Read RD Sharma Solutions Class 9 Chapter 8 Lines and Angles below, students should study RD Sharma class 9 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 9 Mathematics have been prepared by teacher of Grade 9. These RD Sharma class 9 Solutions have been designed as per the latest NCERT syllabus for class 9 and if practiced thoroughly can help you to score good marks in standard 9 Mathematics class tests and examinations

**Exercise 8.1**

**Question 1: Write the complement of each of the following angles:**

**(i) 20°**

**(ii) 35°**

**(iii) 90°**

**(iv) 77°**

**(v) 30°**

**Solution:**

(i) Sum of complement angles = 90°

First angle = 20°

The other angle is = 90° – 20° = 70°

Hence, the other complement angle is 70°.

(ii) Sum of complement angles = 90°

First angle = 35°

The other angle is = 90° – 35° = 55°

Hence, the other complement angle is 55°.

(iii) Sum of complement angles = 90°

First angle = 90°

The other angle is = 90° – 90° = 0°

Hence, the other complement angle is 0°.

(iv) Sum of complement angles = 90°

First angle = 77°

The other angle is = 90° – 77° = 13°

Hence, the other complement angle is 13°.

(v) Sum of complement angles = 90°

First angle = 30°

The other angle is = 90° – 30° = 60°

Hence, the other complement angle is 60°.

**Question 2 : Write the supplement of each of the following angles:**

**(i) 54°**

**(ii) 132°**

**(iii) 138°**

**Solution:**

(i) Sum of supplement angles = 180°.

First angle = 54°

The other angle is = 180° – 54° = 126°

Hence, the other supplementary angle is 126°.

(ii) Sum of supplement angles = 180°.

First angle = 132°

The other angle is = 180° – 132° = 48°

Hence, the other supplementary angle is 48°.

(iii) Sum of supplement angles = 180°.

First angle = 138°

The other angle is = 180° – 138° = 42°

Hence, the other supplementary angle is 42°.

**Question 3: If an angle is 28° less than its complement, find its measure?**

**Solution:**

Let any angle of complementary angles be x°.

The other complementary angle will be (x+28)°

The sum of complementary angle will be 90°

90° = x+(x+28)°

90° = x+x+28°

90°-28° = x+x

62° = 2x

(62°)/2 = x

31° = x

Thus, the angle measured is 31°.

**Question 4: If an angle is 30° more than one half of its complement, find the measure of the angle?**

**Solution:**

Let any angle of complementary angles be x°.

The other complementary angle will be x/2+30°

The sum of complementary angle will be 90°

90° = x+(x/2+30)°

90° = x+x/2+30°

90°-30° = x+x/2

60° = (2x+x)/2

60×2=3x

(120°)/3=x

40°=x

Other angle = 40/2+30

Other angle = 40/2+30

Other angle = 20+30

Other angle = 50°

Thus, the measure of the angle is 50°.

**Question 5: Two supplementary angles are in the ratio 4:5. Find the angles?**

**Solution:**

The sum of supplementary angles will be180°.

It is given that the two supplementary angles are in the ratio 4:5.

Sum of the Ratios = 4 + 5 = 9

First angle = 4/9×180=80°

Second angle = 5/9×180=100°

Thus, the required angles are 80° and 100°.

**Question 6: Two supplementary angles differ by 48°. Find the angles?**

**Solution:**

Let the first angle of supplementary angle be x°.

Second angle of supplementary angle be (180 - x°).

It is given that the difference of two supplementary angles is 48°.

x-(180-x)=48

x-180+x=48

2x=48-180

2x=228

x=114

First angle will be 114°

Second angle will be 180°-114°

Second angle will be 66°

Thus, the two supplementary angles are 66° and 114°.

**Question 7: An angle is equal to 8 times its complement. Determine its measure?**

**Solution:**

Let the first angle of complementary angle be x°.

Second angle of complementary angle be (90 - x°).

It is given that an angle is equal to 8 times.

x°=8(90-x°)

x°=720-8x°

x°+ 8x° = 720

9x° = 720

x° = (720 )/9

x° = 80°

Hence, the required angle is 80°

**Question 8: If the angle (2x-10)° and (x-5)° are complementary angles. Find the value of x°.**

**Solution:**

It is given that,

The first angle of complementary angles is (2x-10)°

The second angle of complementary angles is (x-5)°

(2x-10)°+(x-5)°=90

2x-10+x-5°=90

3x-15=90

3x=90+15

3x=105

x=105/3

x=35

Hence, the value of x is 35°.

**Question 9: If the complement of an angle is equal to supplement of thrice of it find the measure of it.**

**Solution:**

The sum of complementary angles 90°

And the sum of Supplementary angles 180°

Let the first angle of complementary angle will be x°

And the other angle be (90-x°)

Let the first angle of supplementary angle will be 3x°

And the other angle be (180-3x°)

It is given that the complement of an angle is equal to supplement of thrice

(90-x°)=(180-3x°)

90-x°=180-3x°

-x°+3x°=180-90

2x°=90°

x=(90°)/2

x=45°

**Question 10: If an angle differ by is complementary by 10°. Find the angle.**

**Solution:**

Let the first angle of complementary angle is x°

and the second angle of complementary angle is 90-x°

It is given that the difference of complementary angles is 10.

x-(90-x)=10

x-90+x=10

x+x=10+90

2x=100

x=100/2

x=50

**Question 11: If the supplement of an angle is three times its complement. Find the angle.**

**Solution:**

The sum of complementary angles 90°

And the sum of Supplementary angles 180°

Let the first angle of complementary angle is x°

and the second angle of complementary angle is 90-x°

and the second angle of supplementary angle is 180-x°

180-x=3(90-x)

180-x=270-3x

180-270=x-3x

-90=-2x

90/2=x

45°=x

**Question 12: If the supplement of an angle is two third of its Determine the angle and its supplement.**

**Solution:**

The sum of Supplementary angles will be 180°

Let the first angle of supplementary angle be x°.

