Chapter 9 Some Applications of Trigonometry (Concepts)

Having established the fundamental definitions and relationships of trigonometric ratios ($\sin$, $\cos$, $\tan$, etc.) and identities within the theoretical framework of right-angled triangles in the previous chapter, we now pivot to explore the immense practical power of trigonometry. This chapter, often titled "Some Applications of Trigonometry," focuses specifically on solving real-world problems related to determining heights and distances that are often difficult or impossible to measure directly. Imagine needing to find the height of a colossal monument, the width of a raging river, or the altitude of a distant aircraft – trigonometry provides the elegant mathematical tools for such indirect measurements, transforming seemingly inaccessible dimensions into solvable geometric problems.

To navigate these applications effectively, we first need to define some key terminology associated with observation:

• Line of Sight: This is the imaginary straight line drawn from the eye of an observer to the specific point on the object being viewed.
• Angle of Elevation: When the object being viewed is above the horizontal level of the observer's eye, the angle formed between the horizontal line and the line of sight is called the angle of elevation. Imagine looking up at the top of a flagpole; the angle your gaze makes with the horizontal ground is the angle of elevation.
• Angle of Depression: Conversely, when the object being viewed is below the horizontal level of the observer's eye, the angle formed between the horizontal line and the line of sight is the angle of depression. Picture standing on a clifftop and looking down at a boat; the angle your gaze makes with the horizontal line extending outwards from your eye level is the angle of depression. (Note: By alternate interior angles, the angle of depression from the observer to the object equals the angle of elevation from the object back to the observer, assuming parallel horizontal lines).

The core problem-solving methodology in this chapter involves a systematic process:

1. Translate the Word Problem into a Diagram: This is arguably the most crucial step. Carefully read the problem description and draw a clear, labeled geometric diagram (almost always involving one or more right-angled triangles) representing the situation. Mark all known lengths and angles provided in the problem.
2. Identify Knowns and Unknowns: Determine which side lengths or angles are given and which specific height or distance needs to be calculated within the relevant right-angled triangle(s).
3. Select the Appropriate Trigonometric Ratio: Based on the known angle and the relationship between the known side(s) and the unknown side (relative to the known angle – Opposite, Adjacent, Hypotenuse), choose the appropriate trigonometric ratio ($\sin A = \frac{Opp}{Hyp}$, $\cos A = \frac{Adj}{Hyp}$, or $\tan A = \frac{Opp}{Adj}$) that connects these quantities.
4. Set up and Solve the Equation: Formulate an equation using the chosen ratio and the known values. Solve this equation algebraically to find the value of the unknown height or distance. Remember to use the standard angle values ($30^\circ, 45^\circ, 60^\circ$) where applicable.

The complexity of problems can vary. Simpler problems might involve just a single right-angled triangle – for instance, finding the height of a tower given the distance to its base and the angle of elevation to its top. However, many problems involve scenarios requiring the analysis of two or more right-angled triangles. These might share a common side (like the height of a tower observed from two different points) or be linked through related distances. Examples include determining the height of a multi-storey building by observing the angles of elevation to its top and bottom from a point opposite, finding the width of a river by observing an object on the opposite bank from two points, or calculating the height of a flying object using angles of elevation from two different locations. Problems involving angles of depression are solved using similar geometric principles. Success hinges on accurate diagrammatic representation and the careful, correct application of the fundamental trigonometric ratios, showcasing trigonometry's profound utility in measuring our world indirectly.

Introduction to Heights and Distances

One of the significant practical applications of trigonometry is in finding the heights and distances of objects or locations that are difficult or impossible to measure directly. For example, we can determine the height of a tall building, a mountain, or the distance across a river using trigonometric principles without physically climbing the building or crossing the river. These applications primarily rely on forming right-angled triangles and using the trigonometric ratios defined in the previous chapter.

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Key Terms in Heights and Distances

To solve problems involving heights and distances using trigonometry, we need to understand a few specific terms related to observation and angles:

Line of Sight:

The line of sight is an imaginary straight line that connects the eye of an observer to the point on the object that the observer is looking at. It represents the direction in which the observer is looking.

Angle of Elevation:

When an observer looks at an object that is located above the observer's horizontal level, the angle of elevation is the angle formed between the line of sight and the horizontal line (drawn from the observer's eye parallel to the ground). This angle is always measured upwards from the horizontal.

In the figure, if the observer's eye is at A, and the object is at B, with AC being the horizontal line from A, then $\angle \text{BAC}$ is the angle of elevation.

Angle of Depression:

When an observer looks at an object that is located below the observer's horizontal level, the angle of depression is the angle formed between the line of sight and the horizontal line (drawn from the observer's eye parallel to the ground). This angle is always measured downwards from the horizontal.

In the figure, if the observer's eye is at A (at a certain height above B), and the object is at B, with AC being the horizontal line from A, then $\angle \text{BAC}$ is the angle of depression.

Important Relationship: The angle of elevation of an object from an observer's eye is equal to the angle of depression of the observer's eye from the object, assuming the horizontal lines are parallel (which they are in a plane). In the diagram for the angle of depression, if there is a point D on the ground directly below A, and a horizontal line through B, then $\angle \text{BAC} = \angle \text{ABD}$ (alternate interior angles formed by parallel horizontal lines and the line of sight as a transversal).

