---

Cross-multiply:

$ 1 \times (h + 40\sqrt{3}) = \sqrt{3} \times h\sqrt{3} $

$ h + 40\sqrt{3} = (\sqrt{3} \times \sqrt{3})h $

$ h + 40\sqrt{3} = 3h $

---

Now, solve for $h$. Subtract $h$ from both sides:

$ 40\sqrt{3} = 3h - h $

$ 40\sqrt{3} = 2h $

---

Divide both sides by 2:

$ h = \frac{40\sqrt{3}}{2} $

$ h = 20\sqrt{3} $ m

---

The height of the tree is $20\sqrt{3}$ m.

---

Comparing this result with the given options:

(A) 20 m

(B) $20\sqrt{3}$ m

(C) 40 m

(D) $40\sqrt{3}$ m

Our calculated height matches option (B).

---

Therefore, the correct option is (B) $20\sqrt{3}$ m.

Given:

Height of the higher aeroplane from the ground, $H_1 = 1000$ m.

Angle of elevation of the higher aeroplane from a point on the ground = $60^\circ$.

Angle of elevation of the lower aeroplane from the same point on the ground = $45^\circ$.

The two aeroplanes are vertically above each other at the instant of observation.

---

To Find:

The vertical distance between the aeroplanes at that instant.

---

Solution:

Let P be the point on the ground from where the angles of elevation are measured. Let A be the point on the ground directly below the aeroplanes. Let T1 be the position of the higher aeroplane and T2 be the position of the lower aeroplane at the instant they are vertically aligned above A.

The height of the higher aeroplane from the ground is AT1 = $H_1 = 1000$ m.

The height of the lower aeroplane from the ground is AT2 = $H_2$.

The horizontal distance from the observation point P to the point directly below the aeroplanes A is PA = $d$.

The angle of elevation of T1 from P is $\angle APT1 = 60^\circ$.

The angle of elevation of T2 from P is $\angle APT2 = 45^\circ$.

Consider the right-angled triangle $\triangle PAT1$. The angle at A is $90^\circ$.

Using the tangent trigonometric ratio:

$ \tan(\angle APT1) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{AT1}{PA} $

Substitute the given value $H_1 = 1000$ m and $ \tan(60^\circ) = \sqrt{3} $ into equation (i):

$ \sqrt{3} = \frac{1000}{d} $

---

Solve for the horizontal distance $d$:

$ d = \frac{1000}{\sqrt{3}} $ m

---

Now consider the right-angled triangle $\triangle PAT2$. The angle at A is $90^\circ$.

Using the tangent trigonometric ratio:

$ \tan(\angle APT2) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{AT2}{PA} $

Substitute the value $ \tan(45^\circ) = 1 $ into equation (iii):

$ 1 = \frac{H_2}{d} $

---

Substitute the value of $d$ from equation (ii) into equation (iv):

$ H_2 = \frac{1000}{\sqrt{3}} $ m

---

The vertical distance between the aeroplanes is the difference in their heights from the ground, $V = H_1 - H_2$.

$ V = 1000 - \frac{1000}{\sqrt{3}} $

---

Factor out 1000:

$ V = 1000 \left(1 - \frac{1}{\sqrt{3}}\right) $ m

---

The vertical distance between the aeroplanes is $1000 \left(1 - \frac{1}{\sqrt{3}}\right)$ m.

---

Comparing this result with the given options:

(A) $1000(\sqrt{3}-1)$ m

(B) $1000(1-\frac{1}{\sqrt{3}})$ m

(C) $1000(1 - \frac{1}{\sqrt{2}})$ m

(D) $1000(1 + \frac{1}{\sqrt{3}})$ m

Our calculated distance matches option (B).

---

Therefore, the correct option is (B) $1000(1-\frac{1}{\sqrt{3}})$ m.

Given:

Let the height of the tower be $h$.

Let the two points on the ground in the same straight line with the base of the tower be at distances $a = 4$ m and $b = 9$ m from the base.

The angles of elevation from these two points to the top of the tower are complementary.

---

To Find:

The height of the tower, $h$.

---

Solution:

When the angles of elevation of the top of a tower from two points at distances $a$ and $b$ from the base and in the same straight line with it are complementary, the height of the tower $h$ is given by the formula:

$ h = \sqrt{ab} $

---

Substitute the given values of $a = 4$ m and $b = 9$ m into the formula:

$ h = \sqrt{4 \times 9} $

$ h = \sqrt{36} $

$ h = 6 $

---

The height of the tower is 6 m.

---

Comparing this result with the given options:

(A) 4 m

(B) 9 m

(C) 6 m

(D) 13 m

Our calculated height matches option (C).

---

Therefore, the correct option is (C) 6 m.

Given:

Height of the kite above the ground, $h = 60$ m.

Angle of inclination of the string with the ground, $\theta = 60^\circ$.

The string is straight.

---

To Find:

The length of the string.

---

Solution:

Let the position of the kite in the air be K, the point on the ground where the string is tied be P, and the point on the ground directly below the kite be G. The height of the kite is KG = $h = 60$ m. The string is the segment KP. The angle of inclination of the string with the ground is $\angle KPG = \theta = 60^\circ$.

Assuming the ground is horizontal and the height is measured vertically, $\triangle KGP$ is a right-angled triangle with the right angle at G.

In the right-angled triangle $\triangle KGP$, the height of the kite KG is the side opposite to the angle of inclination $\theta = 60^\circ$. The length of the string KP is the hypotenuse.

We can use the sine trigonometric ratio, which relates the opposite side and the hypotenuse:

$ \sin(\theta) = \frac{\text{Opposite side}}{\text{Hypotenuse}} $

$ \sin(60^\circ) = \frac{KG}{KP} $

---

Let the length of the string be $L$. Substitute the value $ \sin(60^\circ) = \frac{\sqrt{3}}{2} $ into equation (i):

$ \frac{\sqrt{3}}{2} = \frac{60}{L} $

---

Now, we solve for $L$. Cross-multiply:

$ L \times \sqrt{3} = 60 \times 2 $

$ L\sqrt{3} = 120 $

---

Divide both sides by $\sqrt{3}$:

$ L = \frac{120}{\sqrt{3}} $ m

---

The length of the string is $\frac{120}{\sqrt{3}}$ m.

Comparing this result with the given options:

(A) 60 m

(B) $60\sqrt{3}$ m

(C) 120 m

(D) $\frac{120}{\sqrt{3}}$ m

Our calculated length matches option (D).

---

Therefore, the correct option is (D) $\frac{120}{\sqrt{3}}$ m.

Given:

The height of a vertical pole is equal to the length of its shadow on the ground.

---

To Find:

The angle of elevation of the sun.

---

Solution:

Let the height of the pole be $h$.

Let the length of the shadow be $s$.

We are given that the height of the pole is equal to the length of its shadow:

$ h = s $

---

Let the angle of elevation of the sun be $\theta$. The pole, its shadow, and the line of sight from the tip of the shadow to the top of the pole form a right-angled triangle.

In this right-angled triangle, the height of the pole is the side opposite to the angle of elevation $\theta$, and the length of the shadow is the side adjacent to the angle of elevation $\theta$.

We can use the tangent trigonometric ratio:

$ \tan(\theta) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

$ \tan(\theta) = \frac{h}{s} $

---

Substitute the given relationship $h = s$ into the equation:

$ \tan(\theta) = \frac{h}{h} $

---

Assuming $h \neq 0$ (which is true for a pole), we can cancel out $h$:

$ \tan(\theta) = 1 $

---

To find the angle $\theta$, we need to find the angle whose tangent is 1.

We know that $ \tan(45^\circ) = 1 $.

Therefore,

$ \theta = 45^\circ $

---

The angle of elevation of the sun is $45^\circ$.

---

Comparing this result with the given options:

(A) $30^\circ$

(B) $45^\circ$

(C) $60^\circ$

(D) $90^\circ$

Our calculated angle matches option (B).

---

Therefore, the correct option is (B) $45^\circ$.

Given:

Let the height of the tower be $h$.

Let the two points on the ground in the same straight line with the base of the tower be P and Q.

The angles of elevation from these two points to the top of the tower are $45^\circ$ and $60^\circ$.

The distances of these points from the base are $x$ and $y$, with $x > y$.

---

To Find:

The height of the tower, $h$.

---

Solution:

Let the top of the tower be T and the foot of the tower be A. So, TA = $h$. Let P and Q be the two points on the ground in line with A. Since the angle of elevation is smaller from a farther point and larger from a nearer point, and we are given angles $45^\circ$ and $60^\circ$ with distances $x$ and $y$ where $x > y$, it implies that the angle of elevation from the point at distance $x$ is $45^\circ$, and the angle of elevation from the point at distance $y$ is $60^\circ$. Thus, the point at distance $y$ is nearer to the base than the point at distance $x$.

Let the distance of the nearer point Q from the base A be QA = $y$, and the angle of elevation from Q be $\angle TQA = 60^\circ$.

Let the distance of the farther point P from the base A be PA = $x$, and the angle of elevation from P be $\angle TPA = 45^\circ$.

---

Consider the right-angled triangle $\triangle TQA$. The angle at A is $90^\circ$.

Using the tangent trigonometric ratio:

$ \tan(\angle TQA) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{TA}{QA} $

From equation (i), we can express $h$ in terms of $y$:

$ h = y \tan(60^\circ) = y\sqrt{3} $

---

Consider the right-angled triangle $\triangle TPA$. The angle at A is $90^\circ$.

Using the tangent trigonometric ratio:

$ \tan(\angle TPA) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{TA}{PA} $

From equation (iii), we can express $h$ in terms of $x$:

$ h = x \tan(45^\circ) = x \times 1 = x $

---

From equations (ii) and (iv), we have two expressions for $h$ in terms of $y$ and $x$ respectively:

$ h = y\sqrt{3} $

$ h = x $

This implies that for this specific geometry to exist with the given angles and distances, the relationship $x = y\sqrt{3}$ must hold.

The question asks for the height of the tower in terms of $x$ and $y$. Based on the derived relationships, the height $h$ is equal to $x$, and also equal to $y\sqrt{3}$. However, the options are not simply $x$ or $y\sqrt{3}$, but combined expressions of $x$ and $y$. This suggests a possible misinterpretation of the problem phrasing or a typo in the options.

Assuming the question intended a different relationship between $x$, $y$, and the distances or angles that leads to one of the options, let's consider a common variant where $x$ is the distance of the nearer point and $y$ is the distance between the two points. In this case, the angle from distance $x$ is $60^\circ$, and the angle from distance $x+y$ is $45^\circ$.

Let nearer distance from base = $x$, angle $= 60^\circ$. Farther distance from base = $x+y$, angle $= 45^\circ$. Condition $x>y$ is given.

$ \tan(60^\circ) = h/x \implies x = h/\sqrt{3} $

$ \tan(45^\circ) = h/(x+y) \implies x+y = h $

Substitute $x=h/\sqrt{3}$ into the second equation:

$ h/\sqrt{3} + y = h $

$ y = h - h/\sqrt{3} = h(1 - 1/\sqrt{3}) = h \left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) $

Solve for $h$:

$ h = y \frac{\sqrt{3}}{\sqrt{3}-1} $

---

This expression for $h$ matches Option (A).

Let's verify the condition $x>y$ with this interpretation. $x = h/\sqrt{3}$ and $y = h(\sqrt{3}-1)/\sqrt{3}$.

$ x/y = (h/\sqrt{3}) / (h(\sqrt{3}-1)/\sqrt{3}) = 1/(\sqrt{3}-1) = (\sqrt{3}+1)/2 \approx (1.732+1)/2 = 2.732/2 = 1.366 $. Since $x/y \approx 1.366 > 1$, $x > y$ is consistent with this interpretation.

Although this interpretation contradicts the literal phrasing "at distances $x$ and $y$ from the base", it is the most plausible way to arrive at one of the provided options consistently with the given angles and the condition $x>y$.

---

Therefore, assuming the intended problem implies that $x$ is the distance of the nearer point from the base ($60^\circ$ angle) and $y$ is the distance between the two points, the height of the tower is $y \frac{\sqrt{3}}{\sqrt{3}-1}$ m.

---

The correct option is (A) $\frac{y \sqrt{3}}{\sqrt{3}-1}$.

Given:

Height of the cliff, $h = 100$ m.

Initial angle of depression of the boat from the top of the cliff = $30^\circ$.

Final angle of depression of the boat from the top of the cliff = $45^\circ$.

The boat is approaching the shore and is in line with the base of the cliff.

