Unit 5: Lines and Angles

## Exercise C: (Multiple Choice Questions and Answers 1-41)

In the Questions 1 to 41, there are four options, out of which only one is correct. Write the correct one.

# Question: 1

The angles between North and West and South and East are

a.    complementary

b.   supplementary

c.    both are acute

d.   both are obtuse

## Solution

(b)

From the figure, it is clear that the angle between North and West is $90°$ and South and East is $90°$.

$\therefore$ Sum of these two angles $=90°+90°=180°$

Hence, the two angles are supplementary, as their sum is $180°$.

# Question: 2

Angles between South and West and South and East are

a.    vertically opposite angles

b.   complementary angles

c.    making a linear pair

## Solution

(c)

From the above figure, we can say that angle between South and West is $90°$ and angle between south and East is $90°.$ So, their sum is $180°.$ Hence, both angles make a linear pair.

# Question: 3

In Fig. 5.9, $\text{PQ}$ is a mirror, $\text{AB}$ is the incident ray and $\text{BC}$ is the reflected ray. If $\angle \text{ABC}=46°$, then $\angle \text{ABP}$ is equal to

a.    $44°$

b.   $67°$

c.    $13°$

d.    $62°$

Fig. 5.9

## Solution

(b)

We know that, the angle of incidence is always equal to the angle of reflection.

$\angle \text{ABP}=\angle \text{CBQ}$

i.e., $a=b$

now, sum of all angles on a straight line is $180°$

[given, $\angle \text{ABC}=46°$ ]

$a+46°+b=180°$

$2a=180°-46°$                   $\left[\because a=b\right]$

$2a=134°$

$2a=\frac{134°}{2}=67°$

$\angle \text{ABP}=67°$

# Question: 4

If the complement of an angle is $79°,$ then the angle will be of

a.    $1°$

b.   $11°$

c.    $79°$

d.   $101°$

## Solution

(b)

Let the angle be $x°$

Then, the complement of $x°$ will be $\left(90-x\right)°$.

Given, complement of $x°$ is $79°$.

$\therefore \left(90-x\right)°=79°$

$⇒x°=90°-79°=11°$

Therefore, the required angle is $11°$.

Note: Sum of the complementary angles is $90°$.

# Question: 5

Angles which are both supplementary and vertically opposite are

a.

b.

c.

d.

## Solution

(b)

Two angles are said to be supplementary, if their sum is $180°$.

Also, if two angles are vertically opposite, then they are equal.

Therefore, angles given in option (b) are supplementary as well as vertically opposite.

# Question: 6

The angle which makes a linear pair with an angle of $61°$ is of

a.    $29°$

b.   $61°$

c.    $122°$

d.   $119°$

## Solution

(d)

Let the required angle be $x°$. It is given that $x°$ makes a linear pair with $61°$

$x°+61°=180°$

$x°=180°-61°=119°$

# Question: 7

The angles $x$ and $90°-x$ are

1. supplementary
2. complementary
3. vertically opposite
4. making a linear pair

## Solution

(b)

Sum of the given angles $=x+90°-x=90°$

Since, the sum of given two angles is $90°$. Hence, they are complementary to each other.

# Question: 8

The angles $x-10°$ and $190°-x$ are

1. interior angles on the same side of the transversal
2. making a linear pair
3. complementary
4. supplementary

## Solution

(d)

Sum of the given angles

$=\left(x-10°\right)+\left(190°-x\right)$

$=x-10°+190°-x$

$=\left(x-x\right)+\left(190°-10°\right)$

$=0+180°=180°$

Since, the sum of given angles is $180°$, Hence, they are supplementary.

# Question: 9

In Fig. 5.10, the value of $x$ is

a.    $110°$

b.   $46°$

c.    $64°$

d.   $150°$

Fig. 5.10

## Solution

(d)

We know that, the sum of all angles around a point is $360°.$

# Question: 10

In Fig. 5.11, if $\text{AB}\parallel \text{CD}$, $\angle \text{APQ}=50°$ and $\angle \text{PRD}=130°,$ then $\angle \text{QPR}$ is

a.    $130°$

b.   $50°$

c.    $80°$

d.   $30°$

Fig. 5.11

## Solution

(c)

$\text{AB}$ and $\text{CD}$ are parallel and $\text{PR}$ is transversal.

$\angle \text{BPR+}\angle \text{PRD}=180°$
[sum of consecutive interior angle is $180°$ ]

$\angle \text{BPR}+130°=180°$

$\angle \text{BPR}=180°-130°$

$\angle \text{BPR}=50°$

Also, $\angle \text{APQ}+\angle \text{QPR}+\angle \text{BPR}=180°$

$50°+\angle \text{QPR}+50°=180°$

$\angle \text{QPR}=180°-100°$

$\angle \text{QPR}=80°$

# Question: 11

In Fig. 5.12, lines $l$ and $m$ intersect each other at a point. Which of the following is false?

Fig. 5.12

a.    $\angle a=\angle b$

b.   $\angle d=\angle c$

c.    $\angle a+\angle d=180°$

d.   $\angle a=\angle d$

## Solution

(d)

In given figure, $\angle a=\angle b$ and $\angle d=\angle c$ [vertically opposite angle]

Also, $\angle a+\angle d=180°$

And $\angle b+\angle c=180°$     [linear pair]

# Question: 12

If angle $\text{P}$ and angle $\text{Q}$ are supplementary and the measure of angle $\text{P}$ is $60°$, then the measure of angle $\text{Q}$ is

a.    $120°$

b.   $60°$

c.    $30°$

d.   $20°$

## Solution

(a)

It is given that, angle $\text{P}$ and angle $\text{O}$ are supplementary. Hence, their sum will be $180°$

$\begin{array}{c}\angle \text{P}+\angle \text{Q}=180°\\ 60+\angle \text{Q}=180°\\ \angle \text{Q}=180°-60°\\ \angle \text{Q}=120°\end{array}$

# Question: 13

In Fig. 5.13, $\text{POR}$ is a line. The value of $a$ is

Fig. 5.13

a.    $40°$

b.   $45°$

c.    $55°$

d.   $60°$

## Solution

(a)

$\angle \text{POQ}+\angle \text{ROQ}=180°$

$\left(3a+5\right)+\left(2a-25\right)=180°$

$3a+5+2a-25=180°$

$5a-20=180°$

$5a=180+20$

$5a=200$

$a=\frac{200}{5}=40$

Hence, the value of $a$ is 40.