Using Trigonometric Ratios to Solve Problems

Problems involving heights and distances can typically be solved by creating a simplified diagram that represents the given situation as a right-angled triangle. The unknown height or distance will be one side of this triangle, and the angle of elevation or depression will be one of the acute angles. Known lengths will be other sides.

We use the trigonometric ratios (sine, cosine, or tangent, and their reciprocals) that relate the given angle, the known side(s), and the unknown side. Once the correct trigonometric ratio is identified and the equation is set up, we can solve for the unknown quantity using the values of trigonometric ratios for specific angles (Section I4).

• If the relationship between the side opposite to the angle and the hypotenuse is relevant, use the sine ratio: $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$.
• If the relationship between the side adjacent to the angle and the hypotenuse is relevant, use the cosine ratio: $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$.
• If the relationship between the side opposite to the angle and the side adjacent to the angle is relevant, use the tangent ratio: $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$.

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Example 1. A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be $60^\circ$. Find the height of the tower.

Answer:

To Find:

The height of the tower.

Given:

Distance from the foot of the tower to the observation point = 15 m.

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

Solution:

Let the tower be represented by the vertical line segment BC, where B is the foot of the tower and C is the top. Let A be the point on the ground 15 m away from the foot of the tower B. So, the distance AB = 15 m.

The angle of elevation of the top of the tower C from point A is $\angle \text{BAC} = 60^\circ$.

Since the tower stands vertically on the ground, $\triangle \text{ABC}$ is a right-angled triangle with the right angle at B ($\angle \text{ABC} = 90^\circ$).

In the right triangle $\triangle \text{ABC}$, we are given the angle $\angle \text{A} = 60^\circ$ and the side adjacent to this angle, AB = 15 m. We need to find the height of the tower, which is the side opposite to angle A, BC. The trigonometric ratio that relates the opposite side and the adjacent side to an angle is the tangent ($\tan$).

$\tan A = \frac{\text{Side opposite to } A}{\text{Side adjacent to } A} = \frac{\text{BC}}{\text{AB}}$

Substitute the given values $\angle \text{A} = 60^\circ$ and AB = 15 m:

$\tan 60^\circ = \frac{\text{BC}}{15}$

... (1)

From the table of trigonometric ratios for specific angles (Section I4), we know that $\tan 60^\circ = \sqrt{3}$.

Substitute this value into equation (1):

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

Solve for BC by multiplying both sides by 15:

$\text{BC} = 15 \times \sqrt{3}$ metres

... (2)

The height of the tower is $15\sqrt{3}$ metres.

If a numerical value is required, using $\sqrt{3} \approx 1.732$:

$\text{BC} \approx 15 \times 1.732 = 25.98$ metres

Answer: The height of the tower is $15\sqrt{3}$ metres (approximately 25.98 m).

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Example 2. An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is $45^\circ$. What is the height of the chimney?

Answer:

To Find:

The height of the chimney.

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 = $45^\circ$.

Solution:

Let the observer be represented by the line segment AB, where A is the position of her eyes and B is the point on the ground below her. So, AB = 1.5 m. Let the chimney be represented by the line segment EF, where E is the foot of the chimney and F is the top. The distance between the observer and the chimney on the ground is BE = 28.5 m.

Draw a horizontal line from the observer's eyes A, parallel to the ground BE. Let this line be AC, meeting the chimney EF at C. So, AC is horizontal, and BE is horizontal. This makes ABEC a rectangle, so AB = EC = 1.5 m and AC = BE = 28.5 m.

The angle of elevation of the top of the chimney F from the observer's eyes A is $\angle \text{FAC} = 45^\circ$.

Consider the right-angled triangle $\triangle \text{ACF}$, right-angled at C ($\angle \text{ACF} = 90^\circ$ because AC is horizontal and CF is vertical part of chimney). In $\triangle \text{ACF}$, we are given the angle $\angle \text{FAC} = 45^\circ$ and the side adjacent to this angle, AC = 28.5 m. We need to find the side opposite to this angle, CF.

Using the tangent ratio in $\triangle \text{ACF}$:

$\tan (\angle \text{FAC}) = \frac{\text{Side opposite to } \angle \text{FAC}}{\text{Side adjacent to } \angle \text{FAC}} = \frac{\text{CF}}{\text{AC}}$

Substitute the given values $\angle \text{FAC} = 45^\circ$ and AC = 28.5 m:

$\tan 45^\circ = \frac{\text{CF}}{28.5}$

... (1)

From the table of trigonometric ratios for specific angles, we know that $\tan 45^\circ = 1$.

Substitute this value into equation (1):

$1 = \frac{\text{CF}}{28.5}$

Solve for CF by multiplying both sides by 28.5:

$\text{CF} = 28.5 \times 1 = 28.5$ metres

... (2)

The total height of the chimney is EF. From the diagram, EF = EC + CF.

Since ABEC is a rectangle, EC = AB (height of the observer) = 1.5 m.

Total height of chimney (EF) = EC + CF

Substitute the values EC = 1.5 m and CF = 28.5 m (from equation 2):

$\text{EF} = 1.5 + 28.5$

$\text{EF} = 30.0$ metres

... (3)

Answer: The height of the chimney is 30 metres.