Second angle of supplementary angle be (180 - x°)

(180-x°)=2/3 x°

3(180-x°)=2x°

540-3x°=2x°

540=2x°+3x°

540=5x°

540/5=x°

108°=x°

The other supplements are 180 –108=72°

**Question 13: An angle is 14° more than its complementary angle. What is its measure?**

**Solution:**

Let the first angle of complementary angle is x°

and the second angle of complementary angle is 90-x°

The sum of complementary angles is 90 °

x°-14=90-x°

x°+x°=90+14

2x°=90+14

2x°=104

x°=(104 )/2

x°=52°

The other supplementary angle is 52°-14=38°

**Question 14: The measure of an angle is twice of its supplement angle. Find its measure.**

**Solution:**

The sum of Supplementary angles will be 180°

Let the first angle of supplementary angle be x°.

Second angle of supplementary angle be (180-x°)

x°=2(180-x°)

x°=360-2x°

x°+2x°=360

3x°=360

x°=360/3

x°=120°

**Exercise 8.2**

**Question 1: In the below Fig. OA and OB are opposite rays:**

**(i) If x = 25°, what is the value of y?**

**(ii) If y = 35°, what is the value of x?**

**Solution:**

(i) It is given that the value of x = 25

∠AOC = 2y + 5 and ∠BOC = 3x

We, know that,

∠AOC + ∠BOC = 180° (Liner Pair)

(2y + 5) + 3x = 180°

Put the value of x in above equation.

(2y + 5) + 3(25) = 180°

2y + 5 + 75 = 180°

2y + 80 = 180°

2y = 100°

y = 100/2

y = 50°

Hence, the value of y = 50°

(ii) It is given that the value of y = 35°

We, know that,

∠AOC + ∠BOC = 180°

(2y + 5) + 3x = 180

Put the value of y in above equation.

(2(35) + 5) + 3x = 180

75 + 3x = 180

3x = 105

x = 35

Hence, the value of x = 35°

**Question 2: In the below figure, write all pairs of adjacent angles and all the linear pairs.**

**Solution:**

Adjacent angles:-

∠AOC, ∠COB;

∠AOD, ∠BOD;

∠AOD, ∠COD;

∠BOC, ∠COD

Linear pair:-

∠AOD + ∠BOD = 180°

∠AOC+ ∠BOC = 180°

**Question 3: In the given figure, find x. Further find ∠BOC, ∠COD and ∠AOD.**

**Solution:**

∠AOD and ∠BOD form a linear pair.

∠AOD + ∠BOC + ∠COD = 180°

It is given that,

∠AOD=(x+10)

∠COD=x

∠BOC=(x + 20)

∠AOD+∠BOC+∠COD=180° (Liner Pair)

(x+10) +x+(x+20)=180

3x+30=180

3x= 180 –30

x=150/3

x=50°

We can say that,

∠AOD=x+10

∠AOD=50 + 10

∠AOD=60

and

∠COD=x

∠COD=50

and

∠BOC=x+20

∠BOC=50+20

∠BOC=70

Thus,∠AOD=60°,∠COD=50° and ∠BOC=70°

**Question 4: In figure, rays OA, OB, OC, OD and OE have the common end point 0. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.**

**Solution:**

It is given that rays the endpoint O is shared by OA, OB, OC, OD, and OE.

By the construction we build a straight line AS by drawing an opposite ray OS to ray OA.

From the construction we know that AS is line.

So, ∠AOB and ∠BOS are linear pair angles

∠BOS = ∠BOC + ∠COS

∠AOB + ∠BOS = 180°

∠AOB + ∠BOC + ∠COS = 180° __________(1)

So, ∠AOE and ∠EOS are linear pair angles

∠EOS = ∠DOE + ∠DOS

∠AOE + ∠EOS =180°

Or, ∠AOE + ∠DOE + ∠DOS = 180° __________(2)

Adding equations (1) and (2)

∠AOB + ∠BOC + ∠COF + ∠AOE + ∠DOE + ∠DOS = 180° + 180°

∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°

Hence, it is proved that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.

**Question 5: In figure, ∠AOC and ∠BOC form a linear pair. If a – 2b = 30°, find a and b?**

**Solution:**

It is given that,

a – 2b = 30° __________(1)

∠AOC and ∠BOC form a linear pair

a + b = 180° __________(2)

By subtracting equation (1) from (2),

a + b – a + 2b = 180 – 30

3b = 150

b = 150/3

b = 50°

Put the value of b in equation.

a – 2b = 30°

a – 2(50) = 30

a – 100 = 30

a = 30 + 100

a = 130°

Hence, the values of a and b are 130° and 50°.