# Question: 14

In Fig. 5.14, $\text{POQ}$ is a line. If $x=30°$, then $\angle \text{QOR}$ is

Fig. 5.14

1. $90°$
2. $30°$
3. $150°$
4. $60°$

## Solution

(a)

It is given that, $\text{POO}$ is a line. Since, sum of all the angles on a straight line is $180°$. Therefore, $x+2y+3y=180°$

$x+5y=180°$

$30+5y=180°$

$5y=180°-30$

$5y=150$

$y=\frac{150}{5}=30$

$\angle \text{QOR}=3y=3×30=90$

# Question: 15

The measure of an angle which is four times its supplement is

a.    $36°$

b.   $144°$

c.    $16°$

d.   $64°$

## Solution

(b)

Let the required angle be $x$.

Then, its supplement will be $\left(180°-x\right)$

It is given that, the angle is four times its supplement.

Therefore, $x=4\left(180°-x\right)$

$x=4×180°-4x$

$x+4x=720$

$5x=720$

$x=\frac{720}{5}=144$

Hence, the required angle is $144$.

# Question: 16

In Fig. 5.15, the value of $y$ is

Fig. 5.15

a.    $30°$

b.   $15°$

c.    $20°$

d.   $22.5°$

## Solution

(c)

Since, sum of all the angles on a straight line is $180°$.

Therefore, $6y+y+2y=180$

$9y=180$

$y=\frac{180}{9}=20$

# Question: 17

In Fig. 5.16, $\text{PA}\parallel \text{BC}\parallel \text{DT}$ and $\text{AB}\parallel \text{DC}$. Then, the values of $a$ and $b$ are respectively.

Fig. 5.16

a.

b.

c.

d.

## Solution

(b)

It is given that, $\text{PA}\parallel \text{BC}$ and  is transversal.

$\angle \text{PAB}=\angle \text{ABC}$

$50=a$

Also, $\text{AB}\parallel \text{DC}$ and $\text{BC}$ is transversal.

$\angle \text{ABC}+\angle \text{DCB}=180°$

$a+\angle \text{DCB}=180°$

$\angle \text{DCB}=180°-a$

$\angle \text{DCB}=180°-50$

$\angle \text{DCB}=130$

Also, $\text{BC}\parallel \text{DT}$ and $\text{DC}$ is transversal.

$\angle \text{CDT}=\angle \text{DCB}$

$b=130$

# Question: 18

The difference of two complementary angles is $30°.$ Then, the angles are,

1.

## Solution

(a)

Let one of the angle be $x$. Since, the difference between the two angles is $30°$, then the other angle will be $\left(x-30°\right)$.

Also, the two angles are complementary, so their sum is equal to $90°$.

$x+\left(x-30\right)=90$

$x+x-30=90$

$2x=90+30$

$2x=120$

$x=\frac{120}{2}=60$

Required angles are $60°$ and $\left(60-30\right)$ i.e., $60$ and $30$.

# Question: 19

In Fig. 5.17,  and . Then, angles $a$ and $b$ are respectively

Fig. 5.17

a.

b.

c.

d.

## Solution

(a)

Given  and $\text{PR}$ is transversal.

$\angle \text{QPR}=\angle \text{SRP}$

$a=20$

Also, $\text{SP}\parallel \text{RQ}$ and $\text{PR}$ is transversal.

$\angle \text{SPR}=\angle \text{QRP}$

$b=50$

# Question: 20

In Fig. 5.18, $a$ and $b$ are

Fig. 5.18

a.    alternate exterior angles

b.   corresponding angles

c.    alternate interior angles

d.   vertically opposite angles

## Solution

(c)

In the given figure, $a$ and $b$ are alternate interior angles as both lie on opposite sides of transverse line.

# Question: 21

If two supplementary angles are in the ratio $1:2$, then the bigger angle is

1. $120°$
2. $125°$
3. $110°$
4. $90°$

## Solution

(a)

It is given that the angles are in the ratio of $1:2$. Let the angles be $x$ and $2x$. Also, the two angles are supplementary, i.e. their sum is equal to $180°$.

$x+2x=180°$

$3x=180°$

$x=\frac{180°}{3}=60$

Hence, the required angles are $60$ & $2x=2×60$ i.e., $60$ & $120$.

Bigger of the two angles is $120$.

# Question: 22

In Fig. 5.19, $\angle \text{ROS}$ is a right angle and $\angle \text{POR}$ and $\angle \text{QOS}$ are in the ratio $1:5$. Then, $\angle \text{QOS}$ measures

Fig. 5.19

a.    $150°$

b.   $75°$

c.    $45°$

d.   $60°$

## Solution

(b)

Since $\angle \text{POR}$ and $\angle \text{QOS}$ are in the ratio $1:5$. Let the angles be and $x$ and $5x$ respectively.

We know that, the sum of angles forming linear pair is $180°$

$\angle \text{PQR}+\angle \text{ROS}+\angle \text{QOS}=180°$

$x+90+5x=180°$

$6x=180°-90$

$x=\frac{90}{6}=15$

$\angle \text{QOS}=5x=5×15=75$

# Question: 23

Statements $a$ and $b$ are as given below:

$a$ : If two lines intersect, then the vertically opposite angles are equal.

$b$ : If a transversal intersects two other lines, then the sum of two interior angles on the same side of the transversal is $180°$.

Then

a.    Both $a$ and $b$ are true

b.   $a$ is true and $b$ is false

c.    $a$ is false and $b$ is true

d.   both $a$ and $b$ are false

## Solution

(b)

Statement I

If lines $l$ & $m$ intersect each other, then $x$ & $y$ are known as vertically opposite angles so formed are equal.

$x$ = $y$

Statement II

If two lines $l$ & $m$ intersected by a transversal $p$, then the sum of two interior angles will be $180$, only if $l$ & $m$ are parallel.

# Question: 24

For Fig. 5.20, statements $p$ and $q$ are given below:

Fig. 5.20

$p$ : $a$ and $b$ are forming a linear pair.

$q$ : $a$ and $b$ are forming a pair of adjacent angles.

Then,

1. both $p$ and $q$ are true
2. $p$ is true and $q$ is false
3. $p$ is false and $q$ is true
4. both $p$ and $q$ are false

## Solution

(a)

Two angles are called adjacent angles, if they have a common vertex and a common arm but no common interior points. A linear pair is a pair of adjacent angles whose non-common sides are opposite rays.

$\therefore$$a$ and $b$ are pair of adjacent angles and form a linear pair.

# Question: 25

In Fig. 5.21, $\angle \text{ }\text{AOC}$ and $\angle \text{BOC}$ form a pair of

1. vertically opposite angles
2. complementary angles
3. alternate interior angles
4. supplementary angles

## Solution

(d)

Since, $\angle \text{ }\text{AOC}$ and $\angle \text{BOC}$ are on the same line $\text{AOB}$ and forming linear pair.