---

To Find:

The distance the boat travelled during the observation.

---

Solution:

Let T be the top of the cliff, C be the base of the cliff (on the shore). The height of the cliff is TC = $h = 100$ m. Let P1 be the initial position of the boat and P2 be the final position of the boat as it approaches the shore. Points C, P2, and P1 are in a straight line on the ground, with P2 being between C and P1.

The angle of depression from T to P1 is $30^\circ$. The angle of elevation from P1 to T is equal to the angle of depression (alternate interior angles).

The angle of depression from T to P2 is $45^\circ$. The angle of elevation from P2 to T is equal to the angle of depression (alternate interior angles).

---

Consider the right-angled triangle $\triangle TCP_1$, where $\angle TCB = 90^\circ$. Let the distance $CP_1 = d_1$.

Using the tangent trigonometric ratio:

$ \tan(\angle TP_1C) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{TC}{CP_1} $

$ \tan(30^\circ) = \frac{100}{d_1} $

We know that $ \tan(30^\circ) = \frac{1}{\sqrt{3}} $.

$ \frac{1}{\sqrt{3}} = \frac{100}{d_1} $

---

Consider the right-angled triangle $\triangle TCP_2$, where $\angle TCP_2 = 90^\circ$. Let the distance $CP_2 = d_2$.

Using the tangent trigonometric ratio:

$ \tan(\angle TP_2C) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{TC}{CP_2} $

$ \tan(45^\circ) = \frac{100}{d_2} $

We know that $ \tan(45^\circ) = 1 $.

$ 1 = \frac{100}{d_2} $

---

The distance the boat travelled is the distance between the two points P1 and P2, which is $P_1P_2$. Since P2 is closer to C than P1, this distance is the difference between their distances from the base C:

$ P_1P_2 = CP_1 - CP_2 $

$ P_1P_2 = d_1 - d_2 $

---

Substitute the values of $d_1$ and $d_2$ from equations (i) and (ii):

$ P_1P_2 = 100\sqrt{3} - 100 $

---

Factor out 100:

$ P_1P_2 = 100(\sqrt{3} - 1) $ m

---

The distance the boat travelled is $100(\sqrt{3}-1)$ m.

---

Comparing this result with the given options:

(A) $100(\sqrt{3}-1)$ m

(B) $100(\sqrt{3}+1)$ m

(C) $100(1-\frac{1}{\sqrt{3}})$ m

(D) $100\sqrt{3}$ m

Our calculated distance matches option (A).

---

Therefore, the correct option is (A) $100(\sqrt{3}-1)$ m.

Given:

Height of the first pole, $h_1 = 10$ m.

Height of the second pole, $h_2 = 18$ m.

The tops of the poles are connected by a wire.

The wire makes an angle of $30^\circ$ with the horizontal.

---

To Find:

The length of the wire.

---

Solution:

Let the two poles be represented by vertical lines. Let the feet of the poles on the ground be A and C, and their tops be B and D respectively. Assume the poles are vertical and the ground AC is horizontal.

Height of the first pole, AB = $h_1 = 10$ m.

Height of the second pole, CD = $h_2 = 18$ m.

The wire connects the tops, so the wire is the segment BD. Let the length of the wire be $L$.

Draw a horizontal line through the top of the shorter pole B, meeting the taller pole CD at point E. The line BE is parallel to the ground AC, and the angle made by the wire BD with the horizontal BE is given as $30^\circ$. So, $\angle DBE = 30^\circ$.

The segment BE is horizontal, and CE is vertical. The length BE is equal to the distance between the bases of the poles AC. The length CE is the difference in the heights of the poles above the level of B.

CE = CD - ED

Since BE is horizontal and BA and CE are vertical, ABEC forms a rectangle. Therefore, ED = AB = 10 m.

CE = 18 m - 10 m = 8 m.

Now consider the right-angled triangle $\triangle BED$. The angle at E is $90^\circ$ (since BE is horizontal and DE is vertical). The angle $\angle DBE = 30^\circ$. The side opposite to this angle is DE = 8 m, and the hypotenuse is BD = $L$ (the length of the wire).

We can use the sine trigonometric ratio, which relates the opposite side and the hypotenuse:

$ \sin(\angle DBE) = \frac{\text{Opposite side}}{\text{Hypotenuse}} $

$ \sin(30^\circ) = \frac{DE}{BD} $

---

We know that the value of $ \sin(30^\circ) $ is $ \frac{1}{2} $.

Substitute this value into equation (i):

$ \frac{1}{2} = \frac{8}{L} $

---

Now, solve for $L$. Cross-multiply:

$ 1 \times L = 8 \times 2 $

$ L = 16 $ m

---

The length of the wire is 16 m.

---

Comparing this result with the given options:

(A) 8 m

(B) 16 m

(C) 18 m

(D) $8 \sin 30^\circ$ m

Our calculated length matches option (B).

Option (D) is incorrect; $8 \sin 30^\circ = 8 \times \frac{1}{2} = 4$ m, which is not the length of the wire.

---

Therefore, the correct option is (B) 16 m.

Given:

Let the height of the tower be $h$.

Let P be a point on the ground. The angle of elevation of the top of the tower from P is $\alpha$.

Let Q be a point on the ground vertically below the top of the tower, and Q is 40 m from P. The angle of elevation from Q is $\beta$.

---

To Find:

The correct equation relating $h, \alpha, \beta,$ and 40.

---

Solution:

Let the top of the tower be T and the foot of the tower on the ground be A. The height of the tower is TA = $h$.

The statement "a point Q on the ground vertically below the top of the tower" means that Q is located at the foot of the tower, A. So, Q is the same point as A.

The statement "and 40 m from P" means the distance between point P and point Q (which is A) is 40 m. Therefore, the distance from point P on the ground to the foot of the tower A is PA = 40 m.

The angle of elevation of the top of the tower (T) from point P is $\alpha$. In the right-angled triangle $\triangle TPA$ (where the angle at A is $90^\circ$, assuming the tower is vertical and the ground is horizontal), we have:

$ \tan(\text{angle of elevation from P}) = \frac{\text{Opposite side (height)}}{\text{Adjacent side (distance)}} $

$ \tan(\alpha) = \frac{TA}{PA} $

---

Solving equation (i) for $h$:

$ h = 40 \tan(\alpha) $

---

Now consider the information about the angle of elevation from point Q. Since Q is the foot of the tower A, the angle of elevation from Q to the top T is $\angle TAQ = \angle TAT$. For a vertical tower, the line segment AT is vertical, and the ground at A is horizontal. The angle between a vertical line and a horizontal plane is $90^\circ$. Thus, the angle of elevation from the foot of the tower to the top is $90^\circ$. So, $\beta = 90^\circ$.

If $\beta = 90^\circ$, then $ \tan(\beta) = \tan(90^\circ) $, which is undefined.

Let's look at the options provided:

(A) $h = 40 \tan \alpha$ - This equation directly matches our derivation from the information about point P and its distance from the base.

(B) $h = 40 \tan \beta$ - If $\beta = 90^\circ$, this equation involves an undefined term $ \tan \beta $. This option is likely incorrect based on the geometry.

(C) $h = 40 (\tan \alpha + \tan \beta)$ - If $\beta = 90^\circ$, this equation involves an undefined term $ \tan \beta $. This option is likely incorrect.

(D) This setup is not standard for complementary or two-point problems on the same line. - While the phrasing might be slightly confusing regarding the angles, the geometry described (a point P, distance from P to the base, angle of elevation from P) is standard.

---

Based on the clear part of the problem statement and the options provided, the equation that relates the height of the tower to the angle of elevation from point P at a distance of 40 m from the base is $h = 40 \tan \alpha$. The information about point Q and angle $\beta$ seems to imply that Q is the base and $\beta = 90^\circ$, which makes options involving $ \tan \beta $ problematic.

Therefore, the correct equation consistent with a standard interpretation of the measurements from point P is $h = 40 \tan \alpha$.

---

The correct option is (A) $h = 40 \tan \alpha$.

Given:

Height of the building, $h = 10$ m.

Angle of depression of a point on the ground from the top of the building, $\delta = 30^\circ$.

---

To Find:

The distance of the point from the foot of the building, $d$.

---

Solution:

Let T be the top of the building, F be the foot of the building, and P be the point on the ground. The height of the building is TF = $h = 10$ m. The distance of the point from the foot of the building is FP = $d$.

The angle of depression from the top of the building (T) to the point (P) is the angle between the horizontal line through T and the line of sight TP. Let the horizontal line through T be TH, parallel to the ground FP. The angle of depression is $\angle HTP = 30^\circ$.

Since the horizontal line TH is parallel to the ground FP, and the line of sight TP is a transversal, the angle of depression $\angle HTP$ is equal to the angle of elevation from the point on the ground to the top of the building, $\angle TPF$ (alternate interior angles).

So, the angle of elevation from point P to the top of the building is $\angle TPF = 30^\circ$.

Now consider the right-angled triangle $\triangle TFP$, where $\angle TFP = 90^\circ$ (assuming the building is vertical and the ground is horizontal).

In $\triangle TFP$, we have:

The side opposite to the angle $\angle TPF = 30^\circ$ is the height of the building $TF = h = 10$ m.

The side adjacent to the angle $\angle TPF = 30^\circ$ is the distance of the point from the foot of the building $FP = d$.

We can use the tangent trigonometric ratio:

$ \tan(\angle TPF) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

---

Substitute the given values into equation (i):

$ \tan(30^\circ) = \frac{10}{d} $

We know that the value of $ \tan(30^\circ) $ is $ \frac{1}{\sqrt{3}} $.

$ \frac{1}{\sqrt{3}} = \frac{10}{d} $

---

To solve for $d$, we can cross-multiply:

$ 1 \times d = 10 \times \sqrt{3} $

$ d = 10\sqrt{3} $ m

---

The distance of the point from the foot of the building is $10\sqrt{3}$ m.

---

Comparing this result with the given options:

(A) 10 m

(B) $10 \sqrt{3}$ m

(C) $\frac{10}{\sqrt{3}}$ m

(D) 20 m

Our calculated distance matches option (B).

---

Therefore, the correct option is (B) $10 \sqrt{3}$ m.

Given:

Height of the boy, $h_b = 1.8$ m.

Height of the building, $H = 30$ m.

Initial angle of elevation from the boy's eyes to the top of the building, $\theta_1 = 30^\circ$.

Final angle of elevation from the boy's eyes to the top of the building, $\theta_2 = 60^\circ$.

---

To Find:

The distance the boy walked towards the building.

---

Solution:

Let the height of the building above the boy's eye level be $h$.

$h = H - h_b = 30 \text{ m} - 1.8 \text{ m} = 28.2$ m.

Let P be the initial position of the boy and Q be the final position after walking towards the building. Let A be the point on the ground directly below the top of the building.

Let the horizontal distance from the boy's eye level at point P to the building be $d_1$. Let the horizontal distance from the boy's eye level at point Q to the building be $d_2$. The distance the boy walked is $d_1 - d_2$.

---

Consider the right-angled triangle formed by the initial position (eye level), the point directly below the top of the building at eye level, and the top of the building. The angle of elevation is $30^\circ$, the opposite side is the height above eye level ($h$), and the adjacent side is $d_1$.

Using the tangent trigonometric ratio:

$ \tan(\theta_1) = \frac{h}{d_1} $

$ \tan(30^\circ) = \frac{28.2}{d_1} $

From equation (i), solve for $d_1$:

$ d_1 = 28.2 \sqrt{3} $ m

---

Consider the right-angled triangle formed by the final position (eye level), the point directly below the top of the building at eye level, and the top of the building. The angle of elevation is $60^\circ$, the opposite side is the height above eye level ($h$), and the adjacent side is $d_2$.

Using the tangent trigonometric ratio:

$ \tan(\theta_2) = \frac{h}{d_2} $

$ \tan(60^\circ) = \frac{28.2}{d_2} $

From equation (ii), solve for $d_2$:

$ d_2 = \frac{28.2}{\sqrt{3}} $ m

---

The distance the boy walked towards the building is the difference between the initial and final horizontal distances:

Distance walked = $d_1 - d_2$

Distance walked = $28.2\sqrt{3} - \frac{28.2}{\sqrt{3}}$

---

Factor out 28.2:

Distance walked = $28.2 \left(\sqrt{3} - \frac{1}{\sqrt{3}}\right)$ m

---

Comparing this result with the given options:

(A) $28.2 \sqrt{3}$ m

(B) $\frac{28.2}{\sqrt{3}}$ m

(C) $28.2 (\sqrt{3} - \frac{1}{\sqrt{3}})$ m

(D) $28.2 (\sqrt{3} + \frac{1}{\sqrt{3}})$ m

Our calculated distance matches option (C).

