RD Sharma Solutions Class 9 Lines And Angles Exercise 8.2

RD Sharma Class 9 Solutions Chapter 8 Ex 8.2 Free Download

RD Sharma Solutions Class 9 Chapter 8 Ex 8.2

Q 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?

1

Solution:

(i) Given to us here;

the value of x = 25

As you can see, ∠AOC and ∠BOC  form a linear pair

∠AOC and ∠BOC = 180°

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

∠AOC + ∠BOC = 180°

(2y + 5) + 3x = 180

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

2y + 5 + 75 = 180

2y + 80 = 180

2y = 180 – 80  = 100

y = 100/2 = 50

Therefore, y = 50

(ii) Given to us here;

the value of y = 35

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

(2y+5)+3x  = 180

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

(70+5) + 3x = 180

3x = 180 – 75

3x  = 105

Therefore, x = 35

 

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

2

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

(i)∠AOC, ∠COB

(ii)∠AOD, ∠BOD

(iii)∠AOD, ∠COD

(iv)∠BOC, ∠COD

And Linear pair of angles are AOD,∠BOD,∠AOC,∠BOC.

 

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

3

Solution: As,AOD and ∠BOD forms a line pair,

Therefore, ∠AOD+ ∠BOD=180°

Or, ∠AOD+∠BOC+∠COD=180°

In the figure,

∠AOD=(x+10)° ,∠COD=x° ,∠BOC=(x+20)°

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

3x + 30 = 180

3x = 180 – 30

3x = 150/3

x = 50

Now,

AOD=(x+10) =50 + 10 = 60

∠COD = x = 50°

∠BOC = (x+20) = 50 + 20 = 70

Hence, ∠AOD=60°, ∠COD=50°, ∠BOC=70°

 

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

4

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 AOB and ∠BOX are linear pair angles, therefore,

∠AOB+∠BOX = 180°

Or, ∠AOB + ∠BOC + ∠COF = 180° —–—–(1)

Also,

∠AOE and ∠EOX are linear pair angles, therefore,

∠AOE+∠EOF=180°

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

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

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

∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 180°

Hence, proved.

 

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

5

Solution: Given to us here;

∠AOC and ∠BOC form a linear pair

Now, if a – 2b = 30°, we have to find the value of a and b.

From the figure, we can see,

∠AOC = a° ,∠BOC = b°

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

Already given,  a – 2b = 30°    —–(2)

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

a + b – a + 2b = 180 – 30

3b = 150

b = 150/3

b = 50°

Since, a – 2b = 30°

a – 2(50) = 30

a = 30 + 100

a = 130°

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

 

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

Solution: 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.

class 9 exercise 8.2 question no.6

The 4 pair of adjacent angles are :

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

 

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

6

Solution: Number of Pairs of 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

Hence, there are 10 pair of adjacent angles.

 

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

7

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

3x + 3x + 150 + x = 360°

7x = 360° – 150°

7x = 210°

x = 210/7

x = 30°

Hence, the value of x is 30°.

 

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

8

Solution: Since, from the figure we can see, AOB and ∠BOC are linear pairs,

∠AOB +∠BOC =180°

70 + 2x = 180

2x = 180 – 70

2x = 110

x = 110/2

x = 55

Therefore, the value of x is 55°.

 

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

9

Solution: As from the figure, we can see, POQ and ∠QOS are linear pairs. Therefore,

∠POQ + ∠QOS=180°

∠POQ + ∠QOR+∠SOR=180°

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 DCA = 5x and ∠DCB = 4x. Find the value of x?

10

Solution: As you can see, ACD and ∠BCD form a linear pair.