$\therefore \angle \text{AOC}+\angle \text{BOC}=180°$

Hence, $\angle \text{ }\text{AOC}$ and $\angle \text{BOC}$ are supplementary angles.

# Question: 26

In Fig. 5.22, the value of $a$ is

Fig. 5.22

1. $20°$
2. $15°$
3. $5°$
4. $10°$

## Solution

(d)

From the given figure, we can say that

$\angle \text{BOC}=\angle \text{EOF}$

$40=\angle \text{EOF}$

Since, sum of all angles on a straight line is $180$.

$\angle \text{AOE}+\angle \text{FOE}+\angle \text{EOD}=180°$

$90+40+5a=180°$

$130+5a=180°$

$5a=180°-130$

$a=\frac{50}{5}=10$

# Question: 27

In Fig. 5.23, if $\text{QP}\parallel \text{SR}$, the value of $a$ is

Fig. 5.23

1. $40°$
2. $30°$
3. $90°$
4. $80°$

## Solution

(c)

Draw a line $l$ parallel to $\text{QP}$.

$\angle \text{PQT}=x=60$  [Alternate interior angles]

$\angle \text{RST}=y=30$  [Alternate interior angles]

Now,

$a=x+y$

$a=60+30$

$a=90$

# Question: 28

In which of the following figures, $a$ and $b$ are forming a pair of adjacent angles?

a.

b.

c.

d.

## Solution

(d)

Two angles are called adjacent angles, if they have a common vertex and a common arm but no common interior points.

$\therefore$ In option (d), $a$ and $b$ form a pair of adjacent angles.

# Question: 29

In a pair of adjacent angles, (i) vertex is always common, (ii) one arm is always common, and (iii) uncommon arms are always opposite rays, Then

a.    All (i), (ii) and (iii) are true

b.   (iii) is false

c.    (i) is false but (ii) and (iii) are true

d.   (ii) is false

## Solution

(b)

Two angles are called adjacent angles, if they have a common vertex and a common arm but no common interior points. It is not necessary that uncommon arms must be always opposite rays.

# Question: 30

In Fig. 5.25, lines $\text{PQ}$ and $\text{ST}$ intersect at O. If $\angle \text{POR}=90°$ and $x:y=3:2$, then $z$ is equal to

Fig. 5.25

1. $126°$
2. $144°$
3. $136°$
4. $154°$

## Solution

(b)

Since,  and $\angle \text{TOQ}$ lies on a straight line, $\text{POQ}$ then their sum is equal to $180°$.

$\angle \text{POR}+\angle \text{ROT}+\angle \text{TOQ}=180°$

$90+x+y=180°$

$x+y=180°-90$

$x+y=90$

$x:y=3:2$  [given]

Let $x=3a$ and $y=2a$

$3a+2a=90$

$5a=90$

$a=\frac{90}{5}=18$

Now, $x=3a=3×18=54$ and $y=2a=2×18=36$

Since, $y$ and $z$ forms linear pair.

$y+z=180°$

$36+z=180°$

$z=180°-36$

$z=144$

# Question: 31

In Fig. 5.26, $\text{POQ}$ is a line, then $a$ is equal to

Fig. 5.26

1. $35°$
2. $100°$
3. $80°$
4. $135°$

## Solution

(c)

Since, $\text{POQ}$ is a line

Here, $\angle \text{POR}$ and $\angle \text{QOR}$ form a linear pair.

$\angle \text{POR}+\angle \text{QOR}=180°$

$100+a=180°$

$a=180°-100$

$a=80$

# Question: 32

Vertically opposite angles are always

1. supplementary
2. complementary
4. equal

## Solution

(d)

When two lines intersect, then vertically opposite angles so formed are equal.

# Question: 33

In Fig. 5.27, $a=40°$. The value of $b$ is

Fig. 5.27

1. $20°$
2. $24°$
3. $36°$
4. $120°$

## Solution

(a)

$2a+5b=180°$

$2×40+5b=180°$

$80+5b=180°$

$5b=180°-80$

$b=\frac{100}{5}=20$

# Question: 34

If an angle is $60°$ less than two times of its supplement, then the greater angle is

1. $100°$
2. $80°$
3. $60°$
4. $120°$

## Solution

(a)

Let the angle be $x$, then its supplement will be $\left(180°-x\right)$. Given, the angle $60$ less than $2$ times of its supplement.

Then, $2\left(180°-x\right)-x=60$

$360-2x-x=60$

$360-3x=60$

$360-60=3x$

$300=3x$

$x=\frac{300}{3}=100$

If $x=100$ then its supplement is $180°-x=180°-100=80$

So, the greater angle is $100$.

# Question: 35

In Fig. 5.28, $\text{PQ}\parallel \text{RS}$. If $\angle 1=\left(2a+b\right)°$ and $\angle 6=\left(3a-b\right)°$, then the measure of $\angle 2$ in terms of $b$ is

Fig. 5.28

a.    $\left(2+b\right)°$

b.   $\left(3+b\right)°$

c.    $\left(108-b\right)°$

d.   $\left(180-b\right)°$

## Solution

(c)

$\angle 1=\angle 5$

$\angle 5=\left(2a+b\right)$

Also, $\angle 5+\angle 6=180°$

$\left(2a+b\right)+\left(3a-b\right)=180°$

$\left(2a+3a\right)+\left(b-b\right)=180°$

$5a=180°$

$a=\frac{180}{5}=36$

Now, $\angle 1+\angle 2=180°$

$\angle 2=180°-\angle 1$

$\angle 2=180°-\left(2a+b\right)$

$\angle 2=180°-2a-b$

$\angle 2=180°-2×36-b$

$\angle 2=180°-72-b$

$\angle 2=\left(108°-b\right)$

# Question: 36

In Fig. 5.29, $\text{PQ}\parallel \text{RS}$ and $a:b=3:2$, then $f$ is equal to

Fig. 5.29

1. $36°$
2. $108°$
3. $72°$
4. $144°$

## Solution

(b)

We have $a:b=3:2$

Let $a=3x$ and $b=2x$

Since, $a$ and $b$ form a linear pair.

$a+b=180°$

$3x+2x=180°$

$5x=180°$

$x=\frac{180}{5}=36$

$a=3x=3×36=108$

$f=a$   [Corresponding angles]

$f=108$

# Question: 37

In Fig. 5.30, line $l$ intersects two parallel lines $\text{PQ}$ and $\text{RS}$. Then, which one of the following is not true?

Fig. 5.30

a.    $\angle 1=\angle 3$

b.   $\angle 2=\angle 4$

c.    $\angle 6=\angle 7$

d.   $\angle 4=\angle 8$

## Solution

(d)

$\text{PQ}\parallel \text{RS}$ and $l$ is transversal.