---

Therefore, the correct option is (C) $28.2 (\sqrt{3} - \frac{1}{\sqrt{3}})$ m.

Given:

Height of the building, $h_b = 20$ m.

Angle of elevation of the top of the tower from the top of the building = $30^\circ$.

Angle of elevation of the top of the tower from the foot of the building = $60^\circ$.

---

To Find:

The height of the tower, $h_t$.

---

Solution:

Let the top of the building be B and the foot of the building be F. So, BF = 20 m. Let the top of the tower be T and the foot of the tower be A. Assume the building and the tower are vertical and the ground FA is horizontal.

The height of the tower is TA = $h_t$. The horizontal distance between the foot of the building and the foot of the tower is FA = $d$.

The angle of elevation of the top of the tower (T) from the foot of the building (F) is $\angle TFA = 60^\circ$.

Consider the right-angled triangle $\triangle TFA$, where $\angle TAF = 90^\circ$.

Using the tangent trigonometric ratio:

$ \tan(\angle TFA) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{TA}{FA} $

From equation (i), we can express $h_t$ in terms of $d$:

$ \sqrt{3} = \frac{h_t}{d} \implies h_t = d\sqrt{3} $

---

Draw a horizontal line from the top of the building B parallel to the ground, meeting the tower at point E. The height of the segment of the tower above the level of the top of the building is TE = $h_t - BE$. Since BAEF is a rectangle, BE = AF = $d$ and AE = BF = 20 m. So, TE = $h_t - 20$.

The angle of elevation of the top of the tower (T) from the top of the building (B) is $\angle TBE = 30^\circ$. Consider the right-angled triangle $\triangle TBE$, where $\angle TEB = 90^\circ$.

Using the tangent trigonometric ratio:

$ \tan(\angle TBE) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{TE}{BE} $

Substitute the value $ \tan(30^\circ) = \frac{1}{\sqrt{3}} $ into equation (ii):

$ \frac{1}{\sqrt{3}} = \frac{h_t - 20}{d} $

---

From equation (i), we have $d = \frac{h_t}{\sqrt{3}}$. Substitute this expression for $d$ into the above equation:

$ \frac{1}{\sqrt{3}} = \frac{h_t - 20}{\frac{h_t}{\sqrt{3}}} $

---

Simplify the right side of the equation:

$ \frac{1}{\sqrt{3}} = (h_t - 20) \times \frac{\sqrt{3}}{h_t} $

$ \frac{1}{\sqrt{3}} = \frac{\sqrt{3}(h_t - 20)}{h_t} $

---

Cross-multiply:

$ 1 \times h_t = \sqrt{3} \times \sqrt{3}(h_t - 20) $

$ h_t = (\sqrt{3})^2 (h_t - 20) $

$ h_t = 3 (h_t - 20) $

---

Distribute 3 on the right side:

$ h_t = 3h_t - 60 $

---

Rearrange the equation to solve for $h_t$. Subtract $h_t$ from both sides and add 60 to both sides:

$ 60 = 3h_t - h_t $

$ 60 = 2h_t $

---

Divide both sides by 2:

$ h_t = \frac{60}{2} $

$ h_t = 30 $ m

---

The height of the tower is 30 m.

---

Comparing this result with the given options:

(A) 30 m

(B) 40 m

(C) 50 m

(D) 60 m

Our calculated height matches option (A).

---

Therefore, the correct option is (A) 30 m.

Given:

Horizontal distance between two poles = 15 m.

Height of the second pole, $h_2 = 24$ m.

Angle of depression of the top of the first pole as seen from the top of the second pole = $30^\circ$.

---

To Find:

The height of the first pole, $h_1$.

---

Solution:

Let the first pole be represented by a vertical line segment AB (height $h_1$) and the second pole by a vertical line segment CD (height $h_2 = 24$ m). Let A and C be the feet of the poles on the ground. The horizontal distance between the poles is AC = 15 m.

The angle of depression of the top of the first pole (B) as seen from the top of the second pole (D) is $30^\circ$.

Draw a horizontal line from the top of the second pole D, meeting the vertical line extended upwards from the top of the first pole at point E. The line DE is horizontal, and the angle of depression is the angle between the horizontal line DE and the line of sight DB. So, $\angle EDB = 30^\circ$.

The horizontal distance between the poles is AC = 15 m. Since DE is parallel to AC and AE is vertical, ACED forms a rectangle. Thus, DE = AC = 15 m. Also, AE = CD = 24 m is incorrect, it should be EA = CD if A and C are at the same level, and E is vertically above B.

Let's adjust the diagram. Let the first pole be AB (height $h_1$) and the second pole be CD (height $h_2 = 24$ m), with A and C on the ground. The horizontal distance AC = 15 m.

Draw a horizontal line from the top of the first pole B, meeting the second pole CD at point E. The segment BE is horizontal, and CE is vertical. Since ABEC forms a rectangle, BE = AC = 15 m and AE = BC (not useful here).

The angle of depression of the top of the first pole (B) from the top of the second pole (D) is $30^\circ$. This means the first pole is shorter than the second pole ($h_1 < h_2$).

Draw a horizontal line from D, meeting the vertical line from B at point E'. The angle of depression is the angle between the horizontal line from D and the line of sight DB. Let the horizontal line from D be DH, parallel to the ground AC. The angle of depression of B from D is $\angle HDB = 30^\circ$.

The horizontal distance between the poles is the horizontal distance between D and B. This horizontal distance is equal to AC = 15 m. Draw a perpendicular from B to CD, meeting CD at E. Then BE = AC = 15 m.

The height of the second pole is CD = 24 m. The height of the first pole is AB = $h_1$. The segment BE is horizontal, so CE = AB = $h_1$. The length DE = CD - CE = $24 - h_1$.

Consider the right-angled triangle $\triangle DEB$. The angle at E is $90^\circ$. The horizontal distance BE = 15 m (adjacent to the angle of depression from D, which is equal to $\angle EDB$ by alternate interior angles if we consider the horizontal line through D and the line through B parallel to the ground) or using the angle of depression directly in $\triangle DEB$.

Let's use the alternate interior angle. The horizontal line from D is parallel to the ground. The line DB is the transversal. The angle of depression at D is $30^\circ$. The angle of elevation from B to D would be $30^\circ$ if looking from B to D. However, the angle of depression from D to B is given as $30^\circ$. This angle is formed between the horizontal line through D and the line DB. The horizontal line through D and the line BE are parallel. The line DB is the transversal. The angle of depression $\angle HDB$ is outside the triangle DEB. The alternate interior angle to the angle of depression $\angle HDB$ is the angle formed by the line of sight DB and the horizontal line through B parallel to the ground. This angle is $\angle DBE = 30^\circ$.

Consider the right-angled triangle $\triangle DEB$, where $\angle DEB = 90^\circ$. We have $\angle DBE = 30^\circ$. The adjacent side is BE = 15 m. The opposite side is DE = $24 - h_1$.

Using the tangent trigonometric ratio:

$ \tan(\angle DBE) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

$ \tan(30^\circ) = \frac{DE}{BE} $

Substitute the value $ \tan(30^\circ) = \frac{1}{\sqrt{3}} $ into equation (i):

$ \frac{1}{\sqrt{3}} = \frac{24 - h_1}{15} $

---

Cross-multiply:

$ 1 \times 15 = \sqrt{3} \times (24 - h_1) $

$ 15 = 24\sqrt{3} - h_1\sqrt{3} $

---

Rearrange the equation to solve for $h_1$: Move $h_1\sqrt{3}$ to the left side and 15 to the right side.

$ h_1\sqrt{3} = 24\sqrt{3} - 15 $

---

Divide both sides by $\sqrt{3}$:

$ h_1 = \frac{24\sqrt{3} - 15}{\sqrt{3}} $

$ h_1 = \frac{24\sqrt{3}}{\sqrt{3}} - \frac{15}{\sqrt{3}} $

$ h_1 = 24 - \frac{15}{\sqrt{3}} $ m

---

The height of the first pole is $24 - \frac{15}{\sqrt{3}}$ m.

---

Comparing this result with the given options:

(A) $24 - 15\sqrt{3}$ m

(B) $24 - \frac{15}{\sqrt{3}}$ m

(C) $24 + 15\sqrt{3}$ m

(D) $24 + \frac{15}{\sqrt{3}}$ m

Our calculated height matches option (B).

---

Therefore, the correct option is (B) $24 - \frac{15}{\sqrt{3}}$ m.

Given:

Distance of the observation point B from the point on the ground directly below the balloon (A), $AB = 50$ m.

Angle of elevation of the balloon from point B at the first observation, $\theta_1 = 30^\circ$.

---

To Find:

The height of the balloon from the ground at the first observation, $h_1$.

---

Solution:

Let A be the point on the ground directly below the balloon, and B be the observation point. Let $K_1$ be the position of the balloon at the first observation. The height of the balloon at the first observation is $AK_1 = h_1$. The horizontal distance from B to A is $AB = 50$ m.

The angle of elevation from B to $K_1$ is $\angle ABK_1 = 30^\circ$.

Consider the right-angled triangle $\triangle AK_1B$. The angle at A is $90^\circ$ (since the balloon is rising vertically from A, and the ground is horizontal).

In $\triangle AK_1B$, we have the angle of elevation $30^\circ$, the side opposite to this angle is $AK_1 = h_1$, and the side adjacent to this angle is $AB = 50$ m.

We can use the tangent trigonometric ratio:

$ \tan(\text{angle of elevation}) = \frac{\text{Opposite side}}{\text{Adjacent side}} $

$ \tan(30^\circ) = \frac{AK_1}{AB} $

---

We know that the value of $ \tan(30^\circ) $ is $ \frac{1}{\sqrt{3}} $.

Substitute this value into equation (i):

$ \frac{1}{\sqrt{3}} = \frac{h_1}{50} $

---

To solve for $h_1$, multiply both sides by 50:

$ h_1 = 50 \times \frac{1}{\sqrt{3}} $

$ h_1 = \frac{50}{\sqrt{3}} $ m

---

The height of the balloon from the ground at the first observation is $\frac{50}{\sqrt{3}}$ m.

---

Comparing this result with the given options:

(A) 50 m

(B) $50\sqrt{3}$ m

(C) $\frac{50}{\sqrt{3}}$ m

(D) 100 m

Our calculated height matches option (C).

---

Therefore, the correct option is (C) $\frac{50}{\sqrt{3}}$ m.

Given:

Distance from the foot of the tower to the point of observation = $15 \text{ m}$.

Angle of elevation of the top of the tower = $60^\circ$.

---

To Find:

The height of the tower.

---

Solution:

Let AB be the height of the tower standing vertically on the ground, and let C be the point on the ground $15 \text{ m}$ away from the foot of the tower (A).

Then, AC = $15 \text{ m}$.

The angle of elevation of the top of the tower (B) from point C is $\angle \text{ACB} = 60^\circ$.

In the right-angled triangle ABC, we have:

Substituting the given values into equation (i):

We know that the value of $\tan(60^\circ)$ is $\sqrt{3}$.

Substituting this value into equation (ii):

$\sqrt{3} = \frac{\text{AB}}{15}$

Multiplying both sides by 15:

$\text{AB} = 15 \times \sqrt{3}$

$\text{AB} = 15\sqrt{3} \text{ m}$.

Therefore, the height of the tower is $15\sqrt{3} \text{ m}$.

---

The final answer is $\boxed{15\sqrt{3} \text{ m}}$.

Given:

Height of the kite from the ground = $75 \text{ m}$.

Inclination of the string with the ground = $45^\circ$.

---

To Find:

The length of the string.

---

Solution:

Let B be the position of the kite and A be the point on the ground directly below the kite. Let C be the point on the ground where the string is tied.

Then, the height of the kite from the ground is AB = $75 \text{ m}$.

The inclination of the string with the ground is $\angle \text{ACB} = 45^\circ$.

The length of the string is BC.

Assuming there is no slack in the string, ABC is a right-angled triangle at A.

In the right-angled triangle ABC, we can use the sine function to relate the opposite side (AB) to the hypotenuse (BC):

Substituting the given values into equation (i):

We know that the value of $\sin(45^\circ)$ is $\frac{1}{\sqrt{2}}$.