**Question 6: How many pairs of adjacent angles are formed when two lines intersect at a point?**

**Solution:**

The 4 pair of adjacent angles are:

∠POS, ∠SOQ

∠SOQ, ∠QOR

∠ROP, ∠POS

∠QOS, ∠ROP

**Question 7: How many pairs of adjacent angles, in all, can you name in figure given?**

**Solution:**

Adjacent angles, from the figure, are:

∠EOC and ∠DOC

∠EOD and ∠DOB

∠DOC and ∠COB

∠EOD and ∠DOA

∠DOC and ∠COA

∠BOC and ∠BOA

∠BOA and ∠BOD

∠BOA and ∠BOE

∠EOC and ∠COA

∠EOC and ∠COB

Thus, the above 10 are adjacent angles.

**Question 8: In figure, determine the value of x.**

**Solution:**

360° is equal to the number of all angles at a point O.

3x+3x+150+x=360°

7x=360°–150°

7x = 210°

x=210/7

x=30°

Hence, the value of x is 30°.

**Question 9: In figure, AOC is a line, find x.**

**Solution:**

It is given that,

∠AOB + ∠BOC = 180° (linear pairs)

70 + 2x = 180

2x = 180 – 70

2x = 110

x = 110/2

x = 55

Hence,the value of x is 55°.

**Question 10: In figure, POS is a line, find x.**

**Solution:**

∠POQ + ∠QOS = 180° (linear pairs)

∠POQ + ∠QOR + ∠SOR = 180°

600 + 4x + 400 = 180°

4x = 180° - 100°

4x = 80°

x = 20°

Thus, the value of x is 20°.

**Question 11: In the below figure, ACB is a line such that ∠DCA = 5x and ∠DCB = 4x. Find the value of x?**

**Solution:**

It is given that,

∠DCA = 5x and ∠DCB = 4x

As, AB is a line

∠ACD + ∠BCD = 180° (linear pairs)

5x+4x=180

9x=180

x=20

Hence, the value of x is 20°

**Question 12: In the given figure, Given ∠POR = 3x and ∠QOR = 2x + 10, Find the value of x for which POQ will be a line?**

**Solution:**

POR is a Straight line.

It is given that,

∠POR = 3x

∠QOR = 2x + 10

∠POR + ∠QOR = 180° (Linear Pair)

3x + 2x + 10 = 180

5x + 10 = 180

5x = 180 – 10

5x = 170

x = 170/5

x = 34

Hence the value of x is 34°

**Question 13: In Fig: a is greater than b by one third of a right angle. Find the value of a and b?**

**Solution:**

AB is a straight line

So, a + b = 180

a = 180 – b _________(1)

It is given that,

a is greater than b by one third of a right angle

a = b + 90/3

a = b + 30_________(2)

Put both the equations Equal.

180 – b = b + 30

180 – 30 = 2b

b = 150/2

b = 75

Put the value of b in equation (1)

a = 180 – b

a = 180 – 75

a = 105

Thus, the values of a and b are 105° and 75°.

**Question 14: What value of y would make AOB a line in the below figure, If ∠AOB = 4y and ∠BOC = (6y + 30) ?**

**Solution:**

AB is a straight line,

∠AOC + ∠BOC = 180° (Linear pairs)

4y + 6y + 30 = 180

10y + 30 = 180

10y = 180 - 30

10y = 150

y = 150/10

y = 15

Thus, the value of y is 15°

**Question 15: If the figure below forms a linear pair, forms a linear pair ∠EOB = ∠FOC = 90° and ∠DOC = ∠FOG = ∠AOB = 30°. Find the measure of ∠FOE, ∠COB and ∠DOE**

**Name all the right angles**

**Name three pairs of adjacent complementary angles**

**Name three pairs of adjacent supplementary angles**

**Name three pairs of adjacent angles**

**Solution:**

(i) It is given that,

Let us assumed that,

∠FOE = x, ∠DOE = y and ∠BOC = z

AG is a straight line,

∠AOF + ∠FOG = 180 (Linear pair)

∠AOF + 30 = 180

∠AOF = 180 - 30

∠AOF = 150

We know that,

∠AOF = ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOF

∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOF = 150

30 + z + 30 + y + x = 150

x + y + z =150 - 30 - 30

x + y + z = 90 ________(1)

It is given that,

∠FOC = 90°

We know that

∠FOC = ∠FOE + ∠EOD + ∠DOC

∠FOE + ∠EOD + ∠DOC = 90°

x + y + 30 = 90

x + y = 90 - 30

x + y = 60 ________ (2)

Put the value of x + y in equation (1)

x + y + z = 90

60 + z = 90

z = 90 - 60

z = 30

It is given that ∠BOE = 90

∠BOC + ∠COD + ∠DOE = 90°

30 + 30 + ∠DOE = 90

∠DOE = 90 - 60 = 30

∠DOE = 30

x = 30

Put the value of x in equation (2)

x + y = 60

y = 60 - x

y = 60 - 30

y = 30

Thus, ∠FOE = 30°, ∠COB = 30° and ∠DOE = 30°

(ii) ∠DOG, ∠COF, ∠BOF, ∠AOD are the Right angles.

(iii) ∠AOB, ∠BOD

∠AOC, ∠COD

∠BOC, ∠COE

Above are the adjacent complementary angles.