$\angle 1=\angle 3$ [Corresponding angles]

$\angle 2=\angle 4$     [Corresponding angles]..(i)

$\angle 5=\angle 6$     [Vertically opposite angles]..(ii)

$\angle 5=\angle 7$     [Corresponding angles]..(iii)

$\angle 6=\angle 7$     [From eq. (ii) and (iii)]

$\angle 2+\angle 8=180°$ [linear pair]

$\angle 4+\angle 8=180°$  [From eq. (i)]

Therefore, $\angle 4\ne \angle 8$.

# Question: 38

In above Fig. 5.30, which one of the following is not true?

a.    $\angle 1+\angle 5=180°$

b.   $\angle 2+\angle 5=180°$

c.    $\angle 3+\angle 8=180°$

d.   $\angle 2+\angle 3=180°$

## Solution

(d)

From the above Fig. 5.30, $\angle 2$ and $\angle 3$ are alternate interior angles. Hence, $\angle 2=\angle 3$

# Question: 39

In Fig. 5.30, which of the following is true?

a.    $\angle 1=\angle 5$

b.   $\angle 4=\angle 8$

c.    $\angle 5=\angle 8$

d.   $\angle 3=\angle 7$

## Solution

(c)

From the above figure, $\angle 5$ and $\angle 8$ are alternate interior angles. Hence, $\angle 5=\angle 8$

# Question: 40

In Fig. 5.31, $\text{PQ}\parallel \text{ST}$. Then, the value of $x+y$

Fig. 5.31

a.    $125°$

b.   $135°$

c.    $145°$

d.   $120°$

## Solution

(b)

Since, $\text{PQ}\parallel \text{ST}$, then $\text{PO}$ will also parallel to $\text{ST}$. Now, $\text{PO}\parallel \text{ST}$ and $\text{OS}$ is transversal

Therefore, $x=85$

Now,   $y+130=180°$

$y=180°-130$

$y=50$

$x+y=85+50=135$

# Question: 41

In Fig. 5.32, if $\text{PQ}\parallel \text{RS}$ and $\text{QR}\parallel \text{TS}$, then the value $a$ is

Fig. 5.32

a.    $95°$

b.   $90°$

c.    $85°$

d.   $75°$

## Solution

(a)

Since, $\text{PQ}\parallel \text{RS}$ and $\text{QR}$ is transversal.

$\angle \text{PQR}=\angle \text{SRQ}$      [alternate interior angles]

$\angle \text{SRQ}=85$

Also, $\text{ST}\parallel \text{QR}$ and $\text{RS}$ is transversal.

$\angle \text{SRQ}=\angle \text{RST}$      [alternate interior angles]

$\angle \text{RST}=85$

Now, $\angle \text{RST}+a=180°$ [Linear pair]

$a=180°-85$

$a=95$

In the Questions 42 to 56, fill in the blanks to make the statements true.

# Question: 42

If sum of measures of two angles is $90°$, then the angles are ____________.

## Solution

Complementary

The sum of two complementary angles is $90°$.

# Question: 43

If the sum of measures of two angles is $180°$, then they are _________.

## Solution

Supplementary

The sum of two supplementary angles is $180°$.

# Question: 44

A transversal intersects two or more than two lines at _________ points.

## Solution

Distinct

A transversal intersects two or more than two lines at distinct points.

If a transversal intersects two parallel lines, then (Q. 45 to 48)

# Question: 45

Sum of interior angles on the same side of a transversal is ____________.

## Solution

$180°$

Sum of interior angles on the same side of a transversal is

In the above figure, $x+y=180°$

# Question: 46

Alternate interior angles have one common ________.

## Solution

Arm

Two alternate interior angles have one common arm.

# Question: 47

Corresponding angles are on the ___________ side of the transversal.

## Solution

Same

Two corresponding angles are of the same side of the transversal.

# Question: 48

Alternate interior angles are on the __________ side of the transversal.

## Solution

Opposite

Two alternate interior angles are on the opposite side of the transversal.

# Question: 49

Two lines in a plane which do not meet at a point anywhere are called ______________ lines.

## Solution

Parallel

If two lines are parallel, then they will never meet each other.

# Question: 50

Two angles forming a __________ pair are supplementary.

## Solution

Linear

If two angles form a linear pair, then their sum will be $180°$. Hence, they are supplementary.

# Question: 51

The supplement of an acute is always __________ angle.

## Solution

Obtuse

If angle is acute angle, then its supplement will be an obtuse angle. As, if we subtract an angle which is less than $90°$ from $180°$, then result will be an angle greater than $90°$.

# Question: 52

The supplement of a right angle is always _________ angle.

## Solution

Right

Let $x$ be the supplement of the right angle. Then

$x+90°=180°⇒x=180°-90°=90°$.

# Question: 53

The supplement of an obtuse angle is always _________ angle.

## Solution

Acute

The supplement of an obtuse angle is always an acute angle. As, if we subtract an obtuse angle from $180°$, then result will be an acute angle, i.e. less than $90°$.

# Question: 54

In a pair of complementary angles, each angle cannot be more than _________.

## Solution

$90°$

Two angles are said to be complementary angles, if their sum is $90°$. Hence, if two angles are complementary, then each angle cannot be more than $90°$.

# Question: 55

## Solution

$45°$

Let $x$ be the required angle.

Then, $x+45°=90°⇒x=90°-45°=45°$

# Question: 56

An angle which is half of its supplement is of __________.

## Solution

$60°$

Let the required angle be $x$.

Then, its supplement will be $\left(180°-x\right)$.

It is given that $x$ is the half of it supplement i.e. $\left(180°-x\right)$.

Therefore, $x=\frac{1}{2}\left(180°-x\right)$.

$2x=180°-x$

$2x+x=180°$

$3x=180°$

$x=\frac{180}{3}=60$

In the Questions 57 to 71, State whether the statements are True or False.

# Question: 57

Two right angles are complementary to each other.

## Solution

False

Measure of right angle is $90°$. So, the sum of two right angles $=90°+90°=180°$.

Complementary angles are those whose sum is equal to $90°$. Hence, two right angles are never be complementary.

# Question: 58

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

## Solution

False

Since, sum of two complementary angles is $90°$, so sum of one obtuse and one acute angles cannot make a pair of complementary angles as obtuse angle is greater than $90°$.

# Question: 59

Two supplementary angles are always obtuse angles.

## Solution

False

If two angles are supplementary angles, then it is impossible that both of them are obtuse angles. e.g. $60°$ and $120°$ are supplementary angles but both are not obtuse.

# Question: 60

Two right angles are always supplementary to each other.

