# R.D. Sharma Solutions Class 9th: Ch 8 Lines and Angle Exercise 8.2

#### Chapter 8 Lines and Angle R.D. Sharma Solutions for Class 9th Math Exercise 8.2

**Exercise 8.2**

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

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

**Solution**

2 . In Fig. 8.32, write all pairs of adjacent angles and all the linear pairs.

**Solution**

3.In Fig. 8.33, find x. Further find ∠BOC, ∠COD and∠AOD

**Solution**

Since,

∠AOD and ∠BOD from a linear pair.

∠AOD + ∠BOD = 180°

∠AOD + ∠BOC+ ∠COD = 180°

Given that,

∠AOD = (x+10°),

∠COD = x° and,

∠BOC = (x+20)°

(x+10) + x + (x+20) = 180°

⇒ 3x + 30 = 180

⇒ 3x = 180 - 30

⇒ 3x = 180 - 30

⇒ 3x = 150

⇒ 3x = 150/3

⇒ x = 50

Therefore,

∠AOD = (x+10°) = 50+10 = 60

∠COD = x = 50

∠BOC = x+20 = 50+20 = 70

Thus, ∠AOD = 60°, ∠COD = 50° and ∠BOC = 70°

4. In Fig. 8.34, rays OA, OB, OC, OD and OE have the common end point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°

**Solution**

Let us draw a straight line, AOX.

5. In the figure given below, it is given that ∠AOC and ∠BOC forms a linear pair.

**Solution**

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

**Solution**

Four pairs of adjacent angles will be formed when two lines intersect at a point.

Considering two lines AB and CD intersecting at O

The 4 pairs are :

(i) ∠AOC = ∠BOC

(ii) ∠BOC = ∠DOB

(iii) ∠AOC = ∠AOD

(iv) ∠DOB = ∠AOD

7. How many pairs of adjacent angles, in all, can you name in Fig. 8.36.

**Solution**

In the given figure,

We have 10 adjacent angle pairs, namely:

(i) ∠AOB and ∠BOC

(ii) ∠AOB and ∠BOD

(iii) ∠AOB and ∠BOE

(iv) ∠BOC and ∠COD

(v) ∠BOC and ∠COE

(vi) ∠COD and ∠DOE

(vii) ∠COD and ∠AOC

(viii) ∠COD and ∠BOC

(ix) ∠AOC and ∠COE

(x) ∠AOD and ∠DOE

8. In Fig. 8.37, determine the value of x.

**Solution**

In the given figure:

Since, the sum of all the angles round a point is equal to 360°.

3x+3x+150+x = 360°

⇒ 7x = 360 - 150

⇒ 7x = 210

⇒ x = 210/7

⇒ x = 30

Value of x is 30°.

9. In Fig. 8.38, AOC is a line, find x.

**Solution**

Since,

∠AOB and ∠BOC are linear pairs,

∠AOB and ∠BOC = 180°

70+2x = 180°

⇒ 2x = 180°-70°

⇒ 2x = 180°-70°

⇒ x = 110°/2

⇒ x = 55°

10. In Fig. 8.39, POS is a line, find x.

**Solution**

The figure is given as follows:

Since ∠POQand∠QOS are linear pairs

∠POQ+∠QOS=180∘

∠POQ+∠QOS=180∘

∠POQ+∠QOR+∠SOR=180∘

60 + 4x +40 = 180

⇒ 4x = 180 -100

⇒ 4x = 80

⇒ x = 20

Hence, Value of x = 20

11. In Fig. 8.40, ACB is a line such that ∠DCA = 5x and ∠DCB = 4x. Find the value of x.

**Solution**

It is given that ACB is a line in the figure given below.

Thus, ∠ACD and ∠BCDform a linear pair.

Therefore, their sum must be equal to 180°.

Or, we can say that

∠ACD + ∠BCD = 180°

Also, ∠ACD = 4x and ∠BCD = 5x

This is simplified as,

4x + 5x = 180°

⇒ 9x = 180°

⇒ x = 180/9

⇒ x = 20

Hence, the value of x is 20°.

∠ACD + ∠BCD = 180°

Also, ∠ACD = 4x and ∠BCD = 5x

This is simplified as,

4x + 5x = 180°

⇒ 9x = 180°

⇒ x = 180/9

⇒ x = 20

Hence, the value of x is 20°.

12. Given ∠POR = 3x and ∠QOR = 2x + 10, find the value of x for which POQ will be a line. (Fig. 8.41).

**Solution**

13. In Fig. 8.42, a is greater than b by one third of a right-angle. Find the values of a and b.

**Solution**

Since a and b are linear

a + b = 180

⇒ a = 180 – b —(1)

From given data, a is greater than b by one third of a right angle

a = b + 90/3

⇒ a = b + 30

⇒ a – b = 30 —(2)

Equating (1) and (2),

180 – b = b + 30

⇒ 180 – 30 = 2b

⇒ b = 150 / 2

⇒ b = 75

From (1)

a = 180 – b

⇒ a = 180 – 75

⇒ a = 105

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

14. What value of y would make AOB a line in Fig. 8.43, if ∠AOC = 4y and ∠BOC = (6y + 30)

**Solution**

Since, ∠AOCand∠BOC are linear pairs ∠AOC+∠BOC=180°

6y + 30 + 4y = 180

⇒ 10y + 30 = 180

⇒ 10y = 180 – 30

⇒ 10y = 150

⇒ y = 150/10

⇒ y = 15

15. If Fig. 8.44, ∠AOF and ∠FOG form a linear pair.∠EOB = ∠FOC = 90° and ∠DOC = ∠FOG = ∠AOB = 30°.

(i) Find the measure of ∠FOE, ∠COB and ∠DOE.

(ii) Name all the right angles.

(iii) Name three pairs of adjacent complementary angles.

(iv) Name three pairs of adjacent supplementary angles.

(v) Name three pairs of adjacent angles.

**Solution**

The given figure is as follows:

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

(iii) Adjacent complementary angles are (∠AOB,∠BOD);(∠AOC,∠COD);(∠BOC,∠COE);

(iv) Adjacent supplementary angles are (∠AOB,∠BOG);(∠AOC,∠COG);(∠AOD,∠DOG);

(v) Adjacent angles are (∠BOC, ∠COD);(∠COD,∠DOE);(∠DOE,∠EOF);

16 . In Fig. 8.45, OP, OQ, OR and OS are four rays. Prove that:∠POQ + ∠QOR + ∠SOR + ∠POS = 360° .

**Solution**

17. In Fig. 8.46, 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**

18. In Fig, 8.47, lines PQ and RS intersect each other at point O. If ∠POR: ∠ROQ = 5 : 7, find all the angles.

**Solution**

19. In Fig. 8.48, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between OP and OR. Prove that ∠ROS = 1/2 (∠QOS - POS)

**Solution**

Given that,

OR is perpendicular to line PQ

To prove,

∠ROS = 1/2(∠QOS – ∠POS)

A/q,

∠POR = ∠ROQ = 90° (Perpendicular)

∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS --- (i)

∠POS = ∠POR - ∠ROS = 90° - ∠ROS --- (ii)

Subtracting (ii) from (i)

∠QOS - ∠POS = 90° + ∠ROS - (90° - ∠ROS)

⇒ ∠QOS - ∠POS = 90° + ∠ROS - 90° + ∠ROS

⇒ ∠QOS - ∠POS = 2∠ROS

⇒ ∠ROS = 1/2(∠QOS – ∠POS)

Hence, Proved.