(iv) ∠AOB, ∠BOG

∠AOC, ∠COG

∠AOD, ∠DOG

Above are the adjacent supplementary angles.

(v) ∠BOC, ∠COD

∠COD, ∠DOE

∠DOE, ∠EOF

Above are the adjacent angles.

**Question 16: In below fig. OP, OQ, OR and OS are four rays. Prove that: ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°.**

**Solution:**

∠XOR+∠ROP=180° (Linear Pair)

∠ROP=∠ROQ+∠QOP

∠XOR+∠ROQ+∠QOP=180° __________(1)

∠XOS+∠SOP=180° (Linear Pair) __________(2)

By addition both the equations

∠XOR+∠ROQ+∠QOP+∠XOS+∠SOP=180°+180°

∠XOR+∠ROQ+∠QOP+∠XOS+∠SOP=360°

We know that,

∠SOR=∠XOR+∠XOS

∠SOR+∠ROQ+∠QOP+∠SOP=360°

Hence Proved.

**Question: 17: In below fig, ray OS stand on a line POQ. Ray OR and ray OT are angle bisectors of ∠POS and ∠SOQ respectively. If ∠POS = x, find ∠ROT?**

**Solution:**

∠POS+∠SOQ=180° (Linear Pair)

∠POR+∠ROS+∠SOT+∠TOQ=180°

x/2+x/2+y/2+y/2=180°

(x+x+y+y)/2=180°

2x+2y=2×180°

2x+2y=360°

2(x+y)=360°

x+y=(360°)/2

x+y=180°

We know that,

∠ROT = x/2+y/2

∠ROT = (x + y )/2

∠ROT = (180° )/2

∠ROT = 90°

Hence, the value of ∠ROT is 90°.

**Question 18: In the below fig, lines PQ and RS intersect each other at point O. If ∠POR: ∠ROQ = 5: 7. Find all the angles.**

**Solution:**

It is given that,

∠POR : ∠ROQ = 5:7

∠POR + ∠ROP = 180° (linear pair)

∠POR = 180 × 5/12

∠POR = 75°

∠ROQ = 180 × 5/12

∠ROQ = 105°

So,

∠POS = ∠ROQ = 105° (Vertically opposite angles)

∠SOQ = ∠POR = 75° (Vertically opposite angles)

Hence, the angles will be ∠POS = ∠ROQ = 105° and ∠SOQ = ∠POR = 75°.

**Question 19: In the below fig. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS − ∠POS).**

**Solution:**

It is given that,

OR perpendicular

∠POS+∠SOR=90°

∠POS+∠SOQ=180° (Linear Pair)

Multiplying by 2 we get,

2(∠POS+∠SOR)=2×90°

2∠POS+2∠SOR=180°

∠POS+∠SOQ=2∠POS+2∠SOR

∠SOQ-2∠SOR=2∠POS-∠POS

∠SOQ-∠POS=2∠ROS

(∠SOQ-∠POS)/2=∠ROS

1/2(∠SOQ-∠POS)=∠ROS

∠ROS=1/2(∠SOQ-∠POS)

Hence Proved.

**Exercise 8.3**

**Question 1: In figure, lines l**

_{1}, and l_{2}intersect at O, forming angles as shown in the figure. If x = 45. Find the values of y, z and u.**Solution:**

It is given that x = 45°

z = x = 45° (vertically opposite angles) ___________(1)

From Line l1 we get,

z + u = 180° (linear pair) _______(2)

Put the value of z in equation (2).

z + u = 180°

45° + u = 180°

u = 180° – 45°

u = 135°

u = y = 135° (vertically opposite angles)

Hence, the angles are y = 135°, u = 135° and z = 45°.

**Question 2: In figure, three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.**

**Solution:**

It is given that,

∠BOD = 90° and ∠DOF = 50°

So,

∠BOD = z = 90° (Vertically opposite angles)

∠DOF = y = 50° (Vertically opposite angles)

AB is a straight line,

x + y + z = 180 (Linear pair)

x + y + z = 180

x + 50 + 90 = 180

x = 180 – 140

x = 40

Thus, the values of x, y, z and u are 40°, 50°, 90° and 40°.

**Question 3: In figure, find the values of x, y and z.**

**Solution:**

y = 25° (Vertically opposite angles)

l1 is a straight line,

x + y = 180° (Linear pair of angles)

x = 180° – 25°

x = 155°

l2 is a straight line,

z + 25° = 180° (Linear pair of angles)

z = 180° – 25°

z = 155°

Hence, the value of x, y and z are 155°, 25° and 155°

**Question 4: In figure, find the value of x.**

**Solution:**

It is given that,

∠BOF = 5x

∠AOE = 3x

∠EOD = 2x

∠AOE = ∠BOF = 5x (Vertically opposite angles)

CD is a straight line,

∠COA + ∠AOE + ∠FOC = 180° [Linear pair]

3x + 5x + 2x = 180

10x = 180

x = 180/10

x = 18°

Hence, the value of x is 18°

**Question 5: Prove that bisectors of a pair of vertically opposite angles are in the same straight line.**

**Solution:**

We know that,

∠BOD = ∠COA (Vertically angles)

∠FOB + ∠BOE = 180° (Linear Pair)

∠FOB + ∠BOC + ∠COE =180° __________(1)

∠EOC + ∠COF = 180° (Linear Pair)

∠ EOC + ∠COB + ∠BOF =180° __________(2)

EOF is a Straight line.

