Chapter 5 Lines & Angles (Class 7 - Maths NCERT Exemplar Solutions)

By observing the cardinal directions in Fig. 5.1, we know that the angle between any two consecutive main directions (North, East, South, West) is a right angle ($90^\circ$).

To determine their relationship, we find their sum:

In geometry, two angles whose sum is $180^\circ$ are known as supplementary angles.

Therefore, the correct option is (b).

Two angles are called supplementary if the sum of their measures is exactly $180^\circ$. Let us check the given pairs:

(a) $48^\circ + 42^\circ = 90^\circ$ (These are complementary angles).

(b) $60^\circ + 60^\circ = 120^\circ$.

(c) $75^\circ + 105^\circ = 180^\circ$ (These are supplementary angles).

(d) $179^\circ + 2^\circ = 181^\circ$.

Therefore, the correct option is (c).

In Fig. 5.2, when a transversal intersects two lines, angles that occupy the same relative position at each intersection are called corresponding angles.

Let us analyze the pairs:

(a) $\angle 1$ and $\angle 2$ are adjacent angles forming a linear pair.

(b) $\angle 3$ and $\angle 6$ are alternate interior angles.

(c) $\angle 3$ and $\angle 5$ are interior angles on the same side of the transversal.

(d) $\angle 3$ and $\angle 7$ are in the same relative position (bottom-right of each intersection). Thus, they are corresponding angles.

Therefore, the correct option is (d).

When a transversal intersects two lines, the region between the two lines is the interior. There are four interior angles in total.

Two of these angles lie on the left side of the transversal, and the other two lie on the right side. This results in exactly 2 pairs of interior angles on the same side of the transversal.

Therefore, the correct option is (b).

Lines in the same flat surface (plane) that are at an equal distance from each other at every point and never intersect, no matter how far they are extended, are known as parallel lines.

Blank: parallel lines

A linear pair consists of two angles that are formed when two lines intersect. By definition, these angles are adjacent (sharing a common arm and vertex) and their non-common arms form a straight line, making them supplementary ($180^\circ$).

Blanks: adjacent, supplementary

According to the properties of adjacent angles:

1. They have a common vertex.

2. They have a common arm (or side).

3. They have no common interior points (their regions do not overlap).

Blanks: arm, interior points

Statement: Sum of two complementary angles is $180^\circ$.

Result: False

Reason: By definition, the sum of two complementary angles is always $90^\circ$. The sum of two supplementary angles is $180^\circ$.

Statement: Sum of two supplementary angles is $180^\circ$.

Result: True

Reason: Two angles are said to be supplementary if the sum of their measures is exactly $180^\circ$.

Statement: Sum of interior angles on the same side of a transversal with two parallel lines is $90^\circ$.

Result: False

Reason: When two parallel lines are cut by a transversal, the sum of the interior angles on the same side of the transversal (co-interior angles) is always $180^\circ$ (they are supplementary), not $90^\circ$.

Statement: Vertically opposite angles are equal.

Result: True

Reason: According to the vertically opposite angles theorem, when two lines intersect, the angles opposite to each other at the vertex are always equal.

Given:

In Fig. 5.3, $PQ \parallel RS$ and $QR \parallel ST$.

Angle between $PQ$ and $QR$ is $39^\circ$, i.e., $\angle PQR = 39^\circ$.

To Find:

Values of $x$ and $y$.

Solution:

For parallel lines $PQ$ and $RS$, $QR$ acts as a transversal.

Now, consider $QR \parallel ST$. Here, $RS$ acts as a transversal.

Therefore, the values are $x = 39$ and $y = 39$.

Given:

Fig. 5.4 showing the letter 'N' with angles $1$ and $2$.

Solution:

No, the angles $1$ and $2$ are not adjacent angles.

Reason: For two angles to be adjacent, they must satisfy three conditions:

1. They must have a common vertex.

2. They must have a common arm.

3. Their non-common arms should be on either side of the common arm.

In the given figure of the letter 'N', angle $1$ and angle $2$ are formed at different vertices. Since they do not share a common vertex, they cannot be adjacent angles.

Given:

Points $A, O, B$ are collinear (forming a straight line $AB$).

Ray $OC \perp$ ray $OD$, which means $\angle COD = 90^\circ$.

$\angle AOD = A^\circ$ and $\angle BOC = 3a^\circ$.

To Check:

(i) Whether $\angle AOD$ and $\angle BOC$ are complementary.

(ii) Whether $\angle AOC$ and $\angle BOC$ are supplementary.

Solution:

Since $AOB$ is a straight line, the sum of all angles on it at point $O$ is $180^\circ$.

(i) For $\angle AOD$ and $\angle BOC$:

Sum $= \angle AOD + \angle BOC = A^\circ + 3a^\circ$.

From equation (i), we know $A^\circ + 3a^\circ = 90^\circ$.

Since their sum is $90^\circ$, $\angle AOD$ and $\angle BOC$ are complementary.

(ii) For $\angle AOC$ and $\angle BOC$:

From the figure, $\angle AOC = \angle AOD + \angle DOC$.

Sum $= \angle AOC + \angle BOC = (A^\circ + 90^\circ) + 3a^\circ$.

Rearranging the terms: Sum $= (A^\circ + 3a^\circ) + 90^\circ$.

Substituting value from (i): Sum $= 90^\circ + 90^\circ = 180^\circ$.

Since their sum is $180^\circ$, $\angle AOC$ and $\angle BOC$ are supplementary.

Given:

In Fig. 5.6, $AB \parallel EF$ and $ED \parallel CB$.

$\angle APE = 39^\circ$.

To Find:

The value of $\angle CQF$.

Solution:

Consider parallel lines $AB$ and $EF$ with transversal $EP$.

Now, consider parallel lines $ED$ and $CB$ with transversal $EF$.

Since $B, Q$ and $F$ lie on the line $EF$, $\angle BQF$ and $\angle CQF$ form a linear pair.

Therefore, the value of $\angle CQF$ is $141^\circ$.

To Find: The measures of two complementary angles where one is two-third of the other.

Solution:

Let one of the complementary angles be $x$.

According to the question, the other angle is $\frac{2}{3}$ of $x$, i.e., $\frac{2}{3}x$.

We know that the sum of two complementary angles is $90^\circ$.

Taking the LCM on the LHS:

Now, finding the second angle:

Thus, the two angles are $54^\circ$ and $36^\circ$.

Given:

Line $CD$ intersects line $AB$ at point $F$.

$\angle CFB = 50^\circ$

$\angle EFA = \angle AFD$

To Find:

Measure of $\angle EFC$.

Solution:

When two lines $AB$ and $CD$ intersect at $F$, vertically opposite angles are equal.

It is given that $\angle EFA = \angle AFD$.

Now, points $A, F$ and $B$ lie on a straight line $AB$. The sum of angles on a straight line at a point is $180^\circ$.

Substituting the known values:

Therefore, the measure of $\angle EFC$ is $80^\circ$.

To Find: Which pair of lines are parallel in Fig. 5.8.

Solution:

1. Checking lines $EF$ and $GH$:

Consider the transversal $CD$. The angles $\angle 1$ and $\angle 2$ are interior angles on the same side of the transversal.

Since the sum of co-interior angles is $180^\circ$, the lines are parallel.

2. Checking lines $GH$ and $KP$:

Consider the transversal $CD$. The angles $\angle 4$ and $\angle 3$ are interior angles on the same side of the transversal.

Since $112^\circ \neq 180^\circ$, the sum of co-interior angles is not supplementary.

3. Checking lines $AB$ and $CD$:

Consider line $GH$ as a transversal. For $AB \parallel CD$, the corresponding angles should be equal.

The angle corresponding to $\angle 5$ ($122^\circ$) on line $CD$ would be the angle above $CD$ and to the right of $GH$. Let this angle be $\theta$.

Since $\theta$ and $\angle 4$ form a linear pair:

Comparing corresponding angles:

Thus, the only pair of parallel lines is $EF \parallel GH$.

Solution:

The angle between the cardinal directions North and West is $90^\circ$.

Similarly, the angle between South and East is also $90^\circ$.

Sum of these two angles $= 90^\circ + 90^\circ = 180^\circ$.

Since the sum of these two angles is $180^\circ$, they are supplementary.

Correct Option: (b)

Solution:

The angle between South and West is $90^\circ$.

The angle between South and East is $90^\circ$.

Both these angles share a common vertex (the center point) and a common arm (South).

The non-common arms (West and East) lie on opposite sides of the common arm and form a straight line.

Sum of angles $= 90^\circ + 90^\circ = 180^\circ$.

