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

**Exercise 8.2**

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

(ii) If y = 35Â°, 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 âˆ BOC, âˆ COD andâˆ AOD

**Solution**

Since,

âˆ AOD and âˆ BOD from a linear pair.

âˆ AOD + âˆ BOD = 180Â°

âˆ AOD + âˆ BOC+ âˆ COD = 180Â°

Given that,

âˆ AOD = (x+10Â°),

âˆ COD = xÂ° and,

âˆ BOC = (x+20)Â°

(x+10) + x + (x+20) = 180Â°

â‡’ 3x + 30 = 180

â‡’ 3x = 180 - 30

â‡’ 3x = 180 - 30

â‡’ 3x = 150

â‡’ 3x = 150/3

â‡’ x = 50

Therefore,

âˆ AOD = (x+10Â°) = 50+10 = 60

âˆ COD = x = 50

âˆ BOC = x+20 = 50+20 = 70

Thus,Â âˆ AOD = 60Â°,Â âˆ COD = 50Â° andÂ âˆ BOC = 70Â°

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

**Solution**

Let us drawÂ a straight line, AOX.

5. In the figure given below, it is given thatÂ âˆ AOCÂ andÂ âˆ BOCÂ 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) âˆ AOC = âˆ BOC

(ii) âˆ BOC = âˆ DOB

(iii) âˆ AOC = âˆ AOD

(iv) âˆ DOB = âˆ AOD

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

**Solution**

In the given figure,

Â We have 10 adjacent angle pairs, namely:

(i) âˆ AOB and âˆ BOC

(ii) âˆ AOB and âˆ BOD

(iii) âˆ AOB and âˆ BOE

(iv) âˆ BOC and âˆ COD

(v) âˆ BOC and âˆ COE

(vi) âˆ COD and âˆ DOE

(vii) âˆ COD and âˆ AOC

(viii) âˆ COD and âˆ BOC

(ix) âˆ AOC and âˆ COE

(x) âˆ AOD and âˆ DOE

8.Â 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 360Â°.

3x+3x+150+x = 360Â°

â‡’ 7x = 360 - 150

â‡’ 7x = 210

â‡’ x = 210/7

â‡’ x = 30

Value of x is 30Â°.

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

**Solution**

Since,

âˆ AOB andÂ âˆ BOC are linear pairs,

âˆ AOB andÂ âˆ BOC = 180Â°

70+2x = 180Â°

â‡’ 2x = 180Â°-70Â°

â‡’ 2x = 180Â°-70Â°

â‡’ x = 110Â°/2

â‡’ x = 55Â°

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

**Solution**

The figure is given as follows:

Since âˆ POQandâˆ QOS are linear pairs

âˆ POQ+âˆ QOS=180âˆ˜

âˆ POQ+âˆ QOS=180âˆ˜

âˆ POQ+âˆ QOR+âˆ SOR=180âˆ˜

60 + 4x +40 = 180

â‡’ 4x = 180 -100

â‡’ 4x = 80

â‡’ x = 20

Hence, Value of x = 20

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

**Solution**

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

Thus, âˆ ACD and âˆ BCDform a linear pair.

Â Therefore, their sum must be equal to 180Â°.

Or, we can say that

âˆ ACD + âˆ BCD = 180Â°

Also, âˆ ACD = 4x and âˆ BCD = 5x

This is simplified as,

4x + 5x = 180Â°

â‡’ 9x = 180Â°

â‡’ x = 180/9

â‡’ x = 20

Hence, the value of x is 20Â°.

âˆ ACD + âˆ BCD = 180Â°

Also, âˆ ACD = 4x and âˆ BCD = 5x

This is simplified as,

4x + 5x = 180Â°

â‡’ 9x = 180Â°

â‡’ x = 180/9

â‡’ x = 20

Hence, the value of x is 20Â°.

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

Â

**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

â‡’ a = 180 â€“ b â€”(1)

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

a = b + 90/3

â‡’ a = b + 30

â‡’ a â€“ b = 30Â â€”(2)

Equating (1) and (2),

180 â€“ b = b + 30

â‡’ 180 â€“ 30 = 2b

â‡’ b = 150 / 2

â‡’ b = 75

From (1)

a = 180 â€“ b

â‡’ a = 180 â€“ 75

â‡’ a = 105

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

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

**Solution**

Since, âˆ AOCandâˆ BOC are linear pairs âˆ AOC+âˆ BOC=180Â°

6y + 30 + 4y = 180

â‡’ 10y + 30 = 180

â‡’ 10y = 180 â€“ 30

â‡’ 10y = 150

â‡’ y = 150/10

â‡’ y = 15

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

(i) Find the measure of âˆ FOE, âˆ COB and âˆ 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 âˆ DOG,âˆ COF,âˆ BOF,âˆ AOD

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

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

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

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

**Solution**

17. In Fig. 8.46, ray OS stand on a line POQ, Ray OR and ray OT are angle bisectors of âˆ POS andâˆ SOQ respectively. If âˆ POS = x, find âˆ ROT.

**Solution**

18. In Fig, 8.47, lines PQ and RS intersect each other at point O. If âˆ POR: âˆ 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 âˆ ROS = 1/2 (âˆ QOS - POS)

**Solution**Â

Given that,

OR is perpendicular to line PQ

To prove,

âˆ ROS = 1/2(âˆ QOS â€“ âˆ POS)

A/q,

âˆ POR = âˆ ROQ = 90Â° (Perpendicular)

âˆ QOS = âˆ ROQ + âˆ ROS = 90Â° + âˆ ROS --- (i)

âˆ POS = âˆ POR - âˆ ROS = 90Â° - âˆ ROS --- (ii)

Subtracting (ii) from (i)

âˆ QOS - âˆ POS = 90Â° + âˆ ROS - (90Â° - âˆ ROS)

â‡’ âˆ QOS - âˆ POS = 90Â° + âˆ ROS - 90Â° + âˆ ROS

â‡’ âˆ QOS - âˆ POS = 2âˆ ROS

â‡’ âˆ ROS = 1/2(âˆ QOS â€“ âˆ POS)

Hence, Proved.