Hence Proved

**Question 6: If two straight lines intersect each other, prove that the ray opposite to the bisector of one of the angles thus formed bisects the vertically opposite angle.**

**Solution:**

It is given that,

Straight lines AB and CD intersect each other at O.

Here, OS is the bisector of ∠AOC.

Straight lines AB, CD and ST are intersect each other at O.

∠AOS = ∠BOT (Vertically opposite angles)

∠COS = ∠DOT (Vertically opposite angles)

OS is the bisector of ∠AOC

So, ∠AOS = ∠COS

Thus, ∠BOT = ∠ DOT

Therefore, OT is the bisector of ∠BOD.

**Question 7: If one of the angles formed by intersecting lines is an angle then show that the each of the angles is a right angle.**

**Solution:**

It is given that,

∠1=90°

∠AOB=90°

∠1+∠2=180° (Linear Pair)

90°+∠2=180°

∠2=180°-90°

∠2=90°

∠COD=180°

∠2+∠3=180° (Linear Pair)

90°+∠3=180°

∠3=180°-90°

∠3=90°

∠1+∠4=180° (Linear Pair)

90°+∠4=180°

∠4=180°-90°

∠4=90°

∠1+∠2+∠3+∠4=90°

Hence, ∠AOC=∠BOC=∠BOD=∠AOD=90°

**Question: 8: In the below fig. rays AB and CD intersect at O.**

**(i) Determine y when x = 60**

**(ii) Determine x when y = 40**

**Solution:**

(i) Determine y when x=60°

∠AOC + ∠COB = 180°(Linear Pair)

2x + y = 180°

2(60) + y = 180°

120° + y = 180°

y=180°-120°

y=60°

(ii) Determine x when y=40°

∠AOC + ∠COB = 180°(Linear Pair)

2x + y = 180°

2x + 40 = 180°

2x=180°-40°

2x=140°

x= (140°)/2

x=70°

**Question 9: In the below fig. lines AB. CD and EF intersect at O. Find the measures of ∠AOC, ∠COF, ∠DOE and ∠BOF.**

**Solution:**

**Solution:**

It is given that,

∠AOE + ∠EOB = 180°(Linear Pair)

∠AOE + ∠DOE + ∠DOB = 180°

∠AOE + ∠DOE + 35° = 180°

40° + ∠DOE + 35° = 180°

∠DOE + 75° = 180°

∠DOE = 180°-75°

∠EOD = 105°

∠AOE = ∠FOB = 40°(Vertically Opposite angle)

∠BOD = ∠AOC = 35°(Vertically Opposite angle)

∠DOE = ∠COF = 105°(Vertically Opposite angle)

Hence, the value of ∠AOC, ∠COF, ∠DOE and ∠BOF are 35°, 105°, 105°, and 40° respectively.

**Question 10: AB, CD and EF are three concurrent lines passing through the point O such that OF bisects BOD. If BOF = 35. Find BOC and AOD.**

**Solution:**

It is given that,

OF bisects BOD

∠BOF = 35°

∠BOF = ∠FOD (OF is bisects BOD)

∠BOF = ∠FOD = 35°

∠BOF + ∠FOD = ∠BOD

35° + 35° = ∠BOD

70° = ∠BOD

∠BOD = ∠AOC = 70° (Vertically opposite angles]

∠BOC + ∠AOC = 180° (Linear Pair)

∠BOC + 70° = 180°

∠BOC = 180° - 70°

∠BOC = 110°

∠AOD = ∠BOC = 110° (Vertically opposite angles)

**Question 11: In below figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE?**

**Solution:**

It is given that,

∠AOC + ∠BOE = 70° and ∠BOD = 40°

∠BOD = ∠AOC (vertically opposite angles)

∠BOD = 40°

∠BOD = ∠AOC = 40°

∠AOC + ∠BOE = 70°,

40° + ∠BOF = 70°

∠BOF = 70° - 40°

∠BOE = 30°

∠AOC + ∠COE + ∠BOE = 180° (Lines pair)

40° + ∠COE + 30° = 180°

∠COE = 180° - 30° - 40°

∠COE = 110°

Hence, Reflex ∠COE = 360° - 110° = 250°.

**Question 12: Which of the following statements are true (T) and which are false (F)?**

**(i) Angles forming a linear pair are supplementary.**

**(ii) If two adjacent angles are equal and then each angle measures 90**

**(iii) Angles forming a linear pair can both acute angles.**

**(iv) If angles forming a linear pair are equal, then each of the angles have a measure of 90**

**Solution:**

(i) True

(ii) False

(iii) False

(iv) true

**Question 13: Fill in blanks so as to make the following statements true:**

**(i) If one angle of a linear pair is acute then its other angle will be______**

**(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is ______**

**(iii) If the sum of two adjacent angles is 180, then the ______ arms of the two angles are opposite rays.**

**Solution:**

(i) If one angle of a linear pair is acute then its other angle will be

**Obtuse angle.**(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is

**180°**(iii) If the sum of two adjacent angles is 180, then the

**Uncommon**arms of the two angles are opposite rays.**Exercise 8.4**

**Question 1: In figure, AB, CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.**

**S**

**olution:**

It is given that,

∠1 and ∠2 are in the ratio 3 : 2

It is assumed that,

∠1=3x

∠2=2x

∠1+∠2=180° (Linear pair)