Therefore, they are forming a linear pair.

Correct Option: (c)

Given:

$\angle ABC = 46^\circ$

$PQ$ is a straight line mirror.

To Find:

$\angle ABP$

Solution:

According to the laws of reflection, the angle of incidence is equal to the angle of reflection.

In the given figure, the angles made with the mirror surface are also equal:

Since $PBQ$ is a straight line, the sum of angles on it is $180^\circ$:

Substituting $\angle CBQ$ with $\angle ABP$ and $\angle ABC$ with $46^\circ$:

Correct Option: (b)

Solution:

Let the required angle be $x$.

We know that the sum of an angle and its complement is $90^\circ$.

The measure of the angle is $11^\circ$.

Correct Option: (b)

Solution:

Let the two angles be $x$ and $y$.

Since they are vertically opposite angles, they must be equal:

Since they are also supplementary angles, their sum must be $180^\circ$:

Substituting $y = x$ into the second equation:

Thus, both angles are $90^\circ$.

Correct Option: (b)

Solution:

Let the required angle be $x$.

Angles forming a linear pair are supplementary, meaning their sum is $180^\circ$.

Correct Option: (d)

Solution:

To determine the relationship, we find the sum of the two given angles.

Since the sum of the angles is exactly $90^\circ$, they are complementary.

Correct Option: (b)

Solution:

To determine the relationship, we find the sum of the two given angles.

Since the sum of the angles is exactly $180^\circ$, they are supplementary.

Correct Option: (d)

Solution:

We know that the sum of all angles around a point is $360^\circ$ (Complete angle).

Correct Option: (d)

Given:

$AB \parallel CD$

$\angle APQ = 50^\circ$

$\angle PRD = 130^\circ$

To Find:

$\angle QPR$

Solution:

Since $AB \parallel CD$ and $PR$ acts as a transversal:

From the figure, we can see that $\angle APR$ is composed of two angles:

Substituting the known values:

Correct Option: (c)

Solution:

When two lines $l$ and $m$ intersect at a point:

1. Vertically opposite angles are equal. Therefore, $\angle a = \angle b$ and $\angle d = \angle c$. Options (a) and (b) are true.

2. Adjacent angles on a straight line are supplementary. Therefore, $\angle a + \angle d = 180^\circ$. Option (c) is true.

3. $\angle a$ and $\angle d$ are adjacent angles forming a linear pair. They are equal only if the lines are perpendicular ($90^\circ$ each), which is not specified. Thus, the statement $\angle a = \angle d$ is generally false.

Correct Option: (d)

Given:

Angle $P$ and $Q$ are supplementary.

$\angle P = 60^\circ$

Solution:

By the definition of supplementary angles:

Correct Option: (a)

Solution:

Since $POR$ is a straight line, the angles on it form a linear pair and their sum is $180^\circ$.

Correct Option: (a)

Given:

$POQ$ is a straight line.

$x = 30^\circ$

Solution:

Since $POQ$ is a line, the sum of all angles at point $O$ on one side is $180^\circ$.

From the figure, the angles are $x$, $2y$, and $3y$.

Substituting $x = 30^\circ$:

Now, we need to find $\angle QOR$:

Correct Option: (a)

Solution:

Let the required angle be $x$.

The supplement of this angle will be $(180^\circ - x)$.

According to the given condition, the angle is four times its supplement:

The measure of the angle is $144^\circ$.

Correct Option: (b)

Solution:

The angles $6y$, $y$, and $2y$ lie on a straight line.

The sum of angles on a straight line is $180^\circ$.

Correct Option: (c)

Given:

$PA \parallel BC \parallel DT$ and $AB \parallel DC$.

$\angle PAB = 50^\circ$.

Solution:

Since $PA \parallel BC$ and $AB$ acts as a transversal:

Now, consider $AB \parallel DC$ and $BC$ as a transversal:

Next, consider $BC \parallel DT$ and $CD$ as a transversal:

Thus, the values are $a = 50^\circ$ and $b = 130^\circ$.

Correct Option: (b)

Solution:

Let the two complementary angles be $x$ and $y$.

According to the definition of complementary angles:

According to the given problem, their difference is $30^\circ$:

Adding equations (i) and (ii):

Substituting $x = 60^\circ$ in equation (i):

The angles are $60^\circ$ and $30^\circ$.

Correct Option: (a)

Given:

$PQ \parallel SR$ and $SP \parallel RQ$.

$\angle SRP = 20^\circ$ and $\angle SPR = 50^\circ$.

Solution:

Since $PQ \parallel SR$ and $PR$ is the transversal:

Since $SP \parallel RQ$ and $PR$ is the transversal:

Thus, the values are $a = 20^\circ$ and $b = 50^\circ$.

Correct Option: (a)

Solution:

In the given figure, line $l$ is a transversal intersecting lines $m$ and $n$.

Angles $a$ and $b$ are located on the inner side of the lines $m$ and $n$ and on opposite sides of the transversal $l$.

Therefore, these angles are called alternate interior angles.

Correct Option: (c)

Solution:

Let the two supplementary angles be $x$ and $2x$.

We know that the sum of supplementary angles is $180^\circ$.

The bigger angle is $2x$:

Correct Option: (a)

Given:

$POQ$ is a straight line.

$\angle ROS = 90^\circ$

$\angle POR : \angle QOS = 1 : 5$

Solution:

Let $\angle POR = x$ and $\angle QOS = 5x$.

Since $POQ$ is a straight line, the sum of angles on it is $180^\circ$.

Now, we find $\angle QOS$:

Correct Option: (b)

Solution:

Statement a: This is a standard theorem in geometry. When two lines intersect, the angles opposite each other at the vertex are always equal. So, a is true.

Statement b: The sum of interior angles on the same side of the transversal is $180^\circ$ only if the two lines are parallel. If the lines are not parallel, the sum will not be $180^\circ$. Since the statement doesn't specify that the lines are parallel, b is false.

Correct Option: (b)

Solution:

Statement p: Angles $a$ and $b$ lie on a straight line $AB$ and are supplementary. Therefore, they form a linear pair. So, p is true.

Statement q: Angles $a$ and $b$ share a common vertex $O$ and a common arm $OC$. Their non-common arms ($OA$ and $OB$) are on opposite sides of the common arm. Therefore, they are adjacent angles. So, q is true.

Since both statements are correct, both p and q are true.

Correct Option: (a)

Solution:

In Fig. 5.21, points $A, O,$ and $B$ lie on a straight line. The ray $OC$ stands on this line.

Angles $\angle AOC$ and $\angle BOC$ are adjacent angles that lie on a straight line.

We know that the sum of angles on a straight line is $180^\circ$. Therefore, these angles are supplementary angles (also known as a linear pair).

Correct Option: (d)

Solution:

Assuming the figure consists of three straight lines intersecting at a common point.

From the figure, the angle vertically opposite to $\angle BOC$ is $\angle FOE$.

Now, consider the straight line $AD$. The sum of angles on one side of this line is $180^\circ$.

Substituting the given values ($\angle AOF = 90^\circ$ as shown by the square symbol):

Correct Option: (d)

Given:

In Fig. 5.23, $QP \parallel SR$.

$\angle PQT = 60^\circ$ and $\angle TSR = 30^\circ$.

To Find:

The value of angle $a$ ($\angle QTS$).

Construction Required:

Through point $T$, draw a line $XY$ such that $XY \parallel QP$.

Solution:

It is given that $QP \parallel SR$. By construction, $XY \parallel QP$.

Now, consider parallel lines $QP$ and $XY$ with $QT$ as the transversal:

Next, consider parallel lines $XY$ and $SR$ with $ST$ as the transversal:

From the figure, we can see that the angle $a$ is the sum of $\angle QTX$ and $\angle STX$:

Substituting values from equations (i) and (ii):

Therefore, the value of $a$ is $90^\circ$.

Correct Option: (c)

Solution:

Two angles are adjacent if they have a common vertex, a common arm, and their non-common arms are on opposite sides of the common arm.

(a) These are vertically opposite angles, not adjacent.

(b) These angles do not share a common vertex.

(c) These angles share a common vertex and arm, but angle $a$ is part of angle $b$ (overlapping), so their non-common arms are not on opposite sides of the common arm.

(d) These angles share a common vertex and a common arm, and the non-common arms are on opposite sides. Thus, they are adjacent.

Correct Option: (d)

Solution:

Statements (i) and (ii) are parts of the fundamental definition of adjacent angles. However, statement (iii) says that uncommon arms are always opposite rays. This is only true for linear pairs of angles. For general adjacent angles, the non-common arms can be at any angle to each other. Therefore, (iii) is false.