Therefore, ACD + ∠BCD = 180°

∠DCA = 5x and ∠DCB = 4x

5x + 4x = 180°

9x = 180°

x = 20

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

 

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

11

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

So, ∠POR + ∠QOR = 180°

Also, given to us here;

∠POR =3x and ∠QOR = 2x + 10

2x + 10 + 3x = 180

5x + 10 = 180

5x = 180 – 10

5x = 170

x = 34

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

 

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

12

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

a + b = 180

a = 180 – b  —–(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 – b = 30 —–(2)

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

180 – b = b + 30

180 – 30 = 2b

b = 150 / 2

b = 75

From eq. (1)

a = 180 – b

a = 180 – 75

a = 105

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

 

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

13

Solution: From the figure, AOC and ∠BOC are linear pairs. Therefore;

∠AOC+ ∠BOC=180°

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

6y + 30 + 4y = 180

10y + 30 = 180

10y = 180 – 30

10y = 150

y = 150/10

y = 15

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

 

Q 15 : If the figure below 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

14

Solution: (i) Let FOE = a, ∠DOE = b and ∠BOC = c

Since AOF and ∠FOG are linear pair, therefore;

AOF  + ∠FOG = 180

∠AOF + 30 = 180               [Given, ∠FOG = 30]

∠AOF = 180 – 30

∠AOF= 150

Or, ∠AOB +∠BOC+∠COD+∠DOE+∠EOF=150

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

a + b + c =150 – 30 – 30

a + b + c =90 ——————–(1)

Given, ∠FOC = 90°

So, ∠FOE +∠EOD+∠DOC =90°

a + b +30 = 90

a + b = 90 – 30

a + b =60 —–(2)

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

a + b + c = 90

60 + c = 90

c = 90 – 60 = 30°

Given, BOE = 90°

∠BOC + ∠COD + ∠DOE = 90°

30 + 30 + DOE = 90°

DOE = 90 – 60 = 30°

DOE = a = 30°

From eq.2 we know,

a + b = 60

b = 60 – a

b = 60 – 30

b = 30°

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

(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)

 

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

class 9 ex8.2 q.16

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 TOP and ∠POQ is a linear pair angle, therefore;

∠TOQ+∠POQ =180°                         —–(1)

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

∠TOS +∠SOQ =180°                       —–(2)

But,

SOQ = ∠SOR + ∠QOR                  —–(3)

Now, equation (2) can be written as;

∠TOS+ ∠SOR +∠OQR =180°

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

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

∠TOP + ∠TOS = ∠POS

Therefore, eqaution 4 becomes;

∠POQ + ∠QOR + ∠SOR + ∠POS = 360°

 

Q 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?

16

Solution: As per the given question,

Ray OS stand on the line POQ.

And Ray OR and Ray OT are the angle bisectors of POS and ∠SOQ, respectively.

Given, ∠POS = x

Since, ∠POS and ∠SOQ is linear pair

Therefore, ∠POS + ∠QOS =180°

x + QOS = 180

Or, ∠QOS = 180 – x

Now, ray OR is the bisector POS, therefore;

∠ROS = 1/2 ∠POS

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

In the same way, ray OT bisects  QOS, therefore,

∠TOS = 1/2 ∠QOS

∠TOS = (180 – x)/2       [ QOS = 180 – x ]

∠TOS = 90 – x/2   ……………..2

Therefore, ROT = ∠ROS + ∠SOT

= x/2 + 90 – x/2      [From eq.1 and eq.2]

= 90

Therefore, ∠ROT = 90°

 

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

17

Solution: Given to us here;

From the figure,

∠POR and ∠ROQ are a linear pair of angles.

Therefore, ∠POR + ∠ROQ = 180°

As per the question is given;

∠POR: ∠ROQ = 5:7

So, POR = (5/12) × 180=75

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

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

Also, SOQ = POR = 75°    [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 ROS = 1/2(∠QOS – ∠POS).

18

Solution: Given to us here;

Since, OR perpendicular to line PQ.

Therefore, POR = 90°

∠POS + ∠SOR = 90                        [ Since, POR = ∠POS+∠SOR]

∠ROS = 90°- ∠POS        ———–(1)

∠QOR = 90                                    (∵ OR ⊥ PQ)

∠QOS – ∠ROS = 90°

∠ROS = ∠QOS -90°

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

2∠ROS = ∠QOS – ∠POS

∠ROS = 1/2(∠QOS – ∠POS)

Hence, proved.

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