3x+2x=180°

5x=180°

x=180/5

x=36°

Thus,

∠1=3x

∠1=3 × 36°

∠1=108°

and

∠2=2x

∠2=2×36°

∠2=72°

∠1 = ∠3 = 108° (vertically opposite angles)

∠2 = ∠4 = 72° (vertically opposite angles)

We also know, if a transversal intersects any parallel lines, then the corresponding angles are equal

∠1 = ∠5 = 108° (corresponding angles)

∠2 = ∠6 = 72° (corresponding angles)

So,

∠5 = ∠7 = 108° (vertically opposite angles)

∠6 = ∠8 = 72° (vertically opposite angles)

Hence, the value of angles ∠1 = 108°, ∠2 = 72°, ∠3 = 108°, ∠4 = 72°, ∠5 = 108°, ∠6 = 72°, ∠7 = 108° and ∠8 = 72°.

**Question 2: In figure, I, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠1, ∠2 and ∠3.**

**S**

**olution:**

As per figure we find that,

∠Y = 120° (Vertical opposite angles)

∠3 + ∠Y = 180° (Linear pair angles)

∠3 = 180° – 120°

∠3 = 60°

As Shown in figure,

Line l is parallel to line m,

∠1 = ∠3 (Corresponding angles)

∠1 = 60°

Line m is parallel to line n,

∠2 = ∠Y (Alternate interior angles)

∠2 = 120°

∠2 = ∠Z (Vertical opposite angles)

∠Z = 120°

Hence, the value of angle ∠1 = 60°, ∠2 = 120° and ∠3 = 60°.

**Question 3: In figure, AB || CD || EF and GH || KL. Find ∠HKL.**

**S**

**olution:**

Extend LK to meet line GF at point R.

It is given that,

∠KHY = 25°

∠KHY = ∠XKH = 25° (Alternate interior angles)

∠KHC = ∠CHG = 180° (Linear Pair)

∠XHC + 60° = 180°

∠XHC = 180°- 60°

∠XHC = 120°

∠CHG = ∠AXH = 60° (Corresponding angles)

∠AXH = ∠XKY = 60° (Corresponding angles)

∠XKY = ∠XKL = 180° (Linear Pair)

60° + ∠XKL = 180°

∠XKL = 180°- 60°

∠XKL = 120°

∠HKL = ∠LKX + ∠XKH

∠HKL = 120° + 25°

∠HKL = 145°

Hence, the value of ∠HKL is 145°.

**Question 4: In figure, show that AB || EF.**

**S**

**olution:**

∠BAC = ∠ACD (Alternate interior angle)

∠BAC = ∠ACD = 57°

∠ACD = ∠ACE + ∠ECD

∠ACD = 22° + 35°

∠ACD = 57°

So, BA||CD___________(1)

∠FEC + ∠ECD = 180° (Interior angles on the same side)

145° + 35° = 180°

So, CD||EF___________(1)

From equation 1 and 2 we get,

AB||CD||EF

Hence we can say that, AB||EF.

Hence Proved

**Question 5 : In figure, if AB || CD and CD || EF, find ∠ACE.**

**S**

**olution:**

It is given that, AB|| EF

AB||EF||CD

∠BAC = ∠ACD (Alternate interior angles)

∠BAC = ∠ACD = 70°

∠FEC = ∠ECD = 180°

130° + ∠ECD = 180°

∠ECD = 180° - 130°

∠ECD = 50°

∠ACD = ∠ACE + ∠ECD

70° = ∠ACE + 50°

∠ACE = 70° + 50°

∠ACE = 20°

Hence, the ∠ACE is 20°.

**Question 6: In figure, PQ || AB and PR || BC. If ∠QPR = 102°, determine ∠ABC. Give reasons.**

**S**

**olution:**

It is given that,

∠QPR = 102°

∠QPX + ∠PXA = 180° (sum of interior angles on the same side)

102° + ∠PXA = 180°

∠PXA = 180°- 102°

∠PXA = 78°

Also given that, PR||BC

∠PXA = ∠ABC = 78° (Alternate Interior angle)

Hence, the value of ∠ABC is 78°.

**Question 7: In figure, state which lines are parallel and why?**

**S**

**olution:**

∠EDC = ∠D (Vertically Opposite angles)

∠EDC = ∠D = 100°

∠ACD = ∠D (Corresponding angles)

∠ACD = ∠D = 100°

Corresponding angles are only made by parallel lines.

Hence, we can say that DE || AC.

**Question 8: In figure, if l||m, n || p and ∠1 = 85°, find ∠2.**

**S**

**olution:**

It is given that,

∠1 = 85°

∠1 = ∠3 = 85° (Corresponding angles)

Again, co-interior angles are supplementary, so

∠3 + ∠2 = 180° (interior angles of the same side of a transversal)

85° + ∠2 =180°

∠2 = 180° – 85°

∠2 = 95°

Hence, the value of ∠2 is 95°.