Correct Option: (b)

Solution:

Since $PQ$ is a straight line and $\angle POR = 90^\circ$, the remaining angle $\angle ROQ$ is also $90^\circ$.

Given $x : y = 3 : 2$. Let $x = 3k$ and $y = 2k$.

So, $x = 3 \times 18^\circ = 54^\circ$ and $y = 2 \times 18^\circ = 36^\circ$.

Since lines $PQ$ and $ST$ intersect at $O$, vertically opposite angles are equal:

From the figure, $\angle POT = \angle POR + \angle ROT = 90^\circ + x$.

Correct Option: (b)

Solution:

Since $POQ$ is a straight line, the sum of angles on the line is $180^\circ$.

Given $\angle POR = 100^\circ$ and $\angle ROQ = a$.

Correct Option: (c)

Solution:

It is a fundamental theorem in geometry that when two straight lines intersect, the pair of angles formed opposite each other at the vertex (vertically opposite angles) are always equal in magnitude.

Correct Option: (d)

Given:

$a = 40^\circ$

Solution:

From Fig. 5.27, we can see that the angles $5b$ and $2a$ lie on a straight line.

Therefore, they form a linear pair and their sum must be $180^\circ$.

Substituting the given value of $a = 40^\circ$:

Correct Option: (a)

Solution:

Let the angle be $x$.

The supplement of this angle will be $(180^\circ - x)$.

According to the question, the angle $x$ is $60^\circ$ less than two times its supplement:

Now, finding the supplement angle:

Comparing $100^\circ$ and $80^\circ$, the greater angle is $100^\circ$.

Correct Option: (a)

Given:

$PQ \parallel RS$

$\angle 1 = (2a + b)^\circ$ and $\angle 6 = (3a - b)^\circ$

Solution:

In the given figure, line $l$ is a transversal.

Since $PQ \parallel RS$, the corresponding angles are equal:

Also, $\angle 1$ and $\angle 2$ form a linear pair on line $PQ$:

Substituting the values of $\angle 1$ and $\angle 2$ (using $\angle 6$):

Now, find the measure of $\angle 2$ using the value of $a$:

Correct Option: (c)

Given:

$PQ \parallel RS$ and $a : b = 3 : 2$

Solution:

Angles $a$ and $b$ form a linear pair on line $l$.

Let $a = 3x$ and $b = 2x$.

So, $a = 3 \times 36^\circ = 108^\circ$.

Since $PQ \parallel RS$, the corresponding angles are equal:

Correct Option: (b)

Given:

In Fig. 5.30, line $l$ intersects two parallel lines $PQ$ and $RS$.

The angles at the intersection with line $PQ$ are: $\angle 1$, $\angle 7$, $\angle 8$, and $\angle 2$.

The angles at the intersection with line $RS$ are: $\angle 3$, $\angle 5$, $\angle 6$, and $\angle 4$.

To Find:

Which of the following statements is NOT true:

(a) $\angle 1 = \angle 3$

(b) $\angle 2 = \angle 4$

(c) $\angle 6 = \angle 7$

(d) $\angle 4 = \angle 8$

Solution:

We evaluate each option based on the properties of parallel lines intersected by a transversal:

(a) Checking $\angle 1 = \angle 3$:

Angle $1$ and Angle $3$ are in the top-left position at their respective intersections. These are called corresponding angles. When lines are parallel, corresponding angles are equal.

(b) Checking $\angle 2 = \angle 4$:

Angle $2$ and Angle $4$ are in the bottom-right position at their respective intersections. These are also corresponding angles. Therefore, they are equal.

(c) Checking $\angle 6 = \angle 7$:

Angle $7$ is an exterior angle on the top-right, and Angle $6$ is an exterior angle on the bottom-left. These are alternate exterior angles. When lines are parallel, alternate exterior angles are equal.

(d) Checking $\angle 4 = \angle 8$:

Angle $4$ is an exterior angle on the bottom-right. Angle $8$ is an interior angle on the bottom-left.

From the figure, Angle $4$ corresponds to Angle $2$ ($\angle 4 = \angle 2$).

Angle $2$ and Angle $8$ lie on the straight line $PQ$, so they form a linear pair.

Substituting $\angle 4$ for $\angle 2$:

Since their sum is $180^\circ$, they are supplementary and generally not equal.

Conclusion: Statement (d) is NOT true.

Correct Option: (d)

Given:

In Fig. 5.30, line $l$ is a transversal intersecting two parallel lines $PQ$ and $RS$.

Position of angles at intersection of $PQ$: $\angle 1$ (Top-Left), $\angle 7$ (Top-Right), $\angle 8$ (Bottom-Left), $\angle 2$ (Bottom-Right).

Position of angles at intersection of $RS$: $\angle 3$ (Top-Left), $\angle 5$ (Top-Right), $\angle 6$ (Bottom-Left), $\angle 4$ (Bottom-Right).

To Find:

Which one of the following statements is NOT true:

(a) $\angle 1 + \angle 5 = 180^\circ$

(b) $\angle 2 + \angle 5 = 180^\circ$

(c) $\angle 3 + \angle 8 = 180^\circ$

(d) $\angle 2 + \angle 3 = 180^\circ$

Solution:

We will examine each option one by one based on the properties of parallel lines:

(a) Evaluating $\angle 1 + \angle 5 = 180^\circ$:

On line $RS$, $\angle 3$ and $\angle 5$ form a linear pair.

Substituting $\angle 1$ for $\angle 3$:

(b) Evaluating $\angle 2 + \angle 5 = 180^\circ$:

On line $RS$, $\angle 4$ and $\angle 5$ form a linear pair.

Substituting $\angle 2$ for $\angle 4$:

(c) Evaluating $\angle 3 + \angle 8 = 180^\circ$:

$\angle 3$ and $\angle 8$ are interior angles on the same side of the transversal (co-interior angles).

Therefore, this statement is True.

(d) Evaluating $\angle 2 + \angle 3 = 180^\circ$:

$\angle 2$ and $\angle 3$ are alternate interior angles. For parallel lines, alternate interior angles are equal.

Since they are equal, their sum would be $180^\circ$ only if both are $90^\circ$. As a general property for parallel lines, they are equal, not supplementary.

Conclusion: Statement (d) is NOT true.

Correct Option: (d)

Given:

In Fig. 5.30, line $l$ is a transversal intersecting two parallel lines $PQ$ and $RS$.

The angles at the intersection with $PQ$ are $\angle 1$, $\angle 7$, $\angle 8$, and $\angle 2$.

The angles at the intersection with $RS$ are $\angle 3$, $\angle 5$, $\angle 6$, and $\angle 4$.

To Find:

Which of the following statements is TRUE:

(a) $\angle 1 = \angle 5$

(b) $\angle 4 = \angle 8$

(c) $\angle 5 = \angle 8$

(d) $\angle 3 = \angle 7$

Solution:

We will examine each statement based on the properties of parallel lines intersected by a transversal:

(a) Evaluating $\angle 1 = \angle 5$:

Angle $1$ is the exterior angle at the top-left, and Angle $5$ is the interior angle at the top-right. These angles do not have a standard equality relationship. In fact, since $\angle 1 = \angle 3$ (corresponding) and $\angle 3 + \angle 5 = 180^\circ$ (linear pair), $\angle 1 + \angle 5 = 180^\circ$.

(b) Evaluating $\angle 4 = \angle 8$:

Angle $4$ is the exterior angle at the bottom-right, and Angle $8$ is the interior angle at the bottom-left. Similar to option (a), these angles are supplementary, not equal.

(c) Evaluating $\angle 5 = \angle 8$:

Angle $8$ and Angle $5$ are located in the interior region between the parallel lines $PQ$ and $RS$, and they are on opposite sides of the transversal $l$.

These are called alternate interior angles. According to the property of parallel lines, alternate interior angles are always equal.

Therefore, this statement is True.

(d) Evaluating $\angle 3 = \angle 7$:

Angle $3$ is the interior angle at the top-left, and Angle $7$ is the exterior angle at the top-right. These angles are supplementary, not equal.

Conclusion: Statement (c) is TRUE.

Correct Option: (c)

Given:

$PQ \parallel ST$

Solution:

1. On the line containing $PQ$, the angle $130^\circ$ and $y$ form a linear pair.

2. Since $PQ \parallel ST$, if we consider $SO$ as a transversal, the alternate interior angle to $\angle TSR$ ($85^\circ$) is $x$.