## Solution

True

Measure of a right angle is $90°$. Then, sum of two right angles will be $180°$. So, two right angles are always supplementary to each other.

# Question: 61

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

## Solution

True

One obtuse angle and one acute angle can make a pair of supplementary angles, e.g. $60°$ and $120°$ are supplementary angles. So, one is $60°$ i.e. acute angle and other is $120°$, i.e. obtuse angle.

# Question: 62

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

## Solution

True

Acute angles are those which are less than $90°$. Both angles of a pair of supplementary angles can never be acute.

# Question: 63

Two supplementary angles always form a linear pair.

## Solution

False

Linear pair is always in a straight line.

# Question: 64

Two angles making a linear pair are always supplementary.

## Solution

True

Because linear pair is always in a straight line and straight lines make $180°$ angle.

# Question: 65

Two angles making a linear pair are always adjacent angles.

## Solution

True

e.g. From the above figure, $\angle 1$ and $\angle 2$ form a linear pair and are adjacent angles.

# Question: 66

Vertically opposite angles form a linear pair.

## Solution

False

Two angles making a linear pair are always adjacent angles.

# Question: 67

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

## Solution

False

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

# Question: 68

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

## Solution

True

Vertically opposite angles are equal. So, if one angle is acute, then other angle will be acute and if one angle is obtuse, then the other will be obtuse.

# Question: 69

A linear pair may have two acute angles.

## Solution

False

A linear pair either have both right angles or one acute and one obtuse angle, because angles forming linear pair is $180°$.

# Question: 70

An angle is more than $45°$. Its complementary angle must be less than $45°$.

## Solution

True

e.g. Let one angle $=50°$

$\therefore$ The other angle $=90-50°=40°<45°$.

# Question: 71

Two adjacent angles always form a linear pair.

## Solution

False

Two adjacent angles do not always form a linear pair, but the angles forming linear pair are always adjacent angles.

# Question: 72

Write down each pair of adjacent angles shown in the following figures:

a.

## Solution

Two angles are called adjacent angles, if they have a common vertex and a common arm but no common interior points. Hence, following are adjacent angles:

(i)       (a)

(b)

(c)

(c)

(ii)    (a)

(b)

(c)

(iii) (a)

(b)

(iv) (a)

(b)

(c)

(c)

# Question: 73

In each of the following figures, write, if any, (i) each pair of vertically opposite angles, and (ii) each linear pair.

a.

b.

c.

d.

## Solution

Vertically opposite angles are the angles, opposite to each other when two lines cross, A linear pair is a pair of adjacent angles whose non-common sides are opposite rays. Following are vertically opposite angles and linear pair in the above figure:

b.

c.

d.

 Fig. Vertically opposite angles Linear pair (a) (b) Nil (c) Nil Nil (d)

# Question: 74

Name the pairs of supplementary angles in the following figures:

a.

b.

c.

## Solution

When the sum of the measures of two angles is $180°$, the angles are called supplementary angles. Linear pair angles are supplementary angles as their sum is $180°$. Following are the pairs of supplementary angles in the above figures:

 Fig. Pair of supplementary angles (i) (ii) (iii)

# Question: 75

In Fig 5.36,  and $\angle \text{PTR}=42°$. Find $\angle \text{QUR}$.

Fig. 5.36

## Solution

Since, $\text{PQ}$ and $\text{RS}$ are parallel and $\text{TR}$ is Transversal.

Therefore, $\angle \text{PTR}=\angle \text{TRU}$ [Alternate interior angles]

$\angle \text{TRU}=42$

Now, $\text{TR}$ is parallel to $\text{QU}$ and $\text{RS}$ is transversal.

Therefore, $\angle \text{TRU}+\angle \text{RUQ}=180°$        [consecutive interior angles]

$42+\angle \text{RUQ}=180°$

$\angle \text{RUQ}=180°-42$

$\angle \text{RUQ}=138$

# Question: 76

The drawings below (Fig. 5.37), show angles formed by the goalposts at different positions of a football player. The greater the angle, the better chance the player has of scoring a goal. For example, the player has a better chance of scoring a goal from Position A than from Position B.

In Parts (a) and (b) given below it may help to trace the diagrams and draw and measure angles.

a.    Seven football players are practicing their kicks. They are lined up in a straight line in front of the goalpost [Fig.(ii)]. Which player has the best (the greatest) kicking angle?

b.   Now the players are lined up as shown in Fig. (iii). Which player has the best kicking angle?

c.    Estimate at least two situations such that the angles formed by different positions of two players are complement to each other.

(i)

(ii)

(iii)

## Solution

a.

Since, angle made by $4$ is greatest. Hence, he has the best kicking angle.

b.

From the above figure, we can say that player $4$ has the best kicking angle, as it is greatest.

c.    Since, the angles are complementary. Hence, two situations are $45°,$ $45°$ and $30°,$ $60°.$

# Question: 77

The sum of two vertically opposite angles is $166°.$ Find each of the angles.

## Solution

When two lines intersect, then vertically opposite angles so formed are equal. Let $x$ be the measure of each vertically opposite angles

Then, $x+x=166$

$2x=166$

$x=\frac{166}{2}=83$

So, the measure if each angle is $83$.

# Question: 78

In Fig. 5.38, $l\parallel m\parallel n$. $\angle \text{QPS}=35°$ and $\angle \text{QRT}=55°$ find $\angle \text{PQR}$.

Fig. 5.38

## Solution

From the given figure,

$\angle 1=35$ [Alternate interior angles]

$\angle 2=55$ [Alternate interior angles]

$\angle \text{PQR}=\angle 1+\angle 2=35+55=90$

# Question: 79

In Fig. 5.39, $\text{P}$, $\text{Q}$ and $\text{R}$ are collinear points and $\text{TQ}\perp \text{PR}$, Name:

a.    pair of complementary angles

b.   two pairs of supplementary angles.

c.    four pairs of adjacent angles.

Fig. 5.39

## Solution

a.    Complementary angles are those whose sum is $90°$.

$\therefore \angle \text{TQS}$ and $\angle \text{SQR}$ are pair of complementary angles, as their sum is $90°$.

b.   Supplementary angles are those whose sum is $180°$.

are pair so supplementary angles.

c.    Two angles are called adjacent angles, if they have a common vertex and a common arm but no common interior points.

# Question: 80

In Fig. 5.40, $\text{OR}\perp \text{OP}$,

Fig. 5.40

(i)       Name all the pairs of adjacent angles.

(ii)    Name all the pairs of complementary angles.

## Solution

By definition of adjacent angles and complementary angles, we can say that following pairs are adjacent angles and complementary angles.