Substituting this value into equation (ii):

$\frac{1}{\sqrt{2}} = \frac{75}{\text{BC}}$

Cross-multiplying to solve for BC:

$\text{BC} = 75 \times \sqrt{2}$

$\text{BC} = 75\sqrt{2} \text{ m}$.

Therefore, the length of the string is $75\sqrt{2} \text{ m}$.

---

The final answer is $\boxed{75\sqrt{2} \text{ m}}$.

Given:

Height of the man's eye level above water level = $10 \text{ m}$.

Angle of elevation of the top of the cliff = $60^\circ$.

Angle of depression of the base of the cliff = $30^\circ$.

---

To Find:

The height of the cliff.

---

Solution:

Let A be the base of the cliff and B be the top of the cliff. Let C be the position of the man's eye on the deck of the ship, and let D be the point on the water level directly below C. So, CD = $10 \text{ m}$.

Draw a horizontal line CE from C parallel to AD, meeting AB at E. Then CE = AD and AE = CD = $10 \text{ m}$.

The height of the cliff is AB = AE + EB.

The angle of elevation of the top of the cliff from C is $\angle \text{BCE} = 60^\circ$.

The angle of depression of the base of the cliff from C is $\angle \text{ECA} = 30^\circ$.

Since CE is parallel to AD, $\angle \text{CAD} = \angle \text{ECA} = 30^\circ$ (alternate interior angles).

In the right-angled triangle CEA, we have:

Substituting the known values into equation (i):

From equation (ii), we get:

$\text{CE} = \frac{10}{\tan(30^\circ)} = \frac{10}{1/\sqrt{3}} = 10\sqrt{3} \text{ m}$.

Now, consider the right-angled triangle BCE:

Substituting the known values and CE from equation (iii) into equation (iv):

From equation (v), we get:

$\text{EB} = \tan(60^\circ) \times 10\sqrt{3} = \sqrt{3} \times 10\sqrt{3} = 10 \times 3 = 30 \text{ m}$.

The total height of the cliff AB is the sum of AE and EB:

$\text{AB} = \text{AE} + \text{EB}$

$\text{AB} = 40 \text{ m}$.

Therefore, the height of the cliff is $40 \text{ m}$.

---

The final answer is $\boxed{40 \text{ m}}$.

Given:

Height of the building = $20 \text{ m}$.

Angle of elevation of the top of the building from point P = $30^\circ$.

Angle of elevation of the top of the flagstaff from point P = $45^\circ$.

---

To Find:

The length of the flagstaff.

---

Solution:

Let AB be the height of the building, where AB = $20 \text{ m}$. Let BC be the height of the flagstaff on top of the building. Let P be the point on the ground.

The angle of elevation of the top of the building (B) from P is $\angle \text{APB} = 30^\circ$.

The angle of elevation of the top of the flagstaff (C) from P is $\angle \text{APC} = 45^\circ$.

Consider the right-angled triangle ABP. We have:

Substituting the given values into equation (i):

From equation (ii), we can find AP:

Now, consider the right-angled triangle ACP. The height of the top of the flagstaff from the ground is AC = AB + BC = $20 + \text{BC}$. We have:

Substituting the given values and AP from equation (iii) into equation (iv):

We know that $\tan(45^\circ) = 1$. Substituting this into equation (v):

$1 = \frac{20 + \text{BC}}{20\sqrt{3}}$

Multiplying both sides by $20\sqrt{3}$:

$20\sqrt{3} = 20 + \text{BC}$

Subtracting 20 from both sides to find BC:

$\text{BC} = 20\sqrt{3} - 20$

Factor out 20:

$\text{BC} = 20(\sqrt{3} - 1) \text{ m}$.

Therefore, the length of the flagstaff is $20(\sqrt{3} - 1) \text{ m}$.

---

The final answer is $\boxed{20(\sqrt{3}-1) \text{ m}}$.

Given:

Initial angle of elevation = $30^\circ$.

Final angle of elevation = $60^\circ$.

Distance moved towards the tower = $20 \text{ m}$.

---

To Find:

The height of the tower.

---

Solution:

Let AB be the height of the tower, where A is the base and B is the top. Let P be the initial observation point and Q be the point after moving $20 \text{ m}$ towards the foot of the tower. So, PQ = $20 \text{ m}$.

The angle of elevation from P to B is $\angle \text{APB} = 30^\circ$.

The angle of elevation from Q to B is $\angle \text{AQB} = 60^\circ$.

Let AB = $h$ and AQ = $x$. Then AP = AQ + QP = $x + 20$.

Consider the right-angled triangle ABQ. We have:

Substituting the values into equation (i):

From equation (ii), we get:

Now, consider the right-angled triangle ABP. We have:

Substituting the values into equation (iv):

Substituting the value of $\tan(30^\circ)$ into equation (v):

From equation (vi), we get:

Now, substitute the value of $x$ from equation (iii) into equation (vii):

$\frac{h}{\sqrt{3}} + 20 = h\sqrt{3}$

Subtract $\frac{h}{\sqrt{3}}$ from both sides:

$20 = h\sqrt{3} - \frac{h}{\sqrt{3}}$

$20 = h \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right)$

$20 = h \left( \frac{(\sqrt{3})^2 - 1}{\sqrt{3}} \right)$

$20 = h \left( \frac{3 - 1}{\sqrt{3}} \right)$

$20 = h \left( \frac{2}{\sqrt{3}} \right)$

Now, solve for $h$:

$h = 20 \times \frac{\sqrt{3}}{2}$

$h = 10\sqrt{3}$.

The height of the tower is $10\sqrt{3} \text{ m}$.

---

The final answer is $\boxed{10\sqrt{3} \text{ m}}$.

Given:

Length of the ladder = $15 \text{ m}$.

Angle made by the ladder with the wall = $60^\circ$.

---

To Find:

The height of the point where the ladder touches the wall.

The distance of the foot of the ladder from the wall.

---

Solution:

Let AB be the wall and A be the ground level at the base of the wall. Let the ladder be represented by BC, where B is the point where the ladder touches the wall and C is the foot of the ladder on the ground. The angle between the ladder and the wall is $\angle \text{ABC} = 60^\circ$. The length of the ladder is BC = $15 \text{ m}$.

Triangle ABC is a right-angled triangle at A ($\angle \text{BAC} = 90^\circ$).

We need to find the height AB and the distance AC.

In $\triangle \text{ABC}$, the side adjacent to the angle $60^\circ$ is AB, and the hypotenuse is BC. We can use the cosine function:

Substitute the given values into equation (i):

We know that $\cos(60^\circ) = \frac{1}{2}$. Substitute this into equation (ii):

$\frac{1}{2} = \frac{\text{AB}}{15}$

Solving for AB:

$\text{AB} = 15 \times \frac{1}{2} = \frac{15}{2} = 7.5 \text{ m}$.

Now, to find the distance AC, which is the side opposite to the angle $60^\circ$. We can use the sine function:

Substitute the given values into equation (iv):

We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2}$. Substitute this into equation (v):

$\frac{\sqrt{3}}{2} = \frac{\text{AC}}{15}$

Solving for AC:

$\text{AC} = 15 \times \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{2} \text{ m}$.

The height of the point where the ladder touches the wall is $7.5 \text{ m}$.

The distance of the foot of the ladder from the wall is $\frac{15\sqrt{3}}{2} \text{ m}$.

---

The final answer is $\boxed{\text{Height} = 7.5 \text{ m}, \text{Distance} = \frac{15\sqrt{3}}{2} \text{ m}}$.

Given:

Height of the building = $7 \text{ m}$.

Angle of elevation of the top of the cable tower from the top of the building = $60^\circ$.

Angle of depression of the foot of the cable tower from the top of the building = $45^\circ$.

---

To Find:

The height of the cable tower.

---

Solution:

Let AB be the building with height $7 \text{ m}$ (A is the base, B is the top). Let CD be the cable tower (C is the foot, D is the top).

From the top of the building B, a horizontal line BE is drawn parallel to the ground AC, meeting CD at E.

The height of the building AB = $7 \text{ m}$. Since BE is parallel to AC and BC is a transversal, the distance AC is equal to BE. Also, since AB is perpendicular to AC and CE is perpendicular to BE (assuming vertical tower and horizontal ground), ABCE forms a rectangle. Thus, CE = AB = $7 \text{ m}$.

The height of the tower CD = CE + ED = $7 + \text{ED}$.

The angle of elevation of the top of the tower (D) from B is $\angle \text{DBE} = 60^\circ$.

The angle of depression of the foot of the tower (C) from B is $\angle \text{EBC} = 45^\circ$.

Since BE is parallel to AC, the angle of depression from B to C ($\angle \text{EBC}$) is equal to the angle of elevation from C to B ($\angle \text{BCA}$).

Consider the right-angled triangle BAC.

Substitute the known values into equation (i):

Since $\tan(45^\circ) = 1$, equation (ii) becomes:

From equation (iii), we get AC = $7 \text{ m}$.

Since BE is parallel to AC and ABCE forms a rectangle, BE = AC.

Now, consider the right-angled triangle BDE.

Substitute the known values and BE from equation (iv) into equation (v):

Since $\tan(60^\circ) = \sqrt{3}$, equation (vi) becomes:

From equation (vii), we get ED = $7\sqrt{3} \text{ m}$.

The height of the tower CD is CE + ED.

Factor out 7:

$\text{CD} = 7(1 + \sqrt{3}) \text{ m}$.

---

The final answer is $\boxed{7(1+\sqrt{3}) \text{ m}}$.

Given:

Initial angle of elevation of the tree from the bank = $60^\circ$.

Distance moved away from the bank = $30 \text{ m}$.

Final angle of elevation from the new position = $30^\circ$.

---

To Find:

The height of the tree.

The width of the river.

---

Solution:

Let AB be the tree on the opposite bank, where A is the base on the bank and B is the top. Let P be the initial position of the person on the bank of the river. The width of the river is AP.

The angle of elevation from P to the top of the tree B is $\angle \text{APB} = 60^\circ$.

The person moves $30 \text{ m}$ away from the bank to a point Q, such that P, A, and Q are collinear and P is between A and Q. So, PQ = $30 \text{ m}$. The distance AQ = AP + PQ.

The angle of elevation from Q to the top of the tree B is $\angle \text{AQB} = 30^\circ$.

Let the height of the tree AB = $h$ meters and the width of the river AP = $x$ meters.

Then, AQ = $x + 30$ meters.

Consider the right-angled triangle ABP:

Substitute the known values into equation (i):

We know that $\tan(60^\circ) = \sqrt{3}$. From equation (ii):

Now, consider the right-angled triangle ABQ:

Substitute the known values into equation (iv):

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. Substitute this into equation (v):

Substitute the value of $h$ from equation (iii) into equation (vi):

$\frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x+30}$

Cross-multiply:

$1 \times (x+30) = \sqrt{3} \times x\sqrt{3}$

$x+30 = 3x$

Subtract $x$ from both sides:

$30 = 3x - x$

$30 = 2x$

Divide by 2:

So, the width of the river is $15 \text{ m}$.

Now, substitute the value of $x$ from equation (vii) into equation (iii) to find $h$:

So, the height of the tree is $15\sqrt{3} \text{ m}$.

---

The height of the tree is $15\sqrt{3} \text{ m}$ and the width of the river is $15 \text{ m}$.

---

The final answer is $\boxed{\text{Height} = 15\sqrt{3} \text{ m, Width} = 15 \text{ m}}$.

Given:

Height of the observer = $1.5 \text{ m}$.

Distance of the observer from the tower = $28.5 \text{ m}$.

Angle of elevation of the top of the tower from the observer's eyes = $45^\circ$.

---

To Find:

The height of the tower.

---

Solution:

Let AB be the height of the tower, where A is the base and B is the top. Let C be the position of the observer's feet on the ground, and D be the position of the observer's eyes, so CD is the height of the observer, $1.5 \text{ m}$.

The distance of the observer from the tower is AC = $28.5 \text{ m}$.

Draw a horizontal line DE from the observer's eyes to the tower, meeting AB at E. Then AE = CD = $1.5 \text{ m}$ and DE = AC = $28.5 \text{ m}$.

The angle of elevation of the top of the tower (B) from the observer's eyes (D) is $\angle \text{BDE} = 45^\circ$.

Consider the right-angled triangle BDE. We have:

Substitute the given values into equation (i):

We know that $\tan(45^\circ) = 1$. Substituting this into equation (ii):

$1 = \frac{\text{BE}}{28.5}$

Solving for BE:

The total height of the tower AB is the sum of AE and BE:

$\text{AB} = \text{AE} + \text{BE}$

$\text{AB} = 30 \text{ m}$.