3. Now, we find the sum:

Correct Option: (b)

Given:

In Fig. 5.32, $PQ \parallel RS$ and $QR \parallel TS$.

The measure of $\angle PQR = 85^\circ$.

To Find:

The value of $a$.

Solution:

First, we consider the parallel lines $PQ$ and $RS$ with $QR$ acting as the transversal.

Next, we consider the parallel lines $QR$ and $TS$ with $RS$ acting as the transversal.

From the figure, it is observed that $\angle TSR$ and angle $a$ lie on a straight line at point $S$. Therefore, they form a linear pair.

Substituting the value of $\angle TSR$ from equation (ii) into equation (iii):

Therefore, the value of $a$ is $95^\circ$.

Conclusion: Comparing the result with the given options, the correct option is (a).

Correct Option: (a)

Answer: complementary

Reason: By definition, if the sum of the measures of two angles is $90^\circ$, they are called complementary angles.

Answer: supplementary

Reason: By definition, if the sum of the measures of two angles is $180^\circ$, they are called supplementary angles.

Answer: distinct

Reason: A transversal is a line that intersects two or more lines at different (distinct) points.

Answer: $180^\circ$ (or supplementary)

Reason: When a transversal intersects two parallel lines, the interior angles on the same side of the transversal (co-interior angles) are supplementary, meaning their sum is $180^\circ$.

Answer: arm

Reason: Alternate interior angles share the segment of the transversal between the two lines as a common arm.

Answer: same

Reason: By definition, corresponding angles are pairs of angles that are in similar positions at each intersection where a straight line crosses two others, and they lie on the same side of the transversal.

Answer: opposite

Reason: Alternate interior angles are located in the interior region between the two lines and lie on opposite sides of the transversal.

Answer: parallel

Reason: Lines in the same plane that never intersect, no matter how far they are extended, are known as parallel lines.

Answer: linear

Reason: By the Linear Pair Axiom, if a ray stands on a line, then the sum of two adjacent angles so formed is $180^\circ$. Angles whose sum is $180^\circ$ are called supplementary.

Answer: obtuse

Reason: An acute angle is less than $90^\circ$. Let the acute angle be $x$, where $x < 90^\circ$.

Its supplement $= 180^\circ - x$.

Since $x < 90^\circ$, then $180^\circ - x > 90^\circ$.

Any angle greater than $90^\circ$ (but less than $180^\circ$) is an obtuse angle.

Answer: right

Reason: A right angle is exactly $90^\circ$.

Its supplement $= 180^\circ - 90^\circ = 90^\circ$.

Therefore, the supplement of a right angle is also a right angle.

Answer: acute

Reason: An obtuse angle is greater than $90^\circ$. Let the obtuse angle be $y$, where $y > 90^\circ$.

Its supplement $= 180^\circ - y$.

Since $y > 90^\circ$, then $180^\circ - y < 90^\circ$.

Any angle less than $90^\circ$ is an acute angle.

Answer: $90^\circ$

Reason: Complementary angles are two angles whose sum is $90^\circ$. If one angle were more than $90^\circ$, the sum would exceed $90^\circ$ (assuming positive measures), which is not possible by definition.

Given:

Angle $= 45^\circ$

To Find:

Its complementary angle.

Solution:

Let the complementary angle be $x$.

We know that the sum of complementary angles is $90^\circ$.

The complementary angle is $45^\circ$.

To Find:

Measure of an angle that is half of its supplement.

Solution:

Let the required angle be $x$.

Then its supplement will be $(180^\circ - x)$.

According to the question:

The angle is $60^\circ$.

Statement: Two right angles are complementary to each other.

Result: False

Reason: A right angle measures $90^\circ$. If we take two right angles, their sum is:

$90^\circ + 90^\circ = 180^\circ$

For angles to be complementary, their sum must be exactly $90^\circ$. Therefore, two right angles are supplementary, not complementary.

Statement: One obtuse angle and one acute angle can make a pair of complementary angles.

Result: False

Reason: An obtuse angle is greater than $90^\circ$. For two angles to be complementary, their sum must be exactly $90^\circ$. Since one angle alone is already more than $90^\circ$, it is impossible for their sum to be $90^\circ$.

Statement: Two supplementary angles are always obtuse angles.

Result: False

Reason: Supplementary angles are two angles whose sum is $180^\circ$. They can be:

1. Two right angles ($90^\circ + 90^\circ$)

2. One acute angle and one obtuse angle (e.g., $60^\circ + 120^\circ$)

Two obtuse angles (both $> 90^\circ$) would sum to more than $180^\circ$.

Statement: Two right angles are always supplementary to each other.

Result: True

Reason: Each right angle measures $90^\circ$. The sum of two right angles is:

$90^\circ + 90^\circ = 180^\circ$

By definition, angles whose sum is $180^\circ$ are supplementary.

Statement: One obtuse angle and one acute angle can make a pair of supplementary angles.

Result: True

Reason: For example, let an acute angle be $70^\circ$ and an obtuse angle be $110^\circ$. Their sum is:

$70^\circ + 110^\circ = 180^\circ$

Since the sum is $180^\circ$, they form a pair of supplementary angles.

Statement: Both angles of a pair of supplementary angles can never be acute angles.

Result: True

Reason: An acute angle is less than $90^\circ$. If both angles are less than $90^\circ$, their sum will always be less than $180^\circ$. For example, even $89^\circ + 89^\circ = 178^\circ$, which is not supplementary.

Statement: Two supplementary angles always form a linear pair.

Result: False

Reason: Supplementary angles only need to have a sum of $180^\circ$; they do not need to be adjacent (sharing a vertex and a common arm). A linear pair, however, must be adjacent. Non-adjacent angles can be supplementary but cannot be a linear pair.

Statement: Two angles making a linear pair are always supplementary.

Result: True

Reason: By the definition of a linear pair, the angles are adjacent and their non-common arms form a straight line. The sum of angles on a straight line is always $180^\circ$, which makes them supplementary.

Statement: Two angles making a linear pair are always adjacent angles.

Result: True

Reason: By definition, a linear pair is a pair of adjacent angles whose non-common arms are opposite rays. Since adjacency is a prerequisite for forming a linear pair, the statement is true.

Statement: Vertically opposite angles form a linear pair.

Result: False

Reason: Vertically opposite angles are formed by two intersecting lines and are opposite to each other at the vertex; they are not adjacent. A linear pair, by definition, must be adjacent and supplementary.

Statement: Interior angles on the same side of a transversal with two distinct parallel lines are complementary angles.

Result: False

Reason: When two parallel lines are cut by a transversal, the co-interior angles (interior angles on the same side) are supplementary ($180^\circ$), not complementary ($90^\circ$).

Statement: Vertically opposite angles are either both acute angles or both obtuse angles.

Result: True

Reason: Vertically opposite angles are always equal. Therefore, if one angle is acute ($< 90^\circ$), its opposite must also be acute. If one is obtuse ($> 90^\circ$), its opposite must also be obtuse. (In the special case where lines are perpendicular, both are right angles).

Statement: A linear pair may have two acute angles.

Result: False

Reason: An acute angle is less than $90^\circ$. If we add two acute angles, the sum will always be less than $180^\circ$.

Since the sum of a linear pair must be exactly $180^\circ$, it cannot consist of two acute angles.

Statement: An angle is more than $45^\circ$. Its complementary angle must be less than $45^\circ$.

Result: True

Reason: Let the angle be $x$ and its complement be $y$.

It is given that:

Subtracting $x$ from both sides of the sum equation:

Since $x > 45^\circ$, subtracting a value larger than $45^\circ$ from $90^\circ$ results in a value smaller than $45^\circ$.

Statement: Two adjacent angles always form a linear pair.

Result: False

Reason: Adjacent angles only need a common vertex and a common arm. They form a linear pair only if their sum is $180^\circ$ (i.e., their non-common arms form a straight line). Adjacent angles can have any sum, such as $30^\circ$ and $40^\circ$ ($70^\circ$ total), which is not a linear pair.

Solution:

Two angles are adjacent if they have a common vertex, a common arm, and their non-common arms are on opposite sides of the common arm.