Adjacent angles: $\angle x,\text{\hspace{0.17em}}\angle y;\text{\hspace{0.17em}}\angle x+\angle y,\text{\hspace{0.17em}}\angle z;\text{\hspace{0.17em}}\angle y,\text{\hspace{0.17em}}\angle z;\text{\hspace{0.17em}}\angle x,\text{\hspace{0.17em}}\angle y+\angle z.$

Complementary angles:

# Question: 81

If two angles have a common vertex and their arms form opposite rays (Fig. 5.41), Then,

Fig. 5.41

a.    How many angles are formed?

b.   How many types of angles are formed?

c.    Write all the pairs of vertically opposite angles.

## Solution

a.    Total 13 angles are formed, namely $\angle \text{AOB},$

b.   Following types of angles are formed:

(i)            Linear pair

(ii)         Supplementary

(iii)      Vertically opposite

c.    Following are the pair of vertically opposite angles:

.

# Question: 82

(i)

(ii)

(iii)

(iv)

## Solution

Two angles are called adjacent angles, if they have a common vertex and a common arm but no common interior points.

Hence, $a$ and $b$ form a pair of adjacent angle only in (i).

# Question: 83

In Fig. 5.43, write all the pairs of supplementary angles.

Fig. 5.43

## Solution

Supplementary angles are those angles whose sum is $180°$. Hence, following are the pairs of supplementary angles:

1.

2.

3.

4.

5.

6.

# Question: 84

What is the type of other angle of a linear pair if

a.    One of its angles is acute?

b.   One of its angles is obtuse?

c.    One of its angles is right?

## Solution

Sum of angles of linear pair is $180°$.

a.    If one angle is acute angle, then other angle will be obtuse. As, if we subtract an acute angle from $180°$, we get an angle which is greater than $90°$.

b.   If one angle is obtuse angle, then other angle will be acute. As, if we subtract an obtuse angle from $180°$, we get an angle which is less than $90°$.

c.    If one angle is right angle, then other angle will also be right angle. As, if we subtract $90°$ from $180°$, we get $90°$.

# Question: 85

## Solution

Acute angles are those angles which are less than $90°$. If we add two angles which are less than $90°$, we get the result less than $180°$, e.g. If we add $60°$ and $70°$, we get $60°+70°=130°<180°$ hence, two acute angles cannot form a pair of supplementary angles.

# Question: 86

Two lines $\text{AB}$ and $\text{CD}$ intersect at $\text{O}$ (Fig. 5.44). Write all the pairs of adjacent angles by taking angles  and $4$ only.

Fig. 5.44

## Solution

Two angles are called adjacent angles, if they have a common vertex and a common arm, but no common interior points.

Hence, following are the pairs of adjacent angles taking  angles only, i.e.

# Question: 87

If the complement of an angle is $62°$, then find its supplement.

## Solution

Let the angle be $x°$. We know that, sum of two complementary angles is $90°$

$x+62=90$

$x=62-90=28$

Supplement of any angle is $\left(180°-\text{angle}\right)$

Supplement of $x=180°-28=152°$

# Question: 88

A road crosses a railway line at an angle of $30°$ as shown in Fig.5.45. Find the values of $a,\text{\hspace{0.17em}}b$ and $c$.

Fig. 5.45

## Solution

Lines $l$ and $m$ are parallel, $\text{P}$ is transversal and $x=30$.

Therefore, $y=30$                 [corresponding angles]

Now, $c+y=180°$               [linear pair]

$c+30=180°$

$c=180°-30=150$              [linear pair]

Now, $\angle 1+c=180°$

$\angle 1+150=180°$

$\angle 1=180°-150=30$

$\angle 1=a$                                 [corresponding angles]

$a=30$                                  [corresponding angles]

Also, $\angle 2+a=180°$            [linear pair]

$\angle 2+30=180°$

$\angle 2=180°-30=150$

Also, $\angle 2=b$                        [corresponding angles]

$b=150$

Hence, $a=30$, $b=150$, $c=150$

# Question: 89

The legs of a stool make an angle of $35°$ with the floor as shown in Fig. 5.46. Find the angles $x$ and $y$.

Fig. 5.46

## Solution

Since $l\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}m$ are parallel lines and $\text{PQ}$ is transversal.

$x=\angle \text{PQR}$                           [alternate interior angles]

$x=35$

$x+y=180°$

$35+y=180°$

$y=180°-35$

$y=145$

# Question: 90

Iron rods  and $f$ are making a design in a bridge as shown in Fig. 5.47, in which  $e\parallel f$. Find the marked angles between

Fig. 5.47

(i)       $b$ and $c$

(ii)    $d$ and $e$

(iii) $d$ and $f$

(iv) $c$ and $f$

## Solution

Since $l,m$ are parallel lines and  and $\text{TU}$ are transversal.

Then, $\angle 4=\angle \text{QPS}$        [Alternate interior angles]

$\angle 4=75$

$\angle 1=\angle \text{QOR}$                  [Vertically opposite angles]

$\angle 1=30$

Also,  and $\text{TU}$ are parallel and $m$ and $l$ are transversal.

Therefore, $\angle 2+\angle \text{QPT}=180°$         [Linear pair]

$\angle 2=180°-75=105$

Also, $\angle 2+\angle 3=180°$

$\angle 3=180°-105$

$\angle 3=75$

Hence, we have,

(i)       $30$

(ii)    $105$

(iii) $75$

(iv) $75$

# Question: 91

Amisha makes a star with the help of line segments  and $f$, in which  and $c\parallel f$. Chhaya marks an angle as $120°$ as shown in Fig. 5.48 and asks Amisha to find the  and $\angle z$. Help Amisha in finding the angles.

Fig. 5.48

## Solution

From the given figure, we have

$\angle p=120$                [vertically opposite angles]

$\angle x+\angle p=180°$

$\angle x=180°-120$

Again, $\angle x=\angle 1$     [Alternate interior angles]

$\therefore \angle 1=60$

$\angle 1+\angle y=180°$      [Linear pair]

$\angle y=180°-60$

$\angle y=120$

$\angle z+\angle a=80$         [Linear pair]

$\angle z=180°-120$

# Question: 92

In Fig. 5.49,  and $\angle \text{FED}=42°$. Find $\angle \text{EFD}$.

Fig. 5.49

## Solution

$\text{AF}$ and $\text{ED}$ are parallel and $\text{EF}$ is transversal.

Then, $\angle \text{AFE}=\angle \text{FED}$     [Alternate interior angles]

$\angle \text{AFE}=42$

$\angle \text{AFC}+\angle \text{AFE}+\angle \text{EFD}=180°$

$68+42+\angle \text{EFD}=180°$

$110+\angle \text{EFD}=180°$

$\angle \text{EFD}=180°-110=70$

# Question: 93

In Fig. 5.50, $\text{OB}$ is perpendicular to $\text{OA}$ and $\angle \text{BOC}=49°$. Find $\angle \text{AOD}$.