Therefore, the height of the tower is $30 \text{ m}$.

---

The final answer is $\boxed{30 \text{ m}}$.

Given:

Height of the flagstaff = $5 \text{ m}$.

Angle of elevation of the bottom of the flagstaff = $30^\circ$.

Angle of elevation of the top of the flagstaff = $60^\circ$.

---

To Find:

The height of the tower.

---

Solution:

Let AB be the height of the tower, where A is the base on the horizontal plane and B is the top of the tower. Let BC be the height of the flagstaff on top of the tower, so BC = $5 \text{ m}$. Let P be the point on the plane from where the angles of elevation are observed.

Let the height of the tower AB = $h$ meters and the distance from the point P to the base of the tower AP = $x$ meters.

The angle of elevation of the bottom of the flagstaff (B) from P is $\angle \text{APB} = 30^\circ$.

The angle of elevation of the top of the flagstaff (C) from P is $\angle \text{APC} = 60^\circ$.

The total height from the base of the tower to the top of the flagstaff is AC = AB + BC = $h + 5$ meters.

Consider the right-angled triangle ABP:

Substitute the known values into equation (i):

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. From equation (ii):

Now, consider the right-angled triangle ACP:

Substitute the known values into equation (iv):

We know that $\tan(60^\circ) = \sqrt{3}$. Substitute this into equation (v):

Substitute the value of $x$ from equation (iii) into equation (vi):

$\sqrt{3} = \frac{h+5}{h\sqrt{3}}$

Multiply both sides by $h\sqrt{3}$:

$\sqrt{3} \times h\sqrt{3} = h+5$

$3h = h+5$

Subtract $h$ from both sides:

$3h - h = 5$

$2h = 5$

Divide by 2:

The height of the tower is $2.5 \text{ m}$.

---

The final answer is $\boxed{2.5 \text{ m}}$.

Given:

Height of the lighthouse = $50 \text{ m}$.

Angle of depression to the first ship = $60^\circ$.

Angle of depression to the second ship = $45^\circ$.

The two ships are on the same side of the lighthouse.

---

To Find:

The distance between the two ships.

---

Solution:

Let AB be the lighthouse, with height AB = $50 \text{ m}$. A is the base of the lighthouse on the ground, and B is the top.

Let C and D be the positions of the two ships on the same side of the lighthouse. Assume C is closer to the base A than D, corresponding to the larger angle of depression.

Let BE be a horizontal line from the top of the lighthouse B.

The angle of depression to ship C from B is $\angle \text{EBC} = 60^\circ$.

The angle of depression to ship D from B is $\angle \text{EBD} = 45^\circ$.

Since BE is parallel to the ground AD (assuming the ground is horizontal), we have the alternate interior angles:

Consider the right-angled triangle ABC.

Substitute the known values into equation (i):

We know that $\tan(60^\circ) = \sqrt{3}$. From equation (ii), we get:

Solving for AC:

Now, consider the right-angled triangle ABD.

Substitute the known values into equation (v):

We know that $\tan(45^\circ) = 1$. From equation (vi), we get:

Solving for AD:

The distance between the two ships C and D is the difference between the distances AD and AC, since they are on the same side of the lighthouse.

Substitute the values of AD from equation (viii) and AC from equation (iv) into equation (ix):

To simplify the expression, we can rationalize the term $\frac{50}{\sqrt{3}}$:

$\frac{50}{\sqrt{3}} = \frac{50 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{50\sqrt{3}}{3}$

Substitute this back into the expression for CD:

$\text{CD} = 50 - \frac{50\sqrt{3}}{3}$

Combine the terms by finding a common denominator:

$\text{CD} = \frac{50 \times 3}{3} - \frac{50\sqrt{3}}{3}$

$\text{CD} = \frac{150 - 50\sqrt{3}}{3}$

Factor out 50 from the numerator:

$\text{CD} = \frac{50(3 - \sqrt{3})}{3} \text{ m}$.

Therefore, the distance between the two ships is $\frac{50(3 - \sqrt{3})}{3} \text{ m}$.

---

The final answer is $\boxed{\frac{50(3-\sqrt{3})}{3} \text{ m}}$.

Given:

Height of the statue = $1.6 \text{ m}$.

Angle of elevation of the top of the statue from a point on the ground = $60^\circ$.

Angle of elevation of the top of the pedestal from the same point on the ground = $45^\circ$.

---

To Find:

The height of the pedestal.

---

Solution:

Let AB be the pedestal with A on the ground and B at the top. Let BC be the statue on top of the pedestal, with C being the top of the statue. The height of the statue is BC = $1.6 \text{ m}$.

Let P be the point on the ground from where the angles of elevation are observed.

Let the height of the pedestal AB = $h$ meters and the distance from the point P to the base of the pedestal AP = $x$ meters.

The angle of elevation of the top of the pedestal (B) from P is $\angle \text{APB} = 45^\circ$.

The angle of elevation of the top of the statue (C) from P is $\angle \text{APC} = 60^\circ$.

The total height from the ground to the top of the statue is AC = AB + BC = $h + 1.6$ meters.

Consider the right-angled triangle ABP:

Substitute the known values into equation (i):

We know that $\tan(45^\circ) = 1$. From equation (ii):

Now, consider the right-angled triangle ACP:

Substitute the known values and AC = $h + 1.6$ into equation (iv):

We know that $\tan(60^\circ) = \sqrt{3}$. Substitute this and the value of $x$ from equation (iii) into equation (v):

Multiply both sides by $h$:

$h\sqrt{3} = h+1.6$

Subtract $h$ from both sides:

$h\sqrt{3} - h = 1.6$

Factor out $h$:

$h(\sqrt{3} - 1) = 1.6$

Solve for $h$:

$h = \frac{1.6}{\sqrt{3} - 1}$

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, $(\sqrt{3} + 1)$:

$h = \frac{1.6}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)}$

$h = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2}$

$h = \frac{1.6(\sqrt{3} + 1)}{3 - 1}$

$h = \frac{1.6(\sqrt{3} + 1)}{2}$

$h = 0.8(\sqrt{3} + 1) \text{ m}$.

The height of the pedestal is $0.8(\sqrt{3} + 1) \text{ m}$.

---

The final answer is $\boxed{0.8(\sqrt{3}+1) \text{ m}}$.

Given:

Height of the girl = $1.5 \text{ m}$.

Height of the balloon from the ground = $88.5 \text{ m}$.

Initial angle of elevation of the balloon from the girl's eyes = $60^\circ$.

Final angle of elevation of the balloon from the girl's eyes = $30^\circ$.

---

To Find:

The distance travelled by the balloon during the interval.

---

Solution:

Let G be the position of the girl's eyes, which is $1.5 \text{ m}$ above the ground. Let A and B be the initial and final positions of the balloon, respectively.

The balloon is moving in a horizontal line at a height of $88.5 \text{ m}$ from the ground. The height of the balloon above the girl's eye level is $(88.5 - 1.5) \text{ m} = 87 \text{ m}$.

Let the horizontal line through the girl's eyes be GP. Let the vertical lines from A and B meet the line GP at C and D respectively.

Then AC = BD = $87 \text{ m}$ (height of the balloon above eye level).

The initial angle of elevation from G to A is $\angle \text{AGC} = 60^\circ$.

The final angle of elevation from G to B is $\angle \text{BGD} = 30^\circ$.

The distance travelled by the balloon is the horizontal distance between A and B, which is CD. Since GCDB forms a rectangle (assuming balloon moves horizontally and GC is perpendicular to CD and DB is perpendicular to GD), CD = GD - GC.

Consider the right-angled triangle AGC:

Substitute the known values into equation (i):

We know that $\tan(60^\circ) = \sqrt{3}$. From equation (ii):

Solving for GC:

Rationalize the denominator:

$\text{GC} = \frac{87}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{87\sqrt{3}}{3} = 29\sqrt{3} \text{ m}$.

Now, consider the right-angled triangle BGD:

Substitute the known values into equation (vi):

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. From equation (vii):

Solving for GD:

The distance travelled by the balloon is CD = GD - GC.

$\text{CD} = (87 - 29)\sqrt{3}$

$\text{CD} = 58\sqrt{3} \text{ m}$.

Therefore, the distance travelled by the balloon during the interval is $58\sqrt{3} \text{ m}$.

---

The final answer is $\boxed{58\sqrt{3} \text{ m}}$.

Given:

Initial angle of depression of the car = $30^\circ$.

Time taken to travel from the first observed point to the second observed point = $6$ seconds.

Final angle of depression of the car = $60^\circ$.

The car is moving with a uniform speed towards the foot of the tower.

---

To Find:

The time taken by the car to reach the foot of the tower from the second observed point.

---

Solution:

Let AB be the tower, where A is the foot of the tower on the highway and B is the top of the tower. Let the height of the tower be $h$ meters.

Let C be the initial position of the car and D be the position of the car after 6 seconds. The car moves from C to D and then from D to A along the highway AC.

Let BE be a horizontal line from the top of the tower B.

The angle of depression of the car at C is $\angle \text{EBC} = 30^\circ$. Since BE is parallel to the highway AC, the alternate interior angle is $\angle \text{BCA} = 30^\circ$.

The angle of depression of the car at D is $\angle \text{EBD} = 60^\circ$. Since BE is parallel to the highway AC, the alternate interior angle is $\angle \text{BDA} = 60^\circ$.

Let the uniform speed of the car be $v$ meters per second.

The distance travelled by the car from C to D in 6 seconds is CD = $v \times 6 = 6v$ meters.

Let the time taken by the car to reach the foot of the tower from D be $t$ seconds. The distance covered from D to A is DA = $v \times t = vt$ meters.

The total distance CA = CD + DA = $6v + vt$ meters.

Consider the right-angled triangle ABD:

Substitute the known values into equation (i):

We know that $\tan(60^\circ) = \sqrt{3}$. From equation (ii), we get:

Now, consider the right-angled triangle ABC:

Substitute the known values into equation (iv):

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. Substitute this into equation (v):

From equation (vi), we get:

Substitute the value of $h$ from equation (iii) into equation (vii):

$v(6+t) = (vt\sqrt{3})\sqrt{3}$

$6v+vt = 3vt$

Subtract $vt$ from both sides:

$6v = 3vt - vt$

$6v = 2vt$

Since $v$ is the speed of the car, $v \neq 0$. We can divide both sides by $2v$:

$\frac{6v}{2v} = t$

Therefore, the time taken by the car to reach the foot of the tower from point D is 3 seconds.

---

The final answer is $\boxed{3 \text{ seconds}}$.

Given:

Distance of the first point from the base of the tower = $4 \text{ m}$.

Distance of the second point from the base of the tower = $9 \text{ m}$.

The two points are in the same straight line with the base of the tower.

The angles of elevation of the top of the tower from these two points are complementary.

---

To Find:

The height of the tower.

---

Solution:

Let AB be the height of the tower, where A is the base on the ground and B is the top. Let the height of the tower be AB = $h$ meters.

Let C and D be the two points on the ground such that they are in the same straight line with the base A. Let AC = $4 \text{ m}$ and AD = $9 \text{ m}$. Assume C is closer to the tower than D.

Let the angle of elevation from point C be $\angle \text{BCA} = \theta$, and the angle of elevation from point D be $\angle \text{BDA} = \phi$.

According to the problem, the angles of elevation are complementary, which means their sum is $90^\circ$.

From equation (i), we can write $\phi = 90^\circ - \theta$ or $\theta = 90^\circ - \phi$.

Consider the right-angled triangle ABC:

Substituting the known values into equation (ii):

Now, consider the right-angled triangle ABD:

Substituting the known values into equation (iv):

Using the complementary angle relationship $\phi = 90^\circ - \theta$, substitute this into equation (v):

Using the trigonometric identity $\tan(90^\circ - \theta) = \cot(\theta)$, equation (vi) becomes:

We also know that $\cot(\theta) = \frac{1}{\tan(\theta)}$. Substituting the expression for $\tan(\theta)$ from equation (iii):

Equating the two expressions for $\cot(\theta)$ from equation (vii) and equation (viii):

Cross-multiplying equation (ix):

$h \times h = 9 \times 4$

$h^2 = 36$

Taking the square root of both sides:

Since the height of the tower must be positive:

$h = 6 \text{ m}$.

Therefore, the height of the tower is $6 \text{ m}$.

---

The final answer is $\boxed{6 \text{ m}}$.

Given:

Angles of depression of two consecutive kilometre stones from the top of a hill are $30^\circ$ and $45^\circ$.