(i) Pairs of adjacent angles:

1. $\angle AOB$ and $\angle BOC$

2. $\angle BOC$ and $\angle COD$

3. $\angle AOB$ and $\angle BOD$

4. $\angle AOC$ and $\angle COD$

(ii) Pairs of adjacent angles:

1. $\angle PQT$ and $\angle PQR$

2. $\angle PRQ$ and $\angle QRU$

3. $\angle QPR$ and $\angle SPR$

(iii) Pairs of adjacent angles:

1. $\angle TSV$ and $\angle USV$

2. $\angle SVT$ and $\angle SVU$

(iv) Pairs of adjacent angles:

1. $\angle AOC$ and $\angle AOD$

2. $\angle AOD$ and $\angle DOB$

3. $\angle DOB$ and $\angle BOC$

4. $\angle BOC$ and $\angle COA$

Solution:

Figure (i):

(i) Vertically opposite angles: $(\angle 1, \angle 3)$, $(\angle 2, \angle 4)$, $(\angle 5, \angle 7)$, $(\angle 6, \angle 8)$.

(ii) Linear pairs: $(\angle 1, \angle 2)$, $(\angle 2, \angle 3)$, $(\angle 3, \angle 4)$, $(\angle 4, \angle 1)$, $(\angle 5, \angle 6)$, $(\angle 6, \angle 7)$, $(\angle 7, \angle 8)$, $(\angle 8, \angle 5)$.

Figure (ii):

(i) Vertically opposite angles: None.

(ii) Linear pair: $\angle ABD$ and $\angle CBD$ , $\angle ABE$ and $\angle CBE$ (as $ABC$ is a straight line).

Figure (iii):

(i) Vertically opposite angles: None.

(ii) Linear pair: None.

Figure (iv):

(i) Vertically opposite angles: $(\angle POR, \angle QOS)$, $(\angle ROQ, \angle POS)$.

(ii) Linear pairs: $(\angle ROQ, \angle QOS)$, $(\angle QOS, \angle SOP)$, $(\angle SOP, \angle POR)$, $(\angle POR, \angle ROQ)$. Also, if $POQ$ is a line, $(\angle POT, \angle TOQ)$ is a linear pair.

Solution:

Supplementary angles are pairs of angles whose sum is $180^\circ$. Angles forming a linear pair are always supplementary.

Figure (i):

Pairs: $(\angle AOC, \angle COB)$, $(\angle COB, \angle BOD)$, $(\angle BOD, \angle DOA)$, $(\angle DOA, \angle AOC)$.

Figure (ii):

Pairs: $(\angle POS, \angle SOQ)$, $(\angle SOQ, \angle QOR)$, $(\angle QOR, \angle ROP)$, $(\angle ROP, \angle POS)$.

Figure (iii):

Pairs: $(\angle 1, \angle 2)$, $(\angle 3, \angle 4)$, $(\angle 5, \angle 6)$.

Given:

$PQ \parallel RS$

$TR \parallel QU$

$\angle PTR = 42^\circ$

To Find:

$\angle QUR$

Solution:

Consider the parallel lines $PQ$ and $RS$ with $TR$ acting as the transversal.

Now, consider the parallel lines $TR$ and $QU$ with line $RS$ acting as the transversal.

The angles $\angle TRU$ and $\angle QUR$ are interior angles on the same side of the transversal (co-interior angles).

Substituting the value from equation (i):

Therefore, the measure of $\angle QUR$ is $138^\circ$.

Solution:

(a) In Fig. 5.37(ii), the players are lined up in a straight line. The angle formed with the goalposts is greatest when the player is directly in the center, perpendicular to the goal line. Therefore, Player 4 has the best kicking angle.

(b) In Fig. 5.37(iii), the players are lined up diagonally. As the player moves closer to the center and closer to the goal line, the angle increases. By observation/measurement, Player 4 (or the one closest to the central perpendicular path) has the best kicking angle.

(c) Two situations where angles might be complementary (summing to $90^\circ$):

1. If one player is at a position where the angle is $30^\circ$ and another player is at a position where the angle is $60^\circ$.

2. If two players are at symmetric positions such that their individual angles are both $45^\circ$.

Given:

Sum of two vertically opposite angles = $166^\circ$

To Find:

The measure of each angle.

Solution:

We know that vertically opposite angles are always equal.

Let the measure of each angle be $x$.

According to the given condition:

Therefore, each of the vertically opposite angles measures $83^\circ$.

Given:

Lines $l \parallel m \parallel n$.

$\angle QPS = 35^\circ$ and $\angle QRT = 55^\circ$.

To Find:

$\angle PQR$

Solution:

Consider the parallel lines $l$ and $m$. Line $PQ$ acts as a transversal for them.

Now, consider the parallel lines $m$ and $n$. Line $QR$ acts as a transversal for them.

From the figure, we can observe that:

Substituting the values from equations (i) and (ii):

Therefore, the value of $\angle PQR$ is $90^\circ$.

Given:

Points $P, Q, R$ are collinear. Ray $TQ \perp PR$ ($\angle TQR = 90^\circ$ and $\angle TQP = 90^\circ$).

Solution:

(a) Pair of complementary angles:

Complementary angles sum to $90^\circ$. Since $\angle TQR = 90^\circ$, the adjacent angles within it are:

$\angle TQS$ and $\angle SQR$

(b) Two pairs of supplementary angles:

Supplementary angles sum to $180^\circ$. Angles on the straight line $PR$ are:

1. $\angle TQP$ and $\angle TQR$ (both are $90^\circ$)

2. $\angle PQS$ and $\angle SQR$ (forming a linear pair)

(c) Four pairs of adjacent angles:

1. $\angle PQT$ and $\angle TQS$

2. $\angle TQS$ and $\angle SQR$

3. $\angle PQT$ and $\angle TQR$

4. $\angle PQS$ and $\angle SQR$

Given:

$OR \perp OP$, which means $\angle POR = 90^\circ$.

Solution:

(i) All the pairs of adjacent angles:

1. $\angle POQ$ and $\angle QOR$

2. $\angle QOR$ and $\angle ROS$

3. $\angle POQ$ and $\angle QOS$

4. $\angle POR$ and $\angle ROS$

(ii) All the pairs of complementary angles:

Since $\angle POR = 90^\circ$, any pair of adjacent angles that make up this right angle are complementary:

1. $\angle POQ$ and $\angle QOR$

To Find:

(a) Number of angles formed.

(b) Number of types of angles formed.

(c) Pairs of vertically opposite angles.

Solution:

(a) In Fig. 5.41, when two lines $AB$ and $CD$ intersect at point $O$, a total of 4 angles are formed ($\angle 1, \angle 2, \angle 3,$ and $\angle 4$).

(b) There are 2 types of angle relationships formed here:

1. Vertically opposite angles (which are equal).

2. Linear pairs (which are supplementary).

(c) The pairs of vertically opposite angles are:

1. $\angle 1$ and $\angle 3$

2. $\angle 2$ and $\angle 4$

To Determine: Whether the following pairs of angles are adjacent and provide justification.

Solution:

(i) Yes

Justification: The angles $a$ and $b$ have a common vertex, a common arm (the ray pointing upwards to the right), and their non-common arms are on opposite sides of the common arm. Therefore, they are adjacent.

(ii) No

Justification: Although the angles $a$ and $b$ share a common vertex $O$, they do not share a common arm. There is another angle between them; hence, they are not adjacent.

(iii) No

Justification: These angles do not have a common vertex. Adjacent angles must originate from the same point (vertex).

(iv) No

Justification: These angles share a common vertex and one common arm, but their non-common arms are on the same side of the common arm. In this case, angle $a$ is a part of angle $b$ (overlapping), so they are not adjacent.

To Find:

All pairs of supplementary angles in Fig. 5.43.

Solution:

Supplementary angles are pairs of angles whose sum is $180^\circ$. In a geometry diagram, angles forming a linear pair (adjacent angles on a straight line) are always supplementary.

By observing the intersections in Fig. 5.43, we identify the following pairs:

1. At the Top Intersection:

There are four angles formed by the intersection of two straight lines. The adjacent angles on each line are:

2. At the Bottom-Left Intersection:

The angles $\angle 7$ and $\angle 2$ lie on the same straight line (the base of the triangle extended). Therefore, they are adjacent and supplementary.

3. At the Bottom-Right Intersection:

The angles $\angle 1$ and $\angle 8$ lie on the same straight line (the base of the triangle extended). Therefore, they are adjacent and supplementary.

Final List of Supplementary Pairs:

The required pairs are ($\angle 1, \angle 8$), ($\angle 2, \angle 7$), ($\angle 3, \angle 4$), ($\angle 4, \angle 5$), ($\angle 5, \angle 6$), and ($\angle 6, \angle 3$).

Solution:

We know that a linear pair of angles consists of two adjacent angles whose sum is $180^\circ$.

(a) One of its angle is acute:

If one angle is acute (less than $90^\circ$), then the other angle must be greater than $90^\circ$ ($180^\circ - \text{acute angle} > 90^\circ$) to make the sum $180^\circ$.