**Question 9: If two straight lines are perpendicular to the same line, prove that they are parallel to each other.**

**S**

**olution:**

l and m are straight lines.

n ⊥ l

And n ⊥ m

∠1 = ∠2 (Corresponding angles)

∠1 = ∠2 = 90°

l || m

Hence Prove

**Question 10: Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.**

**S**

**olution:**

It is assumed that

The angles be ∠ACB and ∠ABD

AC perpendicular to AB,

CD is perpendicular to BD.

In a quadrilateral,

∠A+ ∠C+ ∠D+ ∠B = 360° (Sum of angles in a quadrilateral is 360°)

180° + ∠C + ∠B = 360°

∠C + ∠B = 360° –180°

Therefore,

∠ACD + ∠ABD = 180°

∠ABD = ∠ACD = 90° (Angles are equal and supplementary)

Hence Proved,

**Question: 11: In the below fig, lines AB and CD are parallel and P is any point as shown in the figure. Show that ∠ABP + ∠CDP = ∠DPB.**

**S**

**olution:**

It is given that AB||CD

Draw a line MN parallel to AB or CD

AB||MN||CD

CD||MN

∠4 = ∠1 (Alternate interior angles)

MN||AB

∠2 = ∠3 (Alternate interior angles)

∠DPB = ∠1 + ∠2

∠DPB = ∠4 + ∠3

∠DPB = ∠CDP + ∠ABP

Hence Prove

**Question: 12: In the below fig, AB ∥ CD and P is any point shown in the figure. Prove that: ∠ABP + ∠BPD + ∠CDP = 360°**

**S**

**olution:**

We can draw a line MN parallel to AB or CD through P.

AB|| MN

∠1 + ∠4 = 180°____________(1) (sum of two sides of same side of transversal)

MN||CD

∠5 + ∠3 = 180°____________(2) (sum of two sides of same side of transversal)

From the fig. we get,

∠2 = ∠4 + ∠5

By adding equation (1) and (2) we get,

∠1 + ∠4 + ∠5 + ∠3 = 180° + 180°

∠1 + ∠2 + ∠3 = 360°

∠ABP + ∠BPD + ∠CDP = 360°

Hence Prove

**Question: 13: Two unequal angles of a parallelogram are in the ratio 2: 3. Find all its angles in degrees.**

**S**

**olution:**

Let us assumed that,

∠A = 2x

∠B = 3x

∠A + ∠B = 180° __________(1) (Co-interior angles are supplementary)

Put the value of ∠A and ∠B in equation (1)

2x + 3x = 180°

5x = 180°

x = (180°)/5

x = 36°

A = 2 × 36 = 72°

B = 3 × 36 = 108°

A = C = 72° (Opposite side angles and sides of a parallelogram are equal)

B = D = 108° (Opposite side angles and sides of a parallelogram are equal)

Hence Prove.

**Question: 14: If each of the two lines is perpendicular to the same line, what kind of lines are they to each other?**

**S**

**olution:**

Let us assumed that,

AB ⊥ MN

∠ABD = 90°_______(i)

CO ⊥ MN

∠CON = 90°_______ (ii)

∠AYX = 90°

∠CDN = 90°

∠1 = ∠2 (Corresponding angle)

AB || CD,

Hence Prove.

**Question: 15: In the below fig, ∠1 = 60° and ∠2 = 2/3rd of a right angle. Prove that l ∥ m.**

**S**

**olution:**

It is given that,

∠1 = 60° and ∠2 = 2/3 rd of a right angle.

According to figure m is parallel to l.

∠2 = (2/3) × 90°

∠2 = 60°

∠1 = ∠2 (Corresponding angles)

∠1 = 60°

Hence, Parallel to m as pair of corresponding angles are equal.

**Question: 16: In the below fig, if l ∥ m ∥ n and ∠1 = 60°. Find ∠2.**

**S**

**olution:**

It is given that l || m ||n

∠1 = 60°

∠1 = ∠3 = 60° (Corresponding angles)

∠3 + ∠4 = 180° (linear pair)

60° + ∠4 = 180°

∠4 = 180° - 60°

∠4 = 120°

m ∥ n and P is the transversal

∠4 = ∠2

∠2 = 120° (Alternative interior angle)

Hence, the value of ∠2 is 120°.

**Question: 17: Prove that the straight lines perpendicular to the same straight line are parallel to one another.**

**S**

**olution:**

It is assumed that,

AB and CD perpendicular to MN

∠ABD = 90° (AB is perpendicular to MN)_________(i)

∠CON = 90° (CD is perpendicular to MN)________(ii)

From the equation we get,

∠ABD = ∠CDN = 90°

Hence, Corresponding angles are equal AB||CD.

**Question: 18: The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 60°. Find the other angles.**

**S**

**olution:**

It is given that,

AB || CD

AD || BC

Co-interior angles are supplementary

∠A + ∠D = 180° (Co-interior angles are supplementary)

60° + ∠D = 180°

∠D = 180° - 60°

∠D = 120°

Co-interior angles are supplementary

∠A + ∠B = 180° (Co-interior angles are supplementary)

60 + ∠B = 180°

∠B = 180° - 60°

∠B = 120°

Hence, the value of ∠A = 60° and ∠B = ∠D = 120°

**Question: 19: Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, find the measures of ∠AOC, ∠COB,**

**∠BOD, ∠DOA**

**S**

**olution:**

It is given that, ∠AOC + ∠COB + ∠BOD = 270°

∠AOC + ∠COB + ∠BOD + ∠AOD = 360° (Complete angle)

270° + ∠AOD = 360°

∠AOD = 360° - 270°

∠AOD = 90°

∠AOD + ∠BOD = 180° (Linear pair)

90° + ∠BOD = 180°

∠BOD = 180° - 90°

∠BOD = 90°

∠AOD = ∠BOC = 90 ° (Vertically opposite angles)

∠BOD = ∠AOC = 90° (Vertically opposite angles)

Hence, the value of ∠AOD, ∠BOC, ∠BOD and ∠AOC is 90°.