The stones are due east and are consecutive kilometre stones, so the distance between them is $1 \text{ km} = 1000 \text{ m}$.

---

To Find:

The height of the hill.

---

Solution:

Let AB be the height of the hill, where A is the base on the ground and B is the top. Let AB = $h$ meters.

Let C and D be the positions of the two consecutive kilometre stones on the ground, due east of A. Since the angle of depression is greater for the nearer object, C is closer to A than D. The distance between C and D is CD = $1000 \text{ m}$.

Let BE be a horizontal line through B. The angles of depression from B to C and D are $\angle \text{EBC} = 45^\circ$ and $\angle \text{EBD} = 30^\circ$ respectively.

Since BE is parallel to the ground AD, the alternate interior angles are equal:

Consider the right-angled triangle ABC:

Substitute the known values into equation (i):

Since $\tan(45^\circ) = 1$, equation (ii) becomes:

Now, consider the right-angled triangle ABD:

Substitute the known values into equation (iv):

Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, equation (v) becomes:

The two stones are consecutive and lie on a straight line from the base of the tower. The distance between them is CD = $1000 \text{ m}$. Since C is between A and D, we have:

Substitute the expressions for AC from equation (iii) and AD from equation (vi) into equation (viii):

Factor out $h$ from the left side of equation (ix):

$h(\sqrt{3} - 1) = 1000$

Solve for $h$:

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, $(\sqrt{3} + 1)$:

$h = \frac{1000}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)}$

$h = \frac{1000(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2}$

$h = \frac{1000(\sqrt{3} + 1)}{3 - 1}$

$h = \frac{1000(\sqrt{3} + 1)}{2}$

$h = 500(\sqrt{3} + 1) \text{ m}$.

Therefore, the height of the hill is $500(\sqrt{3} + 1) \text{ m}$.

---

The final answer is $\boxed{500(\sqrt{3}+1) \text{ m}}$.

Given:

Height of the higher aeroplane from the ground = $300 \text{ m}$.

Angle of elevation of the higher aeroplane from a point on the ground = $60^\circ$.

Angle of elevation of the lower aeroplane from the same point on the ground = $45^\circ$.

The two aeroplanes are vertically above each other at the instant of observation.

---

To Find:

The vertical distance between the two aeroplanes.

---

Solution:

Let P be the point on the ground from where the angles of elevation are observed. Let A be the point on the ground vertically below the two aeroplanes. Let B be the position of the lower aeroplane and C be the position of the higher aeroplane at the instant of observation.

The height of the higher aeroplane from the ground is AC = $300 \text{ m}$.

Let the height of the lower aeroplane from the ground be AB = $h$ meters.

Let the horizontal distance from the observation point to the point below the aeroplanes be AP = $x$ meters.

The angle of elevation of the higher aeroplane (C) from P is $\angle \text{APC} = 60^\circ$.

The angle of elevation of the lower aeroplane (B) from P is $\angle \text{APB} = 45^\circ$.

The vertical distance between the two aeroplanes is BC = AC - AB = $300 - h$ meters.

Consider the right-angled triangle APC. We have:

Substituting the known values into equation (i):

From equation (ii), we get:

Solving for $x$ from equation (iii):

Now, consider the right-angled triangle APB. We have:

Substituting the known values into equation (v):

From equation (vi), we get:

From equation (vii), we have $h = x$.

Substitute the value of $x$ from equation (iv) into the expression for $h$:

The vertical distance between the two aeroplanes is $d = \text{AC} - \text{AB} = 300 - h$.

Substitute the value of $h$ from equation (viii):

Factor out 100:

$d = 100(3 - \sqrt{3}) \text{ m}$.

Therefore, the vertical distance between the two aeroplanes is $100(3 - \sqrt{3}) \text{ m}$.

---

The final answer is $\boxed{100(3-\sqrt{3}) \text{ m}}$.

Given:

Length of the rope = $20 \text{ m}$.

Angle made by the rope with the ground = $30^\circ$.

The rope is stretched and tied from the top of a vertical pole to the ground.

---

To Find:

The height of the pole.

---

Solution:

Let AB be the vertical pole, where A is the foot of the pole on the ground and B is the top of the pole. Let C be the point on the ground where the rope is tied.

The rope is BC, and its length is BC = $20 \text{ m}$.

The angle made by the rope with the ground is $\angle \text{BCA} = 30^\circ$.

Triangle ABC is a right-angled triangle at A ($\angle \text{BAC} = 90^\circ$), since the pole is vertical to the ground.

We need to find the height of the pole AB.

In the right-angled triangle ABC, the side opposite to the angle $30^\circ$ is AB, and the hypotenuse is BC. We can use the sine function:

Substitute the given values into equation (i):

We know that $\sin(30^\circ) = \frac{1}{2}$. Substitute this into equation (ii):

$\frac{1}{2} = \frac{\text{AB}}{20}$

Solving for AB:

$\text{AB} = 20 \times \frac{1}{2} = 10 \text{ m}$.

Therefore, the height of the pole is $10 \text{ m}$.

---

The final answer is $\boxed{10 \text{ m}}$.

Given:

Angle made by the broken part with the ground = $30^\circ$.

Distance between the foot of the tree and the point where the top touches the ground = $8 \text{ m}$.

---

To Find:

The height of the tree before it broke.

---

Solution:

Let the tree be AC, where A is the foot of the tree and C is the top. Due to the storm, the tree breaks at a point B. The broken part BC bends and touches the ground at point D, such that BD represents the broken part of the tree (BC = BD).

The foot of the tree is at A, and the top touches the ground at D. The distance between the foot of the tree and the point where the top touches the ground is AD = $8 \text{ m}$.

The broken part BD makes an angle of $30^\circ$ with the ground, so $\angle \text{BDA} = 30^\circ$.

The height of the tree before it broke is the sum of the unbroken part AB and the broken part BC. Since BC = BD, the height of the tree is AB + BD.

Consider the right-angled triangle ABD (right-angled at A, assuming the tree was vertical).

First, find the length of the unbroken part AB using the tangent function:

Substitute the known values into equation (i):

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. From equation (ii):

Next, find the length of the broken part BD (which is equal to the original broken part BC) using the cosine function:

Substitute the known values into equation (iv):

We know that $\cos(30^\circ) = \frac{\sqrt{3}}{2}$. From equation (v):

The total height of the tree before it broke is AB + BD.

Height of tree = $\frac{8+16}{\sqrt{3}} = \frac{24}{\sqrt{3}}$

Rationalize the denominator:

Height of tree = $\frac{24}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{24\sqrt{3}}{3} = 8\sqrt{3} \text{ m}$.

Therefore, the height of the tree before it broke is $8\sqrt{3} \text{ m}$.

---

The final answer is $\boxed{8\sqrt{3} \text{ m}}$.

Given:

Length of the bridge across the river = $150 \text{ m}$.

Angle made by the bridge with the bank = $45^\circ$.

---

To Find:

The width of the river.

---

Solution:

Let the bridge be represented by the line segment AB, where A is a point on one bank and B is a point on the opposite bank. The length of the bridge is AB = $150 \text{ m}$.

Let the bank be represented by a straight line. The angle made by the bridge with the bank is $45^\circ$.

Let W represent the width of the river, which is the perpendicular distance between the two parallel banks.

Consider a right-angled triangle formed by the bridge (hypotenuse), the bank (adjacent side along the angle), and the width of the river (opposite side).

In this right-angled triangle, the angle between the bridge and the bank is given as $45^\circ$. The width of the river is the side opposite to this angle, and the length of the bridge is the hypotenuse.

We can use the sine function to relate these quantities:

Substitute the given values into equation (i):

We know that the value of $\sin(45^\circ)$ is $\frac{1}{\sqrt{2}}$. Substitute this value into equation (ii):

$\frac{1}{\sqrt{2}} = \frac{W}{150}$

Multiply both sides by 150 to solve for W:

$W = 150 \times \frac{1}{\sqrt{2}} = \frac{150}{\sqrt{2}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$W = \frac{150}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{150\sqrt{2}}{2}$

$W = 75\sqrt{2} \text{ m}$.

Therefore, the width of the river is $75\sqrt{2} \text{ m}$.

---

The final answer is $\boxed{75\sqrt{2} \text{ m}}$.

Given:

Height of the building = $20 \text{ m}$.

Angle of elevation of the bottom of the transmission tower (top of the building) from a point on the ground = $45^\circ$.

Angle of elevation of the top of the transmission tower from the same point on the ground = $60^\circ$.

---

To Find:

The height of the transmission tower.

---

Solution:

Let AB be the height of the building, where A is the base on the ground and B is the top of the building. So, AB = $20 \text{ m}$.

Let BC be the height of the transmission tower fixed at the top of the building, so B is the bottom and C is the top of the tower. Let BC = $h$ meters.

Let P be the point on the ground from where the angles of elevation are observed. Let the distance from the point P to the base of the building A be AP = $x$ meters.

The angle of elevation of the bottom of the tower (B) from P is $\angle \text{APB} = 45^\circ$.

The angle of elevation of the top of the tower (C) from P is $\angle \text{APC} = 60^\circ$.

The total height from the ground to the top of the transmission tower is AC = AB + BC = $20 + h$ meters.

Consider the right-angled triangle ABP. We have:

Substitute the known values into equation (i):

We know that $\tan(45^\circ) = 1$. From equation (ii):

Now, consider the right-angled triangle ACP. We have:

Substitute the known values and AC = $20 + h$ into equation (iv):

We know that $\tan(60^\circ) = \sqrt{3}$. Substitute this and the value of $x$ from equation (iii) into equation (v):

Multiply both sides by 20:

$20\sqrt{3} = 20+h$

Subtract 20 from both sides to solve for $h$:

$h = 20\sqrt{3} - 20$

Factor out 20:

$h = 20(\sqrt{3} - 1) \text{ m}$.

Therefore, the height of the transmission tower is $20(\sqrt{3} - 1) \text{ m}$.

---

The final answer is $\boxed{20(\sqrt{3}-1) \text{ m}}$.

Given:

Height of the vertical pole = $6 \text{ m}$.

Length of the pole's shadow = $4 \text{ m}$.

Length of the tower's shadow = $28 \text{ m}$.

---

To Find:

The height of the tower.

---

Solution:

Let the height of the pole be $h_p = 6 \text{ m}$ and the length of its shadow be $s_p = 4 \text{ m}$.

Let the height of the tower be $h_t$ and the length of its shadow be $s_t = 28 \text{ m}$.

At the same time of the day, the sun makes the same angle of elevation with the ground for both the pole and the tower. This means that the triangle formed by the pole and its shadow is similar to the triangle formed by the tower and its shadow. Both triangles are right-angled triangles (the pole and tower are vertical, and the ground is horizontal).

By AA similarity (Angle of elevation of the sun is the same, and both are right angles), the two triangles are similar.

In similar triangles, the ratio of corresponding sides is equal.

Substituting the given values into equation (i):

Now, we solve for $h_t$ from equation (ii) by cross-multiplication or by multiplying both sides by 28:

$h_t = \frac{6}{4} \times 28$

$h_t = \frac{3}{2} \times 28$

$h_t = 3 \times 14$

Therefore, the height of the tower is $42 \text{ m}$.

---

The final answer is $\boxed{42 \text{ m}}$.

Given:

Let the height of the tower be AB, standing on the ground at point A. Let C and D be two points on the ground, on the same straight line with A, such that AC = $a$ metres and AD = $b$ metres.

The angles of elevation of the top of the tower (B) from C and D are complementary. Let $\angle \text{BCA} = \theta$ and $\angle \text{BDA} = \phi$.

---

To Prove:

The height of the tower AB is $\sqrt{ab}$ metres.

---

Proof:

Let the height of the tower be AB = $h$ metres.

Consider the right-angled triangle ABC (right-angled at A). We have:

Substitute the known values into equation (ii):

Now, consider the right-angled triangle ABD (right-angled at A). We have:

Substitute the known values into equation (iv):

From the complementary condition given in equation (i), we have $\phi = 90^\circ - \theta$. Substitute this into equation (v):

Using the trigonometric identity $\tan(90^\circ - \theta) = \cot(\theta)$, equation (vi) becomes:

We also know that $\cot(\theta) = \frac{1}{\tan(\theta)}$. Substitute the expression for $\tan(\theta)$ from equation (iii) into this identity:

Equating the two expressions for $\cot(\theta)$ from equation (vii) and equation (viii):

Cross-multiplying equation (ix):

$h \times h = a \times b$

$h^2 = ab$

Taking the square root of both sides. Since height must be positive, we take the positive square root:

Thus, the height of the tower is $\sqrt{ab}$ metres.

