Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

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?

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans :

(i) Given that,

x = 25

Since ∠AOCand∠BOC form a linear pair

∠AOC+∠BOC = 180

Given that ∠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

y = 50

(ii) Given that,

y = 35

∠AOC+∠BOC = 180

(2y+5)+3x  = 180

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

(70+5) + 3x = 180

3x = 180 – 75

3x  = 105

x = 35

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

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : Adjacent angles are :

(i)∠AOC,∠COB

(ii)∠AOD∠BOD

(iii)∠AOD,∠COD

(iv)∠BOC,∠COD

Linear pairs : ∠AOD,∠BOD,∠AOC,∠BOC

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

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : Since  ∠AOD and ∠BOD form a line pair,

∠AOD+∠BOD = 180

∠AOD+∠BOC+∠COD = 180

Given that,

∠AOD = (x+10)∘,∠COD = x,∠BOC = (x+20)

  • (x + 10) + x + (x + 20) = 180
  • 3x + 30 = 180
  • 3x = 180 – 30
  • 3x = 150/3
  • x = 50

Therefore, ∠AOD=(x+10)

= 50 + 10 = 60

∠COD = x = 50

∠COD = (x+20)

= 50 + 20 = 70

∠AOD = 60∠COD = 50 ∠BOC = 70

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

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : Given that OA,OB,OD and OE have the common end point O.

A ray opposite to OA is drawn

Since ∠AOB,∠BOF are linear pairs,

∠AOB+∠BOF = 180

∠AOB+∠BOC+∠COF = 180  —–(1)

Also,

∠AOEand∠EOF are linear pairs

∠AOE+∠EOF = 180

∠AOE+∠DOF+∠DOE = 180      —–(2)

By adding (1) and (2) equations 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 ?

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : Given that,

∠AOC and ∠BOC form a linear pair

If a – b = 30

∠AOC = a,∠BOC = b

Therefore,  a +b  =180   —–(1)

Given a – 2b = 30       —–(2)

By subtracting (1) and (2)

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

Hence, the values of a and b are 130∘and50∘ respectively.

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

Ans : 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 :

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

Hence, 4 pairs of adjacent angles are formed when two lines intersect at a point.

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

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : Pairs of adjacent angles are :

∠EOC,∠DOC

∠EOD,∠DOB

∠DOC,∠COB

∠EOD,∠DOA

∠DOC,∠COA

∠BOC,∠BOA

∠BOA,∠BOD

∠BOA,∠BOE

∠EOC,∠COA

∠EOC,∠COB

Hence,  10 pair of adjacent angles.

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

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : 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∘

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

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : Since ∠AOBand∠BOC are linear pairs,

∠AOB+∠BOC = 180

  • 70 + 2x = 180
  • 2x = 180 – 70
  • 2x = 110
  • x = 110/2
  • x = 55

Hence, the value of x is 55

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

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : Since ∠POQand∠QOS are linear pairs

∠POQ+∠QOS = 180

  • ∠POQ+∠QOR+∠SOR = 180
  • 60 + 4x +40 = 180
  • 4x = 180 -100
  • 4x = 80
  • x = 20

Hence, Value of x = 20

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

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans: Here, ∠ACD+∠BCD = 180

[Since they are linear pairs]

∠DCA = 5xand∠DCB = 4x

  • 5x + 4x = 180
  • 9x = 180
  • x = 20

Hence, the value of 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 ?

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : For the case that POR is a line

∠POR and ∠QORarelinearparts

∠POR+∠QOR = 180

Also, given that,

∠POR = 3x and ∠QOR = 2x+10

  • 2x + 10 + 3x = 180
  • 5x + 10 = 180
  • 5x = 180 – 10
  • 5x = 170
  • x = 34

Hence the value of 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 ?

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : 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.

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

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans :      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

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,∠COBand∠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

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans :(i) ∠FOE = x,∠DOE = y and ∠BOC = z

Since ∠AOF,∠FOG is a linear pair

  • ∠AOF + 30 = 180
  • ∠AOF = 180 – 30
  • ∠AOF = 150
  • ∠AOB+∠BOC+∠COD+∠DOE+∠EOF = 150
  • 30 + z + 30 + y + x = 150
  • x + y + z =150 – 30 – 30
  • x + y + z =90 —–(1)

∠FOC = 90

  • ∠FOE+∠EOD+∠DOC = 90
  • x + y +30 = 90
  • x + y = 90 – 30
  • x + y =60 —–(2)