**Question: 20: In the below figure, p is a transversal to lines m and n, ∠2 = 120° and ∠5 = 60°. Prove that m|| n.**

**S**

**olution:**

It is given that, ∠2 = 120° and ∠5 = 60°

∠2 + ∠1 = 180° (Linear pair)

120° + ∠1 = 180°

∠1 = 180° − 120°

∠1 = 60°

∠1 = ∠5 = 60°

Hence, m || n (corresponding angles are equal)

**Question: 21: In the below fig. transversal t intersects two lines m and n, ∠4 = 110° and ∠7 = 65° is m ∥ n?**

**S**

**olution:**

It is given that, ∠4 = 110° and ∠7 = 65°

∠7 = ∠5 = 65° (Vertically opposite angle)

∠4 + ∠5 = 180° (co interior angles is supplementary angle)

110° + 65° = 175°

Hence, here m is not parallel to n as the pair of co interior angles is not supplementary.

**Question: 22: Which pair of lines in the below fig. is parallel? Give reasons.**

**S**

**olution:**

∠A + ∠B = 180°

115° + 65° = 180°

∠B + ∠C = 180°

65° + 115° = 180°

The sum of interior angles are supplementary Hence, AB||CD

**Question: 23: If I, m, n are three lines such that I || m and n perpendicular to l, prove that n perpendicular to m.**

**S**

**olution:**

It is given that l || m, n perpendicular to I

l || m and n intersecting line

∠1 = ∠2 (Corresponding angles)

∠2 = 90° (as per fig.)

Thus, we can say that n is perpendicular to m.

**Question: 24: In the below fig, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC = ∠DEF.**

**S**

**olution:**

It is given that AB ∥ DE and BC ∥ EF

By the construction we get BC to x such that it intersects DE at M.

AB ∥ DE and BX is the transversal

∠ABC = ∠DMX (Corresponding angle)________(i)

BX ∥ EF and DE is the transversal

∠DMX = ∠DEF (Corresponding angles) ________ (ii)

From equation (i) and equation (ii) we get,

∠ABC = ∠DEF

Hence Proved.

**Question: 25: In the below fig, arms BA and BC of ABC are respectively parallel to arms ED and EF of DEF Prove that**

**∠ABC + ∠DEP = 180°**

**S**

**olution:**

It is given that,

AB ∥ DE, BC ∥ EF

By the construction we get BC to intersect DE at M

AB || EM and BL is the transversal

∠ABC = ∠EML (Corresponding angle)__________(i)

EF || ML and EM is the transversal

∠DEF + ∠EML = 180° (co-interior angles are supplementary)________(ii)

From equation (1) and (2) we get,

∠DEF + ∠ABC = 180°

Hence Proved.

**Question: 26: With of the following statements are true (T) and which are false (F)? Give reasons.**

**(i) If two lines are intersected by a transversal, then corresponding angles are equal.**

**(ii) If two parallel lines are intersected by a transversal, then alternate interior angles are equal.**

**(ii) Two lines perpendicular to the same line are perpendicular to each other.**

**(iv) Two lines parallel to the same line are parallel to each other.**

**(v) If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal are equal.**

**S**

**olution:**

(i) False

(ii)True

(iii) False

(iv) True

(v) False

**Question: 27: Fill in the blanks in each of the following to make the statement true:**

**(i) If two parallel lines are intersected by a transversal, then each pair of corresponding angles are ____________**

**(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are _____________**

**(iii) Two lines perpendicular to the same line are _______ to each other**

**(iv) Two lines parallel to the same line are __________ to each other.**

**(v) If a transversal intersects a pair of lines in such a way that a pair of alternate angles we equal. then the lines are ___________**

**(vi) If a transversal intersects a pair of lines in such a way that the sum of interior angles on the seine side of transversal is 180'. then the lines are _____________**

**Solution:**

(i) Equal

(ii) Parallel

(iii) Supplementary

(iv) Parallel

(v) Parallel

(vi) Parallel

**Exercise VSAQs.......................**

**Question 1: Define complementary angles.**

**S**

**olution:**

Two angles are said to be complementary, if the sum of their measures is 90°.

**Question 2: Define supplementary angles.**

**S**

**olution:**

Two angles are said to be complementary, if the sum of their measures is 180°.

**Question 3: Define adjacent angles.**

**S**

**olution:**

Two angles are Adjacent when they have a common side and a common vertex.

**Question 4: The complement of an acute angle is _____.**

**S**

**olution:**

The complement of an acute angle is an acute angle.

**Question 5: The supplement of an acute angle is _____.**

**S**

**olution:**

The supplement of an acute angle is an obtuse angle.

**Question 6: The supplement of a right angle is _____.**

**S**

**olution:**

The supplement of a right angle is a right angle.