Therefore, the other angle is obtuse.

(b) One of its angles is obtuse:

If one angle is obtuse (greater than $90^\circ$), then the other angle must be less than $90^\circ$ ($180^\circ - \text{obtuse angle} < 90^\circ$) to make the sum $180^\circ$.

Therefore, the other angle is acute.

(c) One of its angles is right:

If one angle is a right angle ($90^\circ$), then the other angle must be $180^\circ - 90^\circ = 90^\circ$.

Therefore, the other angle is also a right angle.

Solution:

No, two acute angles cannot form a pair of supplementary angles.

Reason:

An acute angle is defined as an angle whose measure is less than $90^\circ$.

If we take the two largest possible acute angles, say $89.9^\circ$ each, their sum would be:

$89.9^\circ + 89.9^\circ = 179.8^\circ$

Since the sum of two acute angles will always be less than $180^\circ$ ($90^\circ + 90^\circ$), they cannot satisfy the condition for supplementary angles, which requires the sum to be exactly $180^\circ$.

Solution:

Adjacent angles are those that have a common vertex and a common arm but no common interior points.

In the given figure, the pairs of adjacent angles are:

(i) $\angle 1$ and $\angle 2$ (Common arm OA)

(ii) $\angle 2$ and $\angle 3$ (Common arm OD)

(iii) $\angle 3$ and $\angle 4$ (Common arm OB)

(iv) $\angle 4$ and $\angle 1$ (Common arm OC)

Thus, the required pairs are ($\angle 1, \angle 2$), ($\angle 2, \angle 3$), ($\angle 3, \angle 4$), and ($\angle 4, \angle 1$).

Given:

Complement of an angle $= 62^\circ$.

To Find:

The supplement of that angle.

Solution:

Let the required angle be $x$.

We know that the sum of an angle and its complement is $90^\circ$.

$x + 62^\circ = 90^\circ$

$x = 90^\circ - 62^\circ$

$x = 28^\circ$

Now, we need to find the supplement of $x = 28^\circ$.

We know that the sum of an angle and its supplement is $180^\circ$.

Supplement of $28^\circ = 180^\circ - 28^\circ$

Supplement of $28^\circ = 152^\circ$

Therefore, the supplement of the angle is $152^\circ$.

Given:

An angle formed by the road and the lower railway track is $30^\circ$ (bottom-left exterior angle).

The railway tracks are parallel, and the road acts as a transversal.

To Find:

The values of $a, b$ and $c$.

Solution:

Step 1: Finding the value of $c$.

The angle $30^\circ$ and angle $c$ are exterior angles on the same side of the transversal (co-exterior angles).

$c = 180^\circ - 30^\circ$

$c = 150^\circ$

Step 2: Finding the value of $a$.

The angles $c$ and $a$ are exterior angles on the same straight line (the top railway track).

$150^\circ + a = 180^\circ$

$a = 180^\circ - 150^\circ$

$a = 30^\circ$

Step 3: Finding the value of $b$.

Let the vertically opposite angle (VOA) of $a$ be $x$.

$x = 30^\circ$

Now, $x$ and $b$ are interior angles on the same side of the transversal (co-interior angles).

$30^\circ + b = 180^\circ$

$b = 180^\circ - 30^\circ$

$b = 150^\circ$

Final Values:

The values are $a = 30^\circ$, $b = 150^\circ$ and $c = 150^\circ$.

Alternate Solution for $b$:

Angle $b$ and angle $c$ are alternate interior angles. Since $c = 150^\circ$, then $b = 150^\circ$.

Given:

Angle made by the leg with the floor $= 35^\circ$.

The top of the stool is parallel to the floor.

Solution:

In Fig. 5.46, the stool's top and the floor line are parallel. The leg of the stool acts as a transversal line.

Finding $x$:

The angle $x$ and the $35^\circ$ angle are alternate interior angles.

Finding $y$:

Angles $x$ and $y$ lie on the same straight line (the top frame of the stool) and are adjacent, forming a linear pair.

Substituting the value of $x$:

$35^\circ + y = 180^\circ$

$y = 180^\circ - 35^\circ$

$y = 145^\circ$

Final Answer:

The values are $x = 35^\circ$ and $y = 145^\circ$.

Given:

$a \parallel b$, $c \parallel d$, and $e \parallel f$.

Solution:

(i) Angle between $b$ and $c$:

By observing the design and the orientation of the rods:

The angle between rod $b$ and rod $c$ is $30^\circ$.

(ii) Angle between $d$ and $e$:

It is given that the angle between $d$ and $e$ is co-interior to the $75^\circ$ angle.

$\text{Angle between } d \text{ and } e = 180^\circ - 75^\circ$

$\text{Angle between } d \text{ and } e = 105^\circ$

(iii) Angle between $d$ and $f$:

The angle between $d$ and $e$ (found above as $105^\circ$) and the angle between $d$ and $f$ are co-interior.

$\text{Angle between } d \text{ and } f = 180^\circ - 105^\circ$

$\text{Angle between } d \text{ and } f = 75^\circ$

(iv) Angle between $c$ and $f$:

The angle between $d$ and $f$ (found above as $75^\circ$) and the angle between $c$ and $f$ are co-interior.

$\text{Angle between } c \text{ and } f = 180^\circ - 75^\circ$

$\text{Angle between } c \text{ and } f = 105^\circ$

Given:

Line segments $a \parallel d, b \parallel e, c \parallel f$.

One exterior angle is marked as $120^\circ$.

Solution:

Finding the value of $x$ and $z$:

Let the vertically opposite angle (VOA) of $120^\circ$ be $\angle 1$.

The angle $\angle 1$ is co-interior to both $\angle x$ and $\angle z$ due to the parallel lines $c \parallel f$ and $a \parallel d$.

$x = 180^\circ - 120^\circ = 60^\circ$

$z = 180^\circ - 120^\circ = 60^\circ$

Finding the value of $y$:

As per the observation, the co-interior angle of $x$ (or $z$), which is $120^\circ$, is the vertically opposite angle of $y$.

Final Answer:

The values are $\angle x = 60^\circ$, $\angle y = 120^\circ$ and $\angle z = 60^\circ$.

Given:

$AB \parallel CD$, $AF \parallel ED$.

$\angle AFC = 68^\circ$ and $\angle FED = 42^\circ$.

To Find:

$\angle EFD$

Solution:

Since $AF \parallel ED$ and $FE$ is a transversal line intersecting them:

$\angle AFE = 42^\circ$

Now, points $C, F, \text{ and } D$ lie on a straight line $CD$. Therefore, the sum of angles on this straight line at point $F$ is $180^\circ$.

Substituting the known values:

$68^\circ + 42^\circ + \angle EFD = 180^\circ$

$110^\circ + \angle EFD = 180^\circ$

$\angle EFD = 180^\circ - 110^\circ$

$\angle EFD = 70^\circ$

Final Answer:

The value of $\angle EFD$ is $70^\circ$.

Given:

$OB \perp OA$, which means $\angle AOB = 90^\circ$.

$\angle BOC = 49^\circ$.

To Find:

The measure of $\angle AOD$.

Solution:

Since $\angle AOB = 90^\circ$, the angles $\angle BOC$ and $\angle AOC$ are complementary angles.

$49^\circ + \angle AOC = 90^\circ$

$\angle AOC = 90^\circ - 49^\circ = 41^\circ$

From the figure, $CD$ is a straight line, which implies $\angle AOC$ and $\angle AOD$ form a linear pair.

$41^\circ + \angle AOD = 180^\circ$

$\angle AOD = 180^\circ - 41^\circ$

$\angle AOD = 139^\circ$

Therefore, the value of $\angle AOD$ is $139^\circ$.

Given:

Lines AB, CD, and EF intersect at O.

$\angle AOE = 30^\circ$

$\angle DOB = 40^\circ$

To Find:

The measure of $\angle COF$.

Solution:

Since the lines AB and EF intersect at O, the vertically opposite angles are equal.

$\angle FOB = 30^\circ$

Similarly, since the lines AB and CD intersect at O, the vertically opposite angles are equal.

$\angle AOC = 40^\circ$

Now, AB is a straight line. Therefore, the sum of all angles on one side of the line at point O is $180^\circ$.

Substituting the values of $\angle AOC$ and $\angle FOB$:

$40^\circ + \angle COF + 30^\circ = 180^\circ$

$70^\circ + \angle COF = 180^\circ$

$\angle COF = 180^\circ - 70^\circ$

$\angle COF = 110^\circ$

Therefore, the value of $\angle COF$ is $110^\circ$.

