## RD Sharma Solutions Class 9 Chapter 8 Ex 8.2

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

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

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

*Solution:*

**(i) **Given to us here;

the value of x = 25

As you can see, âˆ AOCÂ and âˆ BOCÂ Â form a linear pair

âˆ AOCÂ and âˆ BOC = 180Â°

From the figure, âˆ AOC= 2y + 5Â and âˆ BOC = 3x

âˆ AOC + âˆ BOC = 180Â°

(2y + 5) + 3x = 180

(2y +5) + 3 (25) = 180

2y + 5 + 75 = 180

2y + 80 = 180

2y = 180 â€“ 80Â = 100

y = 100/2 = 50

Therefore, y = 50

**(ii) **Given to us here;

the value of y = 35

Since, âˆ AOC + âˆ BOC = 180Â° (Linear pair angles)

(2y+5)+3x Â = 180

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

(70+5) + 3x = 180

3x = 180 â€“ 75

3x Â = 105

Therefore, x = 35

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

** Solution: **From the given figure, theÂ Adjacent angles are :

(i)âˆ AOC, âˆ COB

(ii)âˆ AOD, âˆ BOD

(iii)âˆ AOD, âˆ COD

(iv)âˆ BOC, âˆ COD

And Linear pair of angles areÂ âˆ AOD,âˆ BOD,âˆ AOC,âˆ BOC.

*Q 3 : In the given below the figure, find x. Further find âˆ COD,âˆ AOD,âˆ BOC.*

** Solution: **As,âˆ AOD and âˆ BODÂ forms a line pair,

Therefore, âˆ AOD+ âˆ BOD=180Â°

Or, âˆ AOD+âˆ BOC+âˆ COD=180Â°

In the figure,

âˆ AOD=(x+10)Â° ,âˆ COD=xÂ° ,âˆ BOC=(x+20)Â°

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

3x + 30 = 180

3x = 180 â€“ 30

3x = 150/3

x = 50

Now,

âˆ AOD=(x+10)Â =50 + 10 = 60

âˆ COD = x = 50Â°

âˆ BOC = (x+20)Â = 50 + 20 = 70

Hence, âˆ AOD=60Â°, âˆ COD=50Â°, âˆ BOC=70Â°

*Q 4 : In the Given below figureÂ rays OA, OB, OC, OD and OE have the common end point 0. Show that âˆ AOB+âˆ BOC+âˆ COD+âˆ DOE+âˆ EOA=360Â°.*

** Solution: **Given to us here

**ray OA, OB, OC, OD and OE have the common endpoint O.**

*,*A ray OX is drawn opposite to OA, as shown in the figure.

Since âˆ AOB and âˆ BOXÂ are linear pair angles, therefore,

âˆ AOB+âˆ BOX = 180Â°

Or, âˆ AOB + âˆ BOC + âˆ COF = 180Â° —–—–(1)

Also,

âˆ AOE and âˆ EOXÂ are linear pair angles, therefore,

âˆ AOE+âˆ EOF=180Â°

Or, âˆ AOE+âˆ DOF+âˆ DOE=180Â° Â Â Â Â Â Â Â Â —–(2)

By adding equations, (1) and (2), we get;

âˆ AOB + âˆ BOC + âˆ COF + âˆ AOE + âˆ DOF + âˆ DOE = 180Â°

âˆ AOB + âˆ BOC + âˆ COD + âˆ DOE + âˆ EOA = 180Â°

Hence, proved.

*Q 5 : In the Below figure, âˆ AOC and âˆ BOCÂ form a linear pair. If a – 2b =Â 30Â°,Â find a and b?*

** Solution: **Given to us here;

âˆ AOC and âˆ BOCÂ form a linear pair

Now, if a â€“ 2b = 30Â°, we have to find the value of a and b.

From the figure, we can see,

âˆ AOC = aÂ° ,âˆ BOC = bÂ°

Thus, a + b = 180Â°Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â —–(1)

Already given,Â a – 2b = 30Â°Â Â —–(2)

On subtracting equation (1) and (2) here, we get;

a + b – a + 2b = 180 – 30

3b = 150

b = 150/3

b = 50Â°

Since, a â€“ 2b = 30Â°

a â€“ 2(50) = 30

a = 30 + 100

a = 130Â°

Therefore, the values of a and b are 130Â° and 50Â° respectively.

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

** Solution:Â **Four pairs of adjacent angles are formed when two lines intersect each other at a single point.

Let, two lines AB and CD intersect at point O.

The 4 pair of adjacent angles are :

(âˆ AOD,âˆ DOB),(âˆ DOB,âˆ BOC),(âˆ COA, âˆ AOD) and (âˆ BOC,âˆ COA).