Hence Proved.

Given:

Height of the lighthouse = $h$ (To be found).

Angles of depression of two ships from the top of the lighthouse = $30^\circ$ and $45^\circ$.

Distance between the two ships = $100 \text{ m}$.

The ships are in a line with the lighthouse.

---

To Find:

The height of the lighthouse.

---

Solution:

Let AB be the lighthouse, where A is the base on the ground and B is the top. Let the height of the lighthouse be AB = $h$ meters.

Let C and D be the positions of the two ships on the ground, in a line with A, such that they are sailing towards the lighthouse. Let C be the position of the ship with the larger angle of depression ($45^\circ$) and D be the position of the ship with the smaller angle of depression ($30^\circ$). Thus, C is closer to A than D.

The distance between the two ships is CD = $100 \text{ m}$.

Let BE be a horizontal line from the top of the lighthouse B.

The angle of depression to ship C from B is $\angle \text{EBC} = 45^\circ$.

The angle of depression to ship D from B is $\angle \text{EBD} = 30^\circ$.

Since BE is parallel to the ground AD (assuming the ground is horizontal), we have the alternate interior angles:

Let the distance from the base of the lighthouse to the closer ship C be AC = $x$ meters.

Then the distance from the base of the lighthouse to the farther ship D is AD = AC + CD = $x + 100$ meters.

Consider the right-angled triangle ABC (right-angled at A).

Substitute the known values into equation (i):

Since $\tan(45^\circ) = 1$, equation (ii) becomes:

Now, consider the right-angled triangle ABD (right-angled at A).

Substitute the known values into equation (iv):

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. Substitute this into equation (v):

From equation (vi), we get:

Substitute the value of $x$ from equation (iii) into equation (vii):

$h + 100 = h\sqrt{3}$

Subtract $h$ from both sides:

$100 = h\sqrt{3} - h$

Factor out $h$:

$100 = h(\sqrt{3} - 1)$

Solve for $h$:

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, $(\sqrt{3} + 1)$:

$h = \frac{100}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)}$

$h = \frac{100(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2}$

$h = \frac{100(\sqrt{3} + 1)}{3 - 1}$

$h = \frac{100(\sqrt{3} + 1)}{2}$

$h = 50(\sqrt{3} + 1) \text{ m}$.

Therefore, the height of the lighthouse is $50(\sqrt{3} + 1) \text{ m}$.

---

The final answer is $\boxed{50(\sqrt{3}+1) \text{ m}}$.

Given:

Height of the girl = $1.2 \text{ m}$.

Height of the lamp above the ground = $3.6 \text{ m}$.

Speed of the girl walking away from the lamp-post = $0.6 \text{ m/s}$.

Time duration = $4$ seconds.

---

To Find:

The length of her shadow after $4$ seconds.

---

Solution:

Let LP be the lamp-post, where L is the lamp and P is the base on the ground. So, LP = $3.6 \text{ m}$.

Let FG be the height of the girl, where F is her feet and G is the top of her head. So, FG = $1.2 \text{ m}$.

The girl walks away from the base P. After $4$ seconds, the distance she walks is:

Distance = Speed $\times$ Time

Distance PF = $0.6 \text{ m/s} \times 4 \text{ s} = 2.4 \text{ m}$.

Let FS be the length of her shadow on the ground. The light source is L, the top of the girl's head is G, and the end of her shadow is S.

Triangle $\triangle \text{LPS}$ and triangle $\triangle \text{GFS}$ are similar (by AA similarity, as $\angle \text{LPS} = \angle \text{GFS} = 90^\circ$ and $\angle \text{LSP} = \angle \text{GSF}$ is the common angle of elevation from the end of the shadow).

In similar triangles, the ratio of corresponding sides is equal:

The distance from the base of the lamp-post to the end of the shadow is PS = PF + FS.

Let the length of the shadow FS be $y$ meters.

Substitute the known values into equation (ii):

Substitute the values of PF and FS into equation (iii):

Simplify the left side of equation (iv):

$3 = \frac{2.4 + y}{y}$

Multiply both sides by $y$:

$3y = 2.4 + y$

Subtract $y$ from both sides:

$3y - y = 2.4$

$2y = 2.4$

Divide by 2:

Thus, the length of the girl's shadow after $4$ seconds is $1.2 \text{ m}$.

---

The final answer is $\boxed{1.2 \text{ m}}$.

Given:

Initial angle of elevation from point A = $60^\circ$.

Time duration of flight between observations = $15$ seconds.

Final angle of elevation from point A = $30^\circ$.

Constant height of the jet fighter = $1500 \sqrt{3} \text{ m}$.

---

To Find:

The speed of the jet fighter.

---

Solution:

Let A be the point on the ground from where the angles of elevation are observed. Let the jet fighter be at positions B and C at the two instants, such that B is the initial position and C is the final position after 15 seconds.

Let the constant height of the jet be $h = 1500 \sqrt{3} \text{ m}$. Let P and Q be the points on the ground vertically below B and C respectively.

Then BP = CQ = $h = 1500 \sqrt{3} \text{ m}$.

The angle of elevation of the jet at B from A is $\angle \text{BAP} = 60^\circ$.

The angle of elevation of the jet at C from A is $\angle \text{CAQ} = 30^\circ$.

The jet is flying horizontally, so the distance travelled by the jet in 15 seconds is BC, which is equal to the horizontal distance PQ.

Let the speed of the jet fighter be $v$ m/s. The distance PQ = BC = $v \times 15 = 15v$ meters.

Consider the right-angled triangle ABP. We have:

Substitute the known values into equation (i):

We know that $\tan(60^\circ) = \sqrt{3}$. From equation (ii):

Solving for AP from equation (iii):

Now, consider the right-angled triangle ACQ. We have:

Substitute the known values into equation (v):

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. From equation (vi):

Solving for AQ from equation (vii):

The distance PQ = AQ - AP.

$\text{PQ} = 3000 \text{ m}$.

The distance travelled by the jet in 15 seconds is PQ = $15v$.

Solving for $v$ from equation (xi):

$v = \frac{3000}{15}$

$v = 200$ m/s.

Therefore, the speed of the jet fighter is 200 m/s.

---

The final answer is $\boxed{200 \text{ m/s}}$.

Given:

Initial angle of depression of the car = $30^\circ$.

Final angle of depression of the car = $45^\circ$.

Time taken for the angle of depression to change from $30^\circ$ to $45^\circ$ = $12$ minutes.

The car is moving at a uniform speed towards the foot of the tower.

---

To Find:

The time taken by the car to reach the foot of the tower from the point where the angle of depression is $45^\circ$.

---

Solution:

Let AB be the vertical tower, where A is the foot of the tower on the ground and B is the top of the tower. Let the height of the tower be AB = $h$ meters.

Let C be the initial position of the car where the angle of depression is $30^\circ$, and D be the position of the car after 12 minutes where the angle of depression is $45^\circ$. The car moves from C to D, and then from D to A along the ground.

Let BE be a horizontal line from the top of the tower B.

The angle of depression to the car at C is $\angle \text{EBC} = 30^\circ$. Since BE is parallel to the ground AC, the alternate interior angle is $\angle \text{BCA} = 30^\circ$.

The angle of depression to the car at D is $\angle \text{EBD} = 45^\circ$. Since BE is parallel to the ground AC, the alternate interior angle is $\angle \text{BDA} = 45^\circ$.

Let the uniform speed of the car be $v$ meters per minute.

The distance travelled by the car from C to D in 12 minutes is CD = $v \times 12 = 12v$ meters.

Let the time taken by the car to reach the foot of the tower from D be $t$ minutes. The distance covered from D to A is DA = $v \times t = vt$ meters.

The total distance CA = CD + DA = $12v + vt$ meters.

Consider the right-angled triangle ABD (right-angled at A).

Substitute the known values into equation (i):

We know that $\tan(45^\circ) = 1$. From equation (ii):

Now, consider the right-angled triangle ABC (right-angled at A).

Substitute the known values into equation (iv):

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. Substitute this into equation (v):

From equation (vi), we get:

Substitute the value of $h$ from equation (iii) into equation (vii):

$v(12+t) = (vt)\sqrt{3}$

$12v + vt = vt\sqrt{3}$

Subtract $vt$ from both sides:

$12v = vt\sqrt{3} - vt$

$12v = vt(\sqrt{3} - 1)$

Since $v$ is the speed of the car, $v \neq 0$. We can divide both sides by $v(\sqrt{3} - 1)$:

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, $(\sqrt{3} + 1)$:

$t = \frac{12}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)}$

$t = \frac{12(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2}$

$t = \frac{12(\sqrt{3} + 1)}{3 - 1}$

$t = \frac{12(\sqrt{3} + 1)}{2}$

$t = 6(\sqrt{3} + 1)$ minutes.

Therefore, the time taken by the car from point D to reach the foot of the tower is $6(\sqrt{3} + 1)$ minutes.

---

The final answer is $\boxed{6(\sqrt{3}+1) \text{ minutes}}$.

Given:

Height of the building = $100 \text{ m}$.

Angles of depression of two cars from the top of the building = $30^\circ$ and $45^\circ$.

The two cars are on the opposite sides of the building.

---

To Find:

The distance between the two cars.

---

Solution:

Let AB be the building, where A is the base on the ground and B is the top. Let the height of the building be AB = $100 \text{ m}$.

Let C and D be the positions of the two cars on the road, such that C is on one side of the building and D is on the opposite side. Assume the road is a straight line passing through A.

Let BE be a horizontal line from the top of the building B.

The angle of depression to car C from B is $\angle \text{EBC} = 30^\circ$.

The angle of depression to car D from B is $\angle \text{EBD} = 45^\circ$.

Since BE is parallel to the ground CD (assuming the ground is horizontal), we have the alternate interior angles:

Consider the right-angled triangle ABC (right-angled at A).

Substitute the known values into equation (i):

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. From equation (ii), we get:

Now, consider the right-angled triangle ABD (right-angled at A).

Substitute the known values into equation (iv):

We know that $\tan(45^\circ) = 1$. From equation (v), we get:

Since the two cars are on the opposite sides of the building, the distance between them is the sum of their distances from the base of the building:

Substitute the values of AC from equation (iii) and AD from equation (vi) into equation (vii):

Factor out 100:

Distance between cars = $100(\sqrt{3} + 1) \text{ m}$.

Therefore, the distance between the two cars is $100(\sqrt{3} + 1) \text{ m}$.

---

The final answer is $\boxed{100(\sqrt{3}+1) \text{ m}}$.

Given:

Height of the flagstaff = $20 \text{ m}$.

Angle of elevation of the top of the tower from a point on the ground = $30^\circ$.

Angle of elevation of the top of the flagstaff from the same point on the ground = $60^\circ$.

---

To Find:

The height of the tower.

---

Solution:

Let AB be the height of the tower, where A is the base on the horizontal plane and B is the top of the tower. Let BC be the height of the flagstaff on top of the tower, so BC = $20 \text{ m}$. Let P be the point on the ground from where the angles of elevation are observed.

Let the height of the tower AB = $h$ meters and the distance from the point P to the base of the tower AP = $x$ meters.

The angle of elevation of the top of the tower (B) from P is $\angle \text{APB} = 30^\circ$.

The angle of elevation of the top of the flagstaff (C) from P is $\angle \text{APC} = 60^\circ$.

The total height from the base of the tower to the top of the flagstaff is AC = AB + BC = $h + 20$ meters.

Consider the right-angled triangle ABP:

Substitute the known values into equation (i):

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. From equation (ii):

Now, consider the right-angled triangle ACP:

Substitute the known values and AC = $h + 20$ into equation (iv):

We know that $\tan(60^\circ) = \sqrt{3}$. Substitute this and the value of $x$ from equation (iii) into equation (v):

Multiply both sides by $h\sqrt{3}$:

$\sqrt{3} \times h\sqrt{3} = h+20$

$3h = h+20$

Subtract $h$ from both sides:

$3h - h = 20$

$2h = 20$

Divide by 2:

The height of the tower is $10 \text{ m}$.

---

The final answer is $\boxed{10 \text{ m}}$.

Given:

Angle of elevation of the top of the building from the foot of the tower = $30^\circ$.

Angle of elevation of the top of the tower from the foot of the building = $60^\circ$.

Height of the tower = $50 \text{ m}$.

---

To Find:

The height of the building.