Fig. 5.50

## Solution

From the given figure,

$\angle \text{DOB}+\angle \text{BOC}=180°$        [Linear pair]

$\angle \text{DOB}=180°-49$

$\angle \text{DOB}=131$

$\angle \text{DOB}+\angle \text{BOA}+\angle \text{AOD}=360$ [Complete angle]

$131+90+\angle \text{AOD}=360$

$\angle \text{AOD}=360-221=139$

# Question: 94

Three lines  and $\text{EF}$ intersect each other at $\text{O}$. If $\angle \text{AOE}=30°$ and $\angle \text{DOB}=40°$ (Fig. 5.51), find $\angle \text{COF}$.

Fig. 5.51

## Solution

$\angle \text{AOE}+\angle \text{EOD}+\angle \text{DOB}=180$      [Linear pair]

$30+\angle \text{EOD}+40=180°$

$\angle \text{EOD}=180°-70=110$

Again, $\angle \text{EOD}=\angle \text{COF}$     [Vertically opposite angles]

$\angle \text{COF}=110$

# Question: 95

Measures (in degrees) of two complementary angles are two consecutive even integers. Find the angles.

## Solution

Let the two consecutive angles be $x$ and $x+2$. Since, both angles are complementary. So, their sum will be $90°$

$x+\left(x+2\right)=90°$

$x+\left(x+2\right)=90°$

$2x=90°-2$

$2x=88$

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

Therefore, the angles are $44°$ and $44°+2=46°$.

# Question: 96

If a transversal intersects two parallel lines, and the difference of two interior angles on the same side of a transversal is $20°$, find the angles.

## Solution

Let the two interior angles on the same side of transversal be $x$ and $y$. Given, their difference is $20°$.

$x-y=20$

$y=x-20$ .(i)

Since, $l$ and $m$ are parallel and $\text{P}$ is transversal.

Then,  $x+y=180°$

$x+x-20=180°$       [From (i)]

$2x=180°+20$

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

$y=x-20$

$y=100-20=80°$

Therefore, the angles are $100°$ and $80°$, respectively.

# Question: 97

## Solution

Let one angle be $x$. It is given that other angle is one-third of first.

So, other angle will be $\frac{1}{3}x$.

Again, both the angles are making a linear pair.

So, their sum will be $180°$.

$x+\frac{1}{3}x=180°$

$\frac{3x+x}{3}=180°$

$\frac{4x}{3}=180°$

$x=\frac{180×3}{4}$

$x=135$

Hence, the angles are $135$ and $\frac{1}{3}×135$ i.e., $135$ and $45$.

# Question: 98

Measures (in degrees) of two supplementary angles are two consecutive odd integers. Find the angles.

## Solution

Let two consecutive odd integers $x$, $x+2$. It is given that both are supplementary angles. So, their sum will be $180°$

$x+\left(x+2\right)=180°$

$2x=180°-2$

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

$x=89$

Hence, the two angles are $89$ and $91$.

# Question: 99

In Fig. 5.52, $\text{AE}\parallel \text{GF}\parallel \text{BD,}$ $\text{AB}\parallel \text{CG}\parallel \text{DF}$ and $\angle \text{CHF}=120°$. Find $\angle \text{ABC}$ and $\angle \text{CDE}$.

Fig. 5.52

## Solution

Since,  and $\text{CG}$ is transversal.

Therefore, $\angle \text{BCH}=\angle \text{EHC}$ [Alternate interior angles]

$\angle \text{BCH}=120$

Again,  and $\text{BD}$ is transversal.

Therefore, $\angle \text{BCH}=\angle \text{CDE}$     [Corresponding angles]

$\angle \text{CDE}=120$

Also,  and $\text{BC}$ is transversal.

Therefore, $\angle \text{ABC}+\angle \text{BCH}=180°$ [Interior angles on same side of transversal]

$\angle \text{ABC}=180-120$

$\angle \text{ABC}=60$.

# Question: 100

In Fig. 5.53, find the value of $\angle \text{BOC}$, if points $\text{A}$, $\text{O}$ and $\text{B}$ are collinear.

Fig. 5.53

## Solution

Since,  and $\text{B}$ are collinear. Then, $\text{AOB}$ will be a straight line and sum of all the angles on a straight line is $180°$.

$\angle \text{AOD}+\angle \text{DOC}+\angle \text{COB}=180°$

$\left(x-10\right)+\left(4x-25\right)+\left(x+5\right)=180°$

$x-10+4x-25+x+5=180°$

$6x-30=180°$

$6x=180°+30$

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

$\angle \text{BOC}=\left(x+5\right)$

$=\left(35+5\right)=40$

# Question: 101

In Fig. 5.54, if $l\parallel m$ find the value of $a$ and $b$.

Fig. 5.54

## Solution

Since, $l,m$ are parallel lines and $t$ is transversal.

Therefore, $\angle \text{EAB}+\angle \text{DBA}=180°$ [Interior angles on same side of transversal]

$b+132=180°$

$b=180°-132$

$b=48$

Again, $l,m$ are parallel lines and $t$ is transversal.

$\angle \text{EAC}+\angle \text{BCA}=180°$ [Interior angles on same side of transversal]

$a+b+65=180°$

$a+48+65=180°$

$a=180°-48-65$

$a=67$

# Question: 102

In Fig. 5.55, if $l\parallel m$ and a line $t$ intersects these lines at $\text{P}$ and $\text{Q}$, respectively. Find the sum $2a+b$.

Fig. 5.55

## Solution

From the figure,

$\angle \text{AQB}=\angle \text{FQP}$

$b=132$

Since, $l,m$ are parallel lines and $t$ is transversal.

$\angle \text{EPD}=\angle \text{PQF}$  [Corresponding angles]

$\therefore a=132$

$b=132$              [Vertically opposite angles]

Now, $2a+b=2×132+132$

$=264+132=396$

# Question: 103

In Fig. 5.56, if $\text{QP}\parallel \text{RS}$. Find the values of $a$ and $b$.

Fig. 5.56

## Solution

Since, $\text{QP}\parallel \text{RS}$ and $\text{PR}$ is transversal.

Therefore, $\angle \text{QPR}=\angle \text{SRP}$ [Alternate interior angles]

$a=65$

Also, $\angle \text{SRT}=\angle \text{PQR}$         [Corresponding angles]

$b=70$

# Question: 104

In Fig. 5.57, $\text{PQ}\parallel \text{RT}$. Find the values of $a+b$.