*Q 7: How many pairs of adjacent angles, in all, can you name in the figure below?*

** Solution: **Number ofÂ Pairs of adjacent angles, from the figure, are :

âˆ EOC and âˆ DOC

âˆ EOD and âˆ DOB

âˆ DOC and âˆ COB

âˆ EOD and âˆ DOA

âˆ DOC and âˆ COA

âˆ BOC and âˆ BOA

âˆ BOA and âˆ BOD

âˆ BOA and âˆ BOE

âˆ EOC and âˆ COA

âˆ EOC and âˆ COB

Hence, there are 10 pair of adjacent angles.

*Q.8: In the below figure, find the value of x?*

** Solution:** As we know, the sum of all the angles around a point O is equal to 360Â°. Therefore,

3x + 3x + 150 + x = 360Â°

7x = 360Â° â€“ 150Â°

7x = 210Â°

x = 210/7

x = 30Â°

Hence, the value of x is 30Â°.

*Q.9: In the below figure, AOC is a line, find x.*

** Solution:Â **Since, from the figure we can see,Â âˆ AOB and âˆ BOCÂ are linear pairs,

âˆ AOB +âˆ BOC =180Â°

70 + 2x = 180

2x = 180 â€“ 70

2x = 110

x = 110/2

x = 55

Therefore, the value of x is 55Â°.

*Q.10: In the below figure, POS is a line, Find x?*

** Solution:Â **As from the figure, we can see,Â âˆ POQ and âˆ QOSÂ are linear pairs. Therefore,

âˆ POQ + âˆ QOS=180Â°

âˆ POQ + âˆ QOR+âˆ SOR=180Â°

60 + 4x +40 = 180

4x = 180 -100

4x = 80

x = 20

Hence, the value of x is 20.

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

*Solution: As you can see,Â âˆ ACD and âˆ BCD form a linear pair.*

Therefore,Â âˆ ACD + âˆ BCD = 180Â°

âˆ DCA = 5x and âˆ DCB = 4x

5x + 4x = 180Â°

9x = 180Â°

x = 20

Hence, the value of unknown variable, x is 20.

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

** Solution: **As you can see, from the figure, POQ is a straight line. Therefore,âˆ POR and âˆ QOR are linear parts.

So, âˆ POR + âˆ QOR = 180Â°

Also,Â given to us here;

âˆ POR =3x and âˆ QOR = 2x + 10

2x + 10 + 3x = 180

5x + 10 = 180

5x = 180 â€“ 10

5x = 170

x = 34

Hence the value of unknown variable, x is 34Â°.

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

** Solution:Â **Since angle a and b are linear pair, from the figure, therefore,

a + b = 180

a = 180 â€“ bÂ —–(1)

As per given question, a is greater than b by one third of a right angle, therefore;

a = b + 90/3

a = b + 30

a â€“ b = 30 —–(2)

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

180 â€“ b = b + 30

180 â€“ 30 = 2b

b = 150 / 2

b = 75

From eq. (1)

a = 180 â€“ b

a = 180 â€“ 75

a = 105

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

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

** Solution:Â **From the figure,Â âˆ AOC and âˆ BOCÂ are linear pairs. Therefore;

âˆ AOC+ âˆ BOC=180Â°

Putting the values of both the angles, as per the given data;

6y + 30 + 4y = 180

10y + 30 = 180

10y = 180 â€“ 30

10y = 150

y = 150/10

y = 15

Hence value of y that will make AOB a line is 15Â°.

*Q 15 : If the figure below forms a linear pair,*

*âˆ EOB = âˆ FOC =90 and âˆ DOC = âˆ FOG =âˆ AOB = 30*

*Find the measure of âˆ FOE,âˆ COB\;and âˆ 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) LetÂ âˆ **FOE = a, âˆ DOE = b and âˆ BOC = c

Since âˆ AOF and âˆ FOGÂ are linear pair, therefore;

âˆ AOFÂ + âˆ FOG = 180

âˆ AOFÂ + 30 = 180Â Â Â Â Â Â Â Â [Given,Â âˆ FOG = 30]

âˆ AOFÂ = 180 â€“ 30

âˆ AOF= 150

Or, âˆ AOB +âˆ BOC+âˆ COD+âˆ DOE+âˆ EOF=150

30 + c + 30 + b + a = 150Â Â Â Â Â [Given,Â âˆ AOB =âˆ COD=30]

a + b + c =150 â€“ 30 â€“ 30

a + b + c =90 ——————–(1)

Given, âˆ FOC = 90Â°

So, âˆ FOE +âˆ EOD+âˆ DOC =90Â°

a + b +30 = 90

a + b = 90 â€“ 30

a + b =60 —–(2)

Putting the value from eq (2) in (1), we get;

a + b + c = 90

60 + c = 90

c = 90 â€“ 60 = 30Â°

Given, âˆ BOE = 90Â°

âˆ BOC + âˆ COD + âˆ DOE = 90Â°

30 + 30 + âˆ DOE = 90Â°

âˆ DOE = 90 â€“ 60 = 30Â°

âˆ DOE = a = 30Â°

From eq.2 we know,

a + b = 60

b = 60 â€“ a

b = 60 – 30

b = 30Â°

Therefore,Â âˆ FOE=30Â°, âˆ COB=30Â° and âˆ DOE=30Â°

(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)

*Q16: In below fig. OP, OQ, OR and OS are four rays. Prove that: âˆ POQ + âˆ QOR+âˆ SOR + âˆ POS = 360Â° *

** Solution:** Given to us here;

The four rays OP, OQ, OR and OS, meet at a single point O.