To Find:

The measure of two complementary angles.

Solution:

Let the first even integer (angle) be $x^\circ$.

Since the angles are consecutive even integers, the second angle will be $(x + 2)^\circ$.

We know that the sum of two complementary angles is $90^\circ$.

$2x + 2 = 90$

$2x = 90 - 2$

$2x = 88$

$x = \frac{88}{2} = 44$

The first angle is $44^\circ$.

The second angle is $x + 2 = 44 + 2 = 46^\circ$.

Therefore, the two angles are $44^\circ$ and $46^\circ$.

Given:

The difference between two interior angles on the same side of a transversal is $20^\circ$.

To Find:

The measure of the two angles.

Solution:

Let the two interior angles on the same side of the transversal be $x$ and $y$.

We know that interior angles on the same side of a transversal (co-interior angles) are supplementary, meaning their sum is $180^\circ$.

According to the question, the difference between these angles is $20^\circ$:

Adding equations (i) and (ii):

$(x + y) + (x - y) = 180^\circ + 20^\circ$

$2x = 200^\circ$

$x = \frac{200^\circ}{2} = 100^\circ$

Substituting the value of $x$ in equation (i):

$100^\circ + y = 180^\circ$

$y = 180^\circ - 100^\circ$

$y = 80^\circ$

Therefore, the required angles are $100^\circ$ and $80^\circ$.

To Find:

The measure of the two angles in the linear pair.

Solution:

Let one of the angles be $x$.

According to the given condition, the other angle is $\frac{1}{3}$ of $x$, i.e., $\frac{x}{3}$.

Since they form a linear pair, their sum is $180^\circ$.

Taking the LCM:

$\frac{3x + x}{3} = 180^\circ$

$\frac{4x}{3} = 180^\circ$

$4x = 180^\circ \times 3$

$4x = 540^\circ$

$x = \frac{540^\circ}{4} = 135^\circ$

The first angle is $135^\circ$.

The second angle $= \frac{135^\circ}{3} = 45^\circ$.

Therefore, the two angles are $135^\circ$ and $45^\circ$.

To Find:

The measure of the two supplementary angles.

Solution:

Let the first odd integer (angle) be $x^\circ$.

Since the angles are consecutive odd integers, the next angle will be $(x + 2)^\circ$.

We know that the sum of two supplementary angles is $180^\circ$.

$2x + 2 = 180$

$2x = 180 - 2$

$2x = 178$

$x = \frac{178}{2} = 89$

The first angle is $89^\circ$.

The second angle is $x + 2 = 89 + 2 = 91^\circ$.

Therefore, the two supplementary angles are $89^\circ$ and $91^\circ$.

Given:

$AE \parallel GF \parallel BD$

$AB \parallel CG \parallel DF$

$\angle CHE = 120^\circ$

To Find:

$\angle ABC$ and $\angle CDE$

Solution:

Step 1: Finding the value of $\angle HCD$.

Considering parallel lines $AE \parallel BD$ and $CG$ as a transversal, $\angle CHE$ and $\angle HCD$ are interior angles on the same side of the transversal (co-interior angles).

$120^\circ + \angle HCD = 180^\circ$

$\angle HCD = 180^\circ - 120^\circ$

$\angle HCD = 60^\circ$

Step 2: Finding the value of $\angle ABC$.

Considering parallel lines $AB \parallel CG$ and $BD$ as a transversal, $\angle ABC$ and $\angle HCD$ are corresponding angles.

$\angle ABC = 60^\circ$

Step 3: Finding the value of $\angle CDE$.

Considering parallel lines $CG \parallel DF$ and $BD$ as a transversal, $\angle HCD$ and $\angle CDE$ are interior angles on the same side of the transversal (co-interior angles).

$60^\circ + \angle CDE = 180^\circ$

$\angle CDE = 180^\circ - 60^\circ$

$\angle CDE = 120^\circ$

Final Answer:

The values are $\angle ABC = 60^\circ$ and $\angle CDE = 120^\circ$.

Given:

Points $A, O$ and $B$ are collinear, meaning $AOB$ is a straight line.

$\angle AOD = (x - 10)^\circ$

$\angle DOC = (4x - 25)^\circ$

$\angle COB = (x + 5)^\circ$

To Find:

The value of $\angle BOC$.

Solution:

Since $AOB$ is a straight line, the sum of all angles on it is $180^\circ$.

$(x - 10) + (4x - 25) + (x + 5) = 180$

$x + 4x + x - 10 - 25 + 5 = 180$

$6x - 30 = 180$

$6x = 180 + 30$

$6x = 210$

$x = \frac{210}{6} = 35$

Now, substitute the value of $x$ to find $\angle BOC$:

$\angle BOC = (x + 5)^\circ$

$\angle BOC = (35 + 5)^\circ = 40^\circ$

Final Answer:

The value of $\angle BOC$ is $40^\circ$.

Given:

$l \parallel m$. Angle at $C$ is $65^\circ$. Angle at $B$ is $132^\circ$.

Solution:

Step 1: Finding $b$.

Based on the orientation, angle $b$ and the $132^\circ$ angle are interior angles on the same side of transversal $t$.

$b = 180^\circ - 132^\circ = 48^\circ$.

Step 2: Finding $a$.

The total angle at the top vertex $(a + b)$ and the $65^\circ$ angle at vertex $C$ are interior angles on the same side of transversal $s$.

Substituting the value of $b$:

$(a + 48^\circ) + 65^\circ = 180^\circ$

$a + 113^\circ = 180^\circ \implies a = 180^\circ - 113^\circ = 67^\circ$.

Final Answer:

The values are $a = 67^\circ$ and $b = 48^\circ$.

Given:

Line $l$ is parallel to line $m$ ($l \parallel m$).

Line $t$ is a transversal intersecting lines $l$ and $m$ at points $P$ and $Q$ respectively.

At intersection $Q$, one of the angles is $132^\circ$.

To Find:

The value of the expression $2a + b$.

Solution:

Step 1: Finding the value of $a$.

By observing Fig. 5.55, angle $a$ at intersection $P$ and the $132^\circ$ angle at intersection $Q$ are in corresponding positions relative to the parallel lines and the transversal.

Step 2: Finding the value of $b$.

Angle $b$ and the angle measuring $132^\circ$ are at the same intersection $Q$ and are opposite to each other.

Step 3: Calculating $2a + b$.

Now, substitute the values of $a$ and $b$ into the given expression:

$2a + b = 2(132^\circ) + 132^\circ$

$2a + b = 264^\circ + 132^\circ$

$2a + b = 396^\circ$

Final Answer:

The sum of $2a + b$ is $396^\circ$.

Given:

$QP \parallel RS$

$\angle PQR = 70^\circ$

$\angle QPR = 65^\circ$

To Find:

The values of $a$ and $b$.

Solution:

Step 1: Finding $a$.

Since $QP \parallel RS$ and $PR$ acts as a transversal line, the angle $\angle QPR$ and $\angle a$ are alternate interior angles.

Step 2: Finding $b$.

Since $QP \parallel RS$ and the line $QR$ (extended) acts as a transversal, the angle $\angle PQR$ and $\angle b$ are corresponding angles.

Final Answer:

The values are $a = 65^\circ$ and $b = 70^\circ$.

Given:

$PQ \parallel RT$

$\angle RPQ = 45^\circ$

$\angle RQP = 55^\circ$

To Find:

The value of $a + b$.

Solution:

In Fig. 5.57, since $PQ \parallel RT$ and $PS$ is a transversal line passing through them:

$a = 45^\circ$

Now, considering $PQ \parallel RT$ and $QR$ as a transversal line:

$b = 55^\circ$

We need to find the sum $a + b$:

$a + b = 45^\circ + 55^\circ$

$a + b = 100^\circ$

Therefore, the value of $a + b$ is $100^\circ$.

Given:

$PQ \parallel RS \parallel UT$

Solution (i):

Given $c = 57^\circ$ and $a = \frac{c}{3}$.

$a = \frac{57^\circ}{3} = 19^\circ$

Since $PQ \parallel UT$ and $PT$ is a transversal line, the alternate interior angles are equal.

$19^\circ + b = 57^\circ$

$b = 57^\circ - 19^\circ = 38^\circ$

Now, since $PQ \parallel RS$ and $PR$ is a transversal line, the interior angles on the same side of the transversal (co-interior angles) are supplementary.

$38^\circ + d = 180^\circ$

$d = 180^\circ - 38^\circ = 142^\circ$

Thus, $d = 142^\circ$.

Solution (ii):

Given $c = 75^\circ$ and $a = \frac{2}{5}c$.