Substituting (2) in (1)

x + y + z = 90

  • 60 + z = 90
  • z = 90 – 60 = 30

Given BOE = 90

  • ∠BOC+∠COD+∠DOE = 90
  • 30 + 30 +DOE = 90
  • DOE = 90 – 60 = 30
  • DOE = x = 30

We also know that,

x + y = 60

  • y = 60 – x
  • y = 60 – 30
  • y = 30

Thus we have ∠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, 0Q, OR and OS are four rays. Prove that: ∠POQ+∠QOR+∠SOR+∠POS=360

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans:   Given that

OP, OQ, OR and OS are four rays

You need to produce any of the ray OP, OQ, OR and OS backwards to a point in the figure.

Let us produce ray OQ backwards to a point T

So that TOQ is a line

Ray OP stands on the TOQ

Since ∠TOP,∠POQ is a linear pair

∠TOP+∠POQ = 180                          —–(1)

Similarly,

Ray OS stands on the line TOQ

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

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

So, eqn (2) becomes

∠TOS+∠SOR+∠OQR = 180°

Now, adding (1) and (3) you get ∠TOP+∠POQ+∠TOS+∠SOR+∠QOR = 360  —–(4)

∠TOP+∠TOS=∠POS

Eqn: (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 ∠POSand∠SOQrespectively. If ∠POS = x, find ∠ROT ?

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : Given,

Ray OS stand on a line POQ

Ray OR and Ray OT are angle bisectors of  ∠POSand∠SOQ respectively

∠POS = x

∠POSand∠SOQ is linear pair

∠POS+∠QOS = 180

x + QOS = 180

  • QOS = 180 – x

Now, ray or bisector POS

∠ROS=1/2∠POS

  • x/2

ROS = x/2            [Since POS = x ]

Similarly ray OT bisector QOS

∠TOS = 1/2∠QOS

= (180 – x)/2       [ QOS = 180 – x ]

= 90 – x/2

Hence, ∠ROT=∠ROS+∠ROT

= x/2 + 90 – x/2

= 90

∠ROT = 180

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.

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : Given

∠POR and ∠ROP is linear pair

∠POR+∠ROP = 180

Given that

∠POR:∠ROQ=5:7

Hence, POR =(5/12)x180=75

Similarly ROQ=(7/7+5) 180 = 105

Now POS = ROQ = 105° [Vertically opposite angles]

Also, SOQ = POR = 75°    [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).

RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : Given that

OR perpendicular

∴ ∠POR=90

∠POS+∠SOR=90         [∵ ∠POR=∠POS+∠SOR]

∠ROS=90°−∠POS         —–(1)

∠QOR = 90 (∵ OR ⊥ PQ)

∠QOS–∠ROS = 90

∠ROS=∠QOS−90

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

2∠ROS =∠QOS–∠POS

∠ROS = 1/2(∠QOS–∠POS)

The document RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on RD Sharma Solutions Ex-8.2, Lines And Angles, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What are lines and angles?
Ans. Lines and angles are fundamental concepts in geometry. A line is a straight path that extends indefinitely in both directions. An angle is formed when two lines or line segments meet at a common point. It is measured in degrees.
2. What are the different types of angles?
Ans. There are several types of angles: - Acute angle: An angle that measures less than 90 degrees. - Right angle: An angle that measures exactly 90 degrees. - Obtuse angle: An angle that measures more than 90 degrees but less than 180 degrees. - Straight angle: An angle that measures exactly 180 degrees. - Reflex angle: An angle that measures more than 180 degrees but less than 360 degrees.
3. How do you classify lines based on their position?
Ans. Lines can be classified based on their position in relation to each other. Some common classifications include: - Intersecting lines: Lines that cross each other at a point. - Parallel lines: Lines that never intersect and are always the same distance apart. - Perpendicular lines: Lines that intersect at a right angle (90 degrees). - Skew lines: Lines that are not in the same plane and do not intersect.
4. How do you find angles in a triangle?
Ans. In a triangle, the sum of the three interior angles is always 180 degrees. To find the measure of a specific angle, you can use various properties and theorems, such as the Angle Sum Property of a Triangle, the Exterior Angle Theorem, or the properties of similar triangles.
5. How can angles help solve real-life problems?
Ans. Angles have various applications in real-life situations. They are used in navigation, architecture, engineering, and many other fields. For example, in navigation, angles are used to determine the direction of travel or the position of an object in relation to another. In architecture and engineering, angles are used to design structures, calculate forces, and ensure stability.
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