Now, produce ray OQ backwards to a point T, to form a ray OT and TOQ is a straight line.

As you can see, ray OP stands on the TOQ.

Since âˆ TOP and âˆ POQÂ is a linear pair angle, therefore;

âˆ TOQ+âˆ POQ =180Â° Â Â Â Â Â Â Â Â Â Â Â Â —–(1)

In the same way, ray OS stands on the line TOQ. Thus,

âˆ TOS +âˆ SOQ =180Â° Â Â Â Â Â Â Â Â Â Â Â —–(2)

But,

âˆ SOQ = âˆ SOR + âˆ QORÂ Â Â Â Â Â Â Â Â —–(3)

Now, equation (2) can be written as;

âˆ TOS+ âˆ SOR +âˆ OQR =180Â°

If we add equation (1) and (3), we get;

âˆ TOP +âˆ POQ +âˆ TOS + âˆ SOR + âˆ QOR = 360Â° Â —–(4)

âˆ TOP + âˆ TOS = âˆ POS

Therefore, eqaution 4 becomes;

âˆ POQ + âˆ QOR + âˆ SOR + âˆ POS = 360Â°

*Q 17 : In below fig, ray OS stand on a line POQ. Ray OR and ray OT are angle bisectors of âˆ Â POS and âˆ SOQÂ respectively. If âˆ POSÂ = x, find âˆ ROT?*

** Solution:** As per the given question,

Ray OS stand on the line POQ.

And Ray OR and Ray OT are the angle bisectors of ** âˆ **POS and âˆ SOQ,Â respectively.

Given, âˆ POS = x

Since, âˆ POS and âˆ SOQÂ is linear pair

Therefore, âˆ POS + âˆ QOS =180Â°

x + âˆ QOS = 180

Or, âˆ QOS = 180 â€“ x

Now, ray OR is the bisector âˆ POS, therefore;

âˆ ROS = 1/2 âˆ POS

âˆ ROS = x/2Â Â …………..1Â Â Â Â Â [Since âˆ POS = x ]

In the same way, ray OT bisectsÂ âˆ QOS, therefore,

âˆ TOS = 1/2 âˆ QOS

âˆ TOSÂ = (180 â€“ x)/2Â Â Â Â Â [ âˆ QOS = 180 â€“ x ]

âˆ TOSÂ = 90 â€“ x/2Â Â ……………..2

Therefore,Â âˆ ROT = âˆ ROS + âˆ SOT

= x/2 + 90 â€“ x/2Â Â Â [From eq.1 and eq.2]

= 90

Therefore, âˆ ROT = 90Â°

** Q 18:**Â

*In the below fig, lines PQ and RS intersect each other at point O. If âˆ POR: âˆ ROQ = 5:7. Find all the angles.*** Solution:Â **Given to us here;

From the figure,

âˆ POR and âˆ ROQ areÂ a linear pair of angles.

Therefore, âˆ POR + âˆ ROQ = 180Â°

As per the question is given;

âˆ POR: âˆ ROQ = 5:7

So,Â âˆ POR = (5/12) Ã— 180=75

In the same way, âˆ ROQ = (7/7+5) 180 = 105

Therefore,Â âˆ POS = âˆ ROQ = 105Â° [Both are Vertically opposite angles]

Also, âˆ SOQ = âˆ POR = 75Â° Â Â [Both are Vertically opposite angles]

*Q 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 âˆ ROSÂ = 1/2(âˆ QOS – âˆ POS). *

** Solution: **Given to us here;

Since, OR perpendicular to line PQ.

Therefore,Â âˆ POR = 90Â°

âˆ POS + âˆ SOR = 90Â Â Â Â Â Â Â Â Â Â Â Â Â [ Since,Â âˆ POR = âˆ POS+âˆ SOR]

âˆ ROS = 90Â°- âˆ POSÂ Â Â Â ———–(1)

âˆ QORÂ = 90Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (âˆµ OR âŠ¥ PQ)

âˆ QOS – âˆ ROS = 90Â°

âˆ ROS = âˆ QOS -90Â°

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

2âˆ ROS = âˆ QOS – âˆ POS

âˆ ROS = 1/2(âˆ QOS – âˆ POS)

Hence, proved.