$a = \frac{2}{5} \times 75^\circ = 2 \times 15^\circ = 30^\circ$

As established earlier, $a + b = c$ because $PQ \parallel UT$ and $PT$ is a transversal.

$30^\circ + b = 75^\circ$

$b = 75^\circ - 30^\circ = 45^\circ$

Thus, $b = 45^\circ$.

Given:

$AB \parallel CD$, $\angle AEF = 34^\circ$, $\angle FGD = 135^\circ$

Construction:

Draw a line $l$ passing through point $F$ such that $l \parallel AB$. Since $AB \parallel CD$, it follows that $l \parallel CD$ as well.

Solution:

Let the interior $\angle EFG$ be divided into $\angle 1$ and $\angle 2$ by the constructed line $l$.

Since $AB \parallel l$ and $EF$ is a transversal:

Since $CD \parallel l$ and $FG$ is a transversal, the co-interior angles are supplementary:

$\angle 2 = 180^\circ - 135^\circ = 45^\circ$

Now, calculate interior $\angle EFG$:

Interior $\angle EFG = \angle 1 + \angle 2$

Interior $\angle EFG = 34^\circ + 45^\circ = 79^\circ$

To find the reflex $\angle EFG$:

Reflex $\angle EFG = 360^\circ - 79^\circ = 281^\circ$

Therefore, the reflex $\angle EFG$ is $281^\circ$.

Given:

Line $l$ is parallel to line $m$ ($l \parallel m$).

Lines $n$ and $p$ are transversal lines.

The interior angles formed by transversal $p$ include $66^\circ$ and $x$.

The interior angles formed by transversal $n$ include $y$ and $48^\circ$.

To Find:

The values of $x$ and $y$.

Solution:

Step 1: Finding the value of $x$.

Since line $l \parallel m$ and line $p$ is the transversal, the angle $x$ and the angle measuring $66^\circ$ are interior angles on the same side of the transversal.

$x = 180^\circ - 66^\circ$

$x = 114^\circ$

Step 2: Finding the value of $y$.

Since line $l \parallel m$ and line $n$ is the transversal, the angle $y$ and the angle measuring $48^\circ$ are interior angles on the same side of the transversal.

$y = 180^\circ - 48^\circ$

$y = 132^\circ$

Final Answer:

The calculated values are $x = 114^\circ$ and $y = 132^\circ$.

Given:

Lines $l, m$ and $n$ are parallel to each other ($l \parallel m \parallel n$).

Lines $p$ and $q$ are parallel to each other ($p \parallel q$).

To Find:

The values of $a$, $b$ and $c$.

Solution:

Step 1: Finding the value of $c$.

Considering parallel lines $p$ and $q$ with line $n$ as a transversal, the angle $4c$ and the angle $120^\circ$ are in corresponding positions.

$c = \frac{120^\circ}{4}$

$c = 30^\circ$

Step 2: Finding the value of $b$.

Considering parallel lines $m$ and $n$ with line $p$ as a transversal, the angle $3b$ and the angle $4c$ are corresponding angles.

Substituting the value of $4c = 120^\circ$ from step 1:

$3b = 120^\circ$

$b = \frac{120^\circ}{3}$

$b = 40^\circ$

Step 3: Finding the value of $a$.

Considering parallel lines $n$ and $l$ with line $q$ as a transversal, the angle $6a$ and the angle $120^\circ$ are corresponding angles.

$a = \frac{120^\circ}{6}$

$a = 20^\circ$

Final Answer:

The values are $a = 20^\circ$, $b = 40^\circ$ and $c = 30^\circ$.

Given:

In Fig. 5.62, an angle of $60^\circ$ is formed at the intersection of line $l$ and line $m$.

An angle of $120^\circ$ is formed at the intersection of line $l$ and line $n$.

To Find:

The pair of lines that are parallel.

Solution:

By observing the figure, line $l$ acts as a transversal for lines $m$ and $n$.

Let the vertically opposite angle of the $120^\circ$ angle be denoted as $\angle 1$.

Now, the angle $\angle 1$ and the given $60^\circ$ angle are interior angles on the same side of the transversal line $l$ (co-interior angles).

Let's calculate their sum:

$\text{Sum} = 60^\circ + \angle 1$

$\text{Sum} = 60^\circ + 120^\circ$

$\text{Sum} = 180^\circ$

Since the sum of the interior angles on the same side of the transversal is supplementary ($180^\circ$), the lines $m$ and $n$ must be parallel to each other.

Final Answer:

The pair of parallel lines is $m \parallel n$.

Solution:

(i) To examine if line EF is parallel to line GH:

Let line CD be the transversal for lines EF and GH.

At the intersection of line EF and transversal CD, an exterior angle of $65^\circ$ is given.

This VOA of $65^\circ$ and the angle $70^\circ$ (at the intersection of GH and CD) are a pair of corresponding angles.

Since the corresponding angles are not equal, lines EF and GH are not parallel.

(ii) To examine if line AB is parallel to line CD:

Let line EF be the transversal for lines AB and CD.

At the intersection of line CD and transversal EF, an exterior angle of $65^\circ$ is given.

At the intersection of line AB and transversal EF, an interior angle of $115^\circ$ is given.

The VOA of $65^\circ$ and the $115^\circ$ angle are interior angles on the same side of the transversal EF (co-interior angles). Let's check their sum:

Since the sum of the interior angles on the same side of the transversal is $180^\circ$, the lines must be parallel.

Therefore, line AB is parallel to line CD (AB $\parallel$ CD).

To Find:

The pair of parallel lines from the given figure.

Solution:

Step 1: Check lines EF and GH.

Consider line CD as the transversal line for EF and GH. The angles measuring $123^\circ$ and $57^\circ$ are interior angles on the same side of the transversal (co-interior angles).

Since the sum of co-interior angles is exactly $180^\circ$, line EF is parallel to line GH (EF $\parallel$ GH).

Step 2: Check lines GH and KP.

Consider line CD as the transversal for lines GH and KP. The angle $57^\circ$ and the given angle $55^\circ$ are in corresponding positions.

Since the corresponding angles are not equal, lines GH and KP are not parallel.

Step 3: Check lines AB and CD.

Consider line GH as the transversal line for lines AB and CD. At the intersection with line CD, the given angle is $57^\circ$. Its linear pair is:

$180^\circ - 57^\circ = 123^\circ$

This $123^\circ$ angle and the given angle $122^\circ$ at the intersection with line AB are corresponding angles.

Since the corresponding angles are not equal, lines AB and CD are not parallel.

Final Answer:

The only pair of parallel lines is EF $\parallel$ GH.

Given:

From the given Fig. 5.65, we observe the following angle measures:

$\angle$ at intersection of AB and EF = $130^\circ$

$\angle$ at intersection of AB and GH = $50^\circ$

$\angle$ at intersection of CD and EF = $50^\circ$

To Prove:

(i) AB $\parallel$ CD

(ii) EF $\parallel$ GH

Proof:

Let the linear pair of the $130^\circ$ angle at the intersection of AB and EF be denoted as $\angle 1$.

(i) For lines AB and CD:

Consider lines AB and CD with line EF acting as the transversal.

We found that the linear pair of $130^\circ$ ($\angle 1$) is $50^\circ$.

The $\angle$ at the intersection of CD and EF is also given as $50^\circ$.

These two angles are in corresponding positions relative to the transversal EF.

Since the corresponding angles are equal, the lines are parallel.

(ii) For lines EF and GH:

Consider lines EF and GH with line AB acting as the transversal.

We found that the linear pair of $130^\circ$ ($\angle 1$) is $50^\circ$.

The $\angle$ at the intersection of AB and GH is given as $50^\circ$.

These two angles are in alternate interior positions relative to the transversal AB.

Since the alternate interior angles are equal, the lines are parallel.

Given:

Line $l$ is parallel to line $m$ ($l \parallel m$).

Line $p$ and line $q$ are transversals.

To Find:

The values of $x$ and $y$.

Solution:

Step 1: Finding the value of $y$.

Consider the parallel lines $l$ and $m$ with line $p$ acting as the transversal. The angles $y$ and $80^\circ$ are interior angles on the same side of the transversal.

$y = 180^\circ - 80^\circ$

$y = 100^\circ$

Step 2: Finding the value of $x$.

Consider the parallel lines $l$ and $m$ with line $q$ acting as the transversal. By observing Fig. 5.66, the angle $x$ and the angle $110^\circ$ are alternate interior angles.

Final Answer:

The determined values are $x = 110^\circ$ and $y = 100^\circ$.