---

Solution:

Let AB be the tower with height AB = $50 \text{ m}$, where A is the base and B is the top. Let CD be the building, where C is the base and D is the top. Assume the tower and the building are on the same horizontal ground, so A and C are on the ground.

Let the distance between the foot of the tower and the foot of the building be AC = $x$ meters.

The angle of elevation of the top of the building (D) from the foot of the tower (A) is $\angle \text{DAC} = 30^\circ$.

The angle of elevation of the top of the tower (B) from the foot of the building (C) is $\angle \text{BCA} = 60^\circ$.

Let the height of the building be CD = $h$ meters.

Consider the right-angled triangle ABC (right-angled at A).

Substitute the known values into equation (i):

We know that $\tan(60^\circ) = \sqrt{3}$. From equation (ii), we get:

Now, consider the right-angled triangle DCA (right-angled at C).

Substitute the known values and CD = $h$ into equation (iv):

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. Substitute this and the value of $x$ from equation (iii) into equation (v):

Simplify the right side of equation (vi):

$\frac{1}{\sqrt{3}} = \frac{h \times \sqrt{3}}{50}$

Cross-multiply:

$1 \times 50 = \sqrt{3} \times h\sqrt{3}$

$50 = 3h$

Solve for $h$:

The height of the building is $\frac{50}{3} \text{ m}$.

---

The final answer is $\boxed{\frac{50}{3} \text{ m}}$.

Given:

Altitude of the aeroplane = $200 \text{ m}$.

Angles of depression of two opposite points on the banks of a river = $45^\circ$ and $60^\circ$.

---

To Find:

The width of the river.

---

Solution:

Let the aeroplane be at point P, and let Q be the point on the ground directly below the aeroplane. The altitude of the aeroplane is PQ = $200 \text{ m}$.

Let A and B be two opposite points on the banks of the river, such that A and B are on a straight line passing through Q (i.e., Q lies between A and B or A=Q or B=Q, assuming the river banks are parallel and perpendicular to the line AQB). Since the angles of depression are different, Q must lie between A and B.

Let PR be a horizontal line through P. The angles of depression to points A and B are $\angle \text{RPA} = 45^\circ$ and $\angle \text{RPB} = 60^\circ$.

Since PR is parallel to the ground AB, the alternate interior angles are equal:

The width of the river is the distance AB = AQ + QB.

Consider the right-angled triangle PQA (right-angled at Q).

Substitute the known values into equation (i):

We know that $\tan(45^\circ) = 1$. From equation (ii), we get:

Now, consider the right-angled triangle PQB (right-angled at Q).

Substitute the known values into equation (iv):

We know that $\tan(60^\circ) = \sqrt{3}$. From equation (v), we get:

The width of the river is AB = AQ + QB.

Combine the terms by finding a common denominator:

$\text{AB} = \frac{200\sqrt{3}}{\sqrt{3}} + \frac{200}{\sqrt{3}} = \frac{200\sqrt{3} + 200}{\sqrt{3}}$

Factor out 200 from the numerator:

$\text{AB} = \frac{200(\sqrt{3} + 1)}{\sqrt{3}}$

Rationalize the denominator:

$\text{AB} = \frac{200(\sqrt{3} + 1)}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{200(\sqrt{3} + 1)\sqrt{3}}{3}$

$\text{AB} = \frac{200(3 + \sqrt{3})}{3} \text{ m}$.

Therefore, the width of the river is $\frac{200(3 + \sqrt{3})}{3} \text{ m}$.

---

The final answer is $\boxed{\frac{200(3+\sqrt{3})}{3} \text{ m}}$.

Given:

Height of the cliff = $50 \text{ m}$.

Angle of depression of the top of the tower from the top of the cliff = $30^\circ$.

Angle of depression of the bottom of the tower from the top of the cliff = $45^\circ$.

---

To Find:

The height of the tower.

---

Solution:

Let AB be the cliff with height AB = $50 \text{ m}$, where A is the base and B is the top. Let CD be the tower, where C is the base and D is the top. Assume the cliff and the tower are on the same horizontal ground, so A and C are on the ground.

From the top of the cliff B, a horizontal line BE is drawn parallel to the ground AC, meeting the vertical line from D at E.

The angle of depression of the top of the tower (D) from B is $\angle \text{EBD} = 30^\circ$.

The angle of depression of the bottom of the tower (C) from B is $\angle \text{EBC} = 45^\circ$.

Since BE is parallel to the ground AC, the alternate interior angles are equal:

Consider the right-angled triangle ABC (right-angled at A). We have:

Substitute the known values into equation (i):

We know that $\tan(45^\circ) = 1$. From equation (ii), we get:

Since ABEC forms a rectangle (BE parallel to AC, AB and CE vertical), AC = BE = $50 \text{ m}$.

Let the height of the tower be CD = $h$ meters. Then ED = AC = $50 \text{ m}$ and CE = AB - ED = $50 - \text{ED}$. No, this is incorrect. CD = h. CE = height of the part of the tower below the horizontal line from B. ED = height of the part of the tower above the horizontal line from B. So, CD = CE + ED.

Since ABEC is a rectangle, AE = BC = $50 \text{ m}$. The height of the tower is CD. Let's draw the diagram correctly. The observer is on the cliff. The tower is below the observer.

Let AB be the cliff (A is base, B is top), AB = 50m. Let CD be the tower (C is base, D is top). A and C are on the ground. B and D are tops. Since angles of depression are observed from B, the tower is lower than the cliff. So CD < AB.

Draw a horizontal line BE from B, meeting the vertical line CD (extended downwards if necessary) at E. However, the angles of depression are of the top and bottom of the tower. So the tower is on the ground.

Let AB be the cliff (A is base, B is top), AB = 50 m. Let CD be the tower (C is base, D is top). A and C are on the same horizontal ground. Draw a horizontal line BE from B parallel to AC, meeting CD at E. This doesn't work as CD is vertical.

Let's redraw the scenario. A is the base of the cliff, B is the top (AB = 50m). C is the base of the tower, D is the top. A and C are on the ground. Draw a horizontal line BE from B parallel to AC. The angles of depression are from B to D and from B to C.

The angle of depression of the bottom of the tower (C) from B is $45^\circ$. So $\angle \text{EBC} = 45^\circ$. Since BE is parallel to AC, $\angle \text{BCA} = \angle \text{EBC} = 45^\circ$ (alternate interior angles).

Consider the right-angled triangle ABC. $\tan(\angle \text{BCA}) = \frac{\text{AB}}{\text{AC}} \implies \tan(45^\circ) = \frac{50}{\text{AC}} \implies 1 = \frac{50}{\text{AC}} \implies \text{AC} = 50 \text{ m}$.

Now, draw a horizontal line BF from B, parallel to AC. Let this line meet the vertical line CD extended upwards or downwards. The tower is below the observer, so BF meets the vertical line from D at point F. Then BF = AC = 50 m. The angle of depression of the top of the tower (D) from B is $\angle \text{FBD} = 30^\circ$.

In right-angled triangle BFD (right-angled at F), $\tan(\angle \text{FBD}) = \frac{\text{BF}}{\text{DF}}$. No, the angle of depression is from the horizontal line downwards. The angle of depression of D from B is $30^\circ$. So $\angle \text{FBD} = 30^\circ$.

In right-angled triangle BFD, we have $\tan(\angle \text{FBD}) = \frac{\text{DF}}{\text{BF}}$.

We know that BF = AC = $50 \text{ m}$ and $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. Substitute these into equation (v):

From equation (vi), we get:

The height of the tower is CD. Since ACFB is a rectangle (AB is vertical, CF is vertical, AC and BF are horizontal), CF = AB = $50 \text{ m}$.

The height of the tower CD = CF - DF.

Combine the terms:

$\text{CD} = 50 \left( 1 - \frac{1}{\sqrt{3}} \right) = 50 \left( \frac{\sqrt{3} - 1}{\sqrt{3}} \right)$

Rationalize the denominator:

$\text{CD} = 50 \left( \frac{\sqrt{3} - 1}{\sqrt{3}} \right) \times \frac{\sqrt{3}}{\sqrt{3}} = 50 \frac{(\sqrt{3} - 1)\sqrt{3}}{3} = 50 \frac{3 - \sqrt{3}}{3} \text{ m}$.

Therefore, the height of the tower is $\frac{50(3 - \sqrt{3})}{3} \text{ m}$.

---

The final answer is $\boxed{\frac{50(3-\sqrt{3})}{3} \text{ m}}$.

Given:

Height of the pole on top of the tower = $5 \text{ m}$.

Angle of elevation of the top of the pole from a point 'A' on the ground = $60^\circ$.

Angle of depression of point 'A' from the top of the tower = $45^\circ$.

---

To Find:

The height of the tower.

---

Solution:

Let TP be the tower, where T is the base on the ground and P is the top of the tower. Let PQ be the pole fixed on the top of the tower, so PQ = $5 \text{ m}$. Let A be the point on the ground from where the angles are observed.

Let the height of the tower be TP = $h$ meters, and the distance from point A to the base of the tower be TA = $x$ meters.

The total height from the ground to the top of the pole is TQ = TP + PQ = $h + 5$ meters.

The angle of depression of point A from the top of the tower (P) is $45^\circ$. Let PL be a horizontal line from P. Then $\angle \text{LPA} = 45^\circ$. Since PL is parallel to the ground TA, the alternate interior angle is $\angle \text{PAT} = \angle \text{LPA} = 45^\circ$.

Consider the right-angled triangle PAT (right-angled at T).

Substitute the known values into equation (i):

We know that $\tan(45^\circ) = 1$. From equation (ii):

The angle of elevation of the top of the pole (Q) from point A is $60^\circ$. So, $\angle \text{TAQ} = 60^\circ$.

Consider the right-angled triangle TAQ (right-angled at T).

Substitute the known values into equation (iv):

We know that $\tan(60^\circ) = \sqrt{3}$. Substitute this and the value of $x$ from equation (iii) into equation (v):

Multiply both sides by $h$:

$h\sqrt{3} = h+5$

Subtract $h$ from both sides:

$h\sqrt{3} - h = 5$

Factor out $h$:

$h(\sqrt{3} - 1) = 5$

Solve for $h$:

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, $(\sqrt{3} + 1)$:

$h = \frac{5}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)}$

$h = \frac{5(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2}$

$h = \frac{5(\sqrt{3} + 1)}{3 - 1}$

$h = \frac{5(\sqrt{3} + 1)}{2} \text{ m}$.

Therefore, the height of the tower is $\frac{5(\sqrt{3} + 1)}{2} \text{ m}$.

---

The final answer is $\boxed{\frac{5(\sqrt{3}+1)}{2} \text{ m}}$.

Given:

Height of the window above the ground = $10 \text{ m}$.

Angle of elevation of the top of the second house from the window = $60^\circ$.

Angle of depression of the foot of the second house from the window = $30^\circ$.

---

To Find:

The height of the second house.

The width of the lane.

---

Solution:

Let AB be the height of the window from the ground in the first house, so AB = $10 \text{ m}$. Let CD be the second house, where C is the foot and D is the top. Let A and C be on the ground.

The distance between the two houses across the lane is AC. Let AC = $x$ meters.

From the window B, draw a horizontal line BE parallel to AC, meeting CD at E. Then ABCE is a rectangle, so BE = AC = $x$ meters and CE = AB = $10 \text{ m}$.

The height of the second house is CD = CE + ED = $10 + \text{ED}$ meters.

The angle of elevation of the top of the second house (D) from the window (B) is $\angle \text{DBE} = 60^\circ$.

The angle of depression of the foot of the second house (C) from the window (B) is $\angle \text{EBC} = 30^\circ$. Since BE is parallel to AC, the alternate interior angle is $\angle \text{BCA} = \angle \text{EBC} = 30^\circ$.

Consider the right-angled triangle BCE (right-angled at E). We have:

Substitute the known values into equation (i):

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. From equation (ii):

The width of the lane is AC = $x = 10\sqrt{3} \text{ m}$.

Now, consider the right-angled triangle BDE (right-angled at E). We have:

Substitute the known values and BE from equation (iii) into equation (iv):

We know that $\tan(60^\circ) = \sqrt{3}$. From equation (v):

The height of the second house is CD = CE + ED.

$\text{CD} = 40 \text{ m}$.

Therefore, the height of the second house is $40 \text{ m}$ and the width of the lane is $10\sqrt{3} \text{ m}$.

---

The final answer is $\boxed{\text{Height of house} = 40 \text{ m, Width of lane} = 10\sqrt{3} \text{ m}}$.