Fig. 5.57

## Solution

Since, $\text{PQ}\parallel \text{RT}$ and $\text{RQ}$ is transversal.

Therefore, $\angle \text{TRQ}=\angle \text{RQP}$     [Alternate interior angles]

$b=55$

$\angle \text{SRT}=\angle \text{SPQ}$             [Corresponding angles]

$a=45$

$a+b=45+55=100$

# Question: 105

In Fig. 5.58,  and $\text{UT}$ are parallel lines

Fig. 5.58

(i)       if $c=57°$ and $a=\frac{c}{3}$, find the value of $d$.

(ii)    if $c=75°$ and $a=\frac{2}{5}c$, find $b$.

## Solution

(i)       Since, $\text{PQ}\parallel \text{UT}$ and $\text{PT}$ is transversal

$\angle \text{QPT}=\angle \text{UTP}$

$a+b=c$

$\frac{c}{3}+b=c$

$b=c-\frac{c}{3}$

$b=\frac{3c-c}{3}$

$b=\frac{2c}{3}=\frac{2}{3}×57$

$b=38$

$\text{PQ}\parallel \text{RS}$ and $\text{PR}$ is transversal

Therefore, $\angle \text{QPR}+\angle \text{PRS}=180°$

$b+d=180°$

$d=180°-b$

$d=180°-38=142$

(ii)    Since, $\text{PQ}\parallel \text{UT}$ and $\text{PT}$ is transversal

Therefore, $\angle \text{QPT}=\angle \text{UTP}$

$a+b=c$

$b=c-a$

$b=c-\frac{2}{5}c$

$b=\frac{5c-2c}{5}$

$b=\frac{3c}{5}$

$b=\frac{3×75}{5}=45$

# Question: 106

In Fig. 5.59, $\text{AB}\parallel \text{CD}$. Find the reflex $\angle \text{EFG}$.

Fig. 5.59

## Solution

Construct a line $l$ parallel to $\text{AB}$, passing through $\text{F}$. $l$ is parallel to both $\text{AB}$ and $\text{CD}$.

Then, $\angle 1=34$     [Alternate interior angles]

And $\angle 2+135=180°$    [Interior angles on same side of transversal]

$\angle 2=180°-135=45$

Now, $\angle \text{EFG}=\angle 1+\angle 2$

$\angle \text{EFG}=34+45$

$\angle \text{EFG}=79$

Reflex of $\angle \text{EFG}=360-\angle \text{EFG}$

$=360-79=281$

# Question: 107

In Fig. 5.60, two parallel lines $l$ and $m$ are cut by two transversals $n$ and $p$. Find the values of $x$ and $y$.

Fig. 5.60

## Solution

Since, lines $l$ and $m$ are parallel $p$ is transversal.

Therefore, $x+66=180°$ [Interior angles on same side of transversal]

$x=180°-66$

$x=114$

Again, lines $l$, $m$ are parallel $n$ is transversal.

Therefore, $y+48=180°$ [Interior angles on same side of transversal]

$y=180°-48$

$y=132$

# Question: 108

In Fig. 5.61, $l$, $m$ are $n$ are parallel lines, and the lines $p$ and $q$ are also parallel. Find the values of  and $c$.

Fig. 5.61

## Solution

Since, lines $l$, $n$ are parallel and $q$ is transversal.

Therefore, $6a=120$  [Corresponding angles]

$a=\frac{120}{6}=20$

Also, lines $p$, $q$ are parallel and $n$ is transversal.

Therefore, $4c=120$

$c=\frac{120}{4}=30$              [Corresponding angles]

Again, lines $m$, $n$ are parallel and $p$ is transversal.

Therefore, $4c=3b$    [Corresponding angles]

$b=\frac{4c}{3}$

$b=\frac{4×30}{3}$

$b=40$

# Question: 109

In Fig. 5.62, state which pair of lines are parallel. Give reason.

Fig. 5.62

## Solution

$x=120$         [Vertically opposite angles]

Now, $x+60=120+60=180°$

Since, the sum of consecutive interior angles is $180°$. Hence, $m$ and $n$ will be parallel.

# Question: 110

In Fig. 5.63, examine whether the following pairs of lines are parallel or not:

Fig. 5.63

## Solution

From the given figure, $x=65$ [Vertically opposite angles]

And $y=180°-70$            [Linear pair]

$y=110$

(i)       now, $x+y=65+110=175\ne 180°$

Hence,  $\text{EF}$ and $\text{GH}$ are not parallel.

(ii)    Also, $x+115=65+115=180°$

Hence, $\text{AB}$ and $\text{CD}$ are parallel.

# Question: 111

In Fig. 5.64, find out which pair of lines are parallel:

Fig. 5.64

## Solution

$\angle 1=180°-123$

$\angle 1=57$

Also, $\angle 2=180-57$

$\angle 2=123$

Now, $\angle 1+\angle 2=57+123=180°$

If sum of the consecutive angles is $180°$, then the lines are parallel.

So, $\text{EF}\parallel \text{GH}$

Now, $\angle 3=180°-\angle 2$   [Linear pair]

$\angle 3=180°-123=57$

Also, $\angle 4=180°-55=125$   [Linear pair]

$\angle 3+\angle 4=57+125=182\ne 180°$

Go, $\text{GH}$ and $\text{KP}$ are not parallel.

Also, $\angle 3+122=57+122=179\ne 180°$

Hence, $\text{AB}$ and $\text{CD}$ are not parallel.

# Question: 112

In Fig. 5.65, show that

Fig. 5.65

(i)       $\text{AB}\parallel \text{CD}$

(ii)    $\text{EF}\parallel \text{GH}$

## Solution

(i)       $\angle \text{CIF+}\angle \text{FIJ}=180°$        [Linear pair]

$\angle \text{CIF}=180°-\angle \text{FIJ}=180°-50=130$

Now, $\angle \text{ALI}=\angle \text{CIF}=130$

$\text{AB}\parallel \text{CD}$ as their corresponding angles are equal.

(ii)    $\angle \text{GKL+}\angle \text{LKJ}=180°$        [Linear pair]

$\angle \text{LKJ}=180°-\angle \text{GKL}=180°-50=130$

Now, $\angle \text{ALI}=\angle \text{LKJ}=130°$

$\text{EF}\parallel \text{GH}$ as their corresponding angles are equal.

# Question: 113

In Fig. 5.66, two parallel lines $l$ and $m$ are cut by two transversals $p$ and $q$. Determine the values of $x$ and $y$.

Fig. 5.66

## Solution

$x=110$ [Alternate interior angles]

$y+80°=180°$

$y=180°-80$

$y=100$