Linear Pair Of Angles

Linear Pair Of Angles

Two adjacent angles are said to form a linear pair of angles, if their non-common arms are two opposite rays.

Linear-Pair-Of-Angles

In the adjoining figure, ∠AOC and ∠BOC are two adjacent angles whose non-common arms OA and OB are two opposite rays, i.e., BOA is a line

∴ ∠AOC and ∠BOC form a linear pair of angles.

Theorem 1:

Prove that the sum of all the angles formed on the same side of a line at a given point on the line is 180°.

Given: AOB is a straight line and rays OC, OD and OE stand on it, forming ∠AOC, ∠COD, ∠DOE and ∠EOB.

Linear-Pair-Of-Angle-theorem-1

To prove: ∠AOC + ∠COD + ∠DOE + ∠EOB = 180°.

Proof: Ray OC stands on line AB.

∴ ∠AOC + ∠COB = 180°

⇒ ∠AOC + (∠COD + ∠DOE + ∠EOB) = 180°

[∵ ∠COB = ∠COD + ∠DOE + ∠EOB]

⇒ ∠AOC + ∠COD + ∠DOE + ∠EOB = 180°.

Hence, the sum of all the angles formed on the same side of line AB at a point O on it is 180°.

Theorem 2:

Prove that the sum of all the angles around a point is 360°.

Given: A point O and the rays OA, OB, OC, OD and OE make angles around O.

To prove: ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°

Construction: Draw a ray OF opposite to ray OA.

Proof: Since ray OB stands on line FA,

Linear-Pair-Of-Angle-theorem-2

we have, ∠AOB + ∠BOF = 180°   [linear pair]

∴ ∠AOB + ∠BOC + ∠COF = 180°             ….(i)

[∵ ∠BOF = ∠BOC + ∠COF]

Again, ray OD stands on line FA.

∴ ∠FOD + ∠DOA = 180° [linear pair]

or ∠FOD + ∠DOE + ∠EOA = 180°               …(ii)

[∵ ∠DOA = ∠DOE + ∠EOA]

Adding (i) and (ii), we get,

∠AOB + ∠BOC + ∠COF + ∠FOD + ∠DOE + ∠EOA = 360°

∴ ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°

[∵ ∠COF + ∠FOD = ∠COD]

Hence, the sum of all the angles around a point O is 360°.

Linear Pair Of Angles Example Problems With Solutions

Example 1:    In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find ∠AOC, ∠COD and ∠BOD.

Linear-Pair-Of-Angles-Example-1

Solution:    (3x + 7)° + (2x – 19)° + x° = 180′ (linear pair)

⇒ 6x – 12) = 180°

⇒ 6x = 192°

⇒ x = 32°

∴ ∠AOC = 3x + 7 = 3(32) + 7 = 96 + 7 = 103°

∠COD = 2x – 19 = 2(32) – 19 = 64 – 19 = 45°

∠BOD = x° = 32°.

Example 2:    In figure, OA, OB are opposite rays and ∠AOC + ∠BOD = 90°. Find ∠COD.

Linear-Pair-Of-Angles-Example-2

Solution:    Since OA and OB are opposite rays. Therefore, AB is a line. Since ray OC stands on line AB.

∴ ∠AOC + ∠COB = 180°

⇒ ∠AOC + ∠COD + ∠BOD = 180°

[∵ ∠COB = ∠COD + ∠BOD]

⇒ (∠AOC + ∠BOD) + ∠COD = 180°

⇒ 90° + ∠COD = 180°

[∵ ∠AOC + ∠BOD = 90° (Given)]

⇒ ∠COD = 180° – 90° = 90°

Example 3:    In figure, OP bisects ∠BOC and OQ, ∠AOC. Show that ∠POQ = 90°.

Linear-Pair-Of-Angles-Example-3

Solution:    According to question, OP is bisector of ∠BOC

Linear-Pair-Of-Angles-Example-3-1

Example 4:    In figure OA and OB are opposite rays :

Linear-Pair-Of-Angles-Example-4

(i) If x = 75, what is the value of y ?

(ii) If y = 110, what is the value of x ?

Solution:    Since ∠AOC and ∠BOC form a linear pair.

Therefore, ∠AOC + ∠BOC = 180º

⇒ x + y = 180º        …(1)

(i) If x = 75, then from (i)

75 + y = 180º

y = 105º.

(ii) If y = 110 then from (i)

x + 110 = 180

⇒ x = 180 – 110 = 70.

Example 5:    In figure ∠AOC and ∠BOC form a linear pair. Determine the value of x.

Linear-Pair-Of-Angles-Example-5

Solution:    Since ∠AOC and ∠BOC form a linear pair.

∴ ∠AOC + ∠BOC = 180º

⇒ 4x + 2x = 180º

⇒ 6x = 180º

⇒ x = 180/6 = 30º

Thus, x = 30º

Example 6:    In figure OA, OB are opposite rays and

∠AOC + ∠BOD = 90º. Find ∠COD.

Linear-Pair-Of-Angles-Example-6

Solution:    Since OA and OB are opposite rays. Therefore, AB is a line. Since ray OC stands on line AB. Therefore,

∠AOC + ∠COB = 180º      [Linear Pairs]

⇒ ∠AOC + ∠COD + ∠BOD = 180º

[∵ ∠COB = ∠COD + ∠BOD]

⇒ (∠AOC + ∠BOD) + ∠COD = 180º

⇒ 90º + ∠COD = 180º

[∵ ∠AOC + ∠BOD = 90º (Given)]

⇒ ∠COD = 180º – 90º

⇒ ∠COD = 90º

Example 7:    In figure ray OE bisects angle ∠AOB and OF is a ray opposite to OE. Show that

∠FOB = ∠FOA.

Linear-Pair-Of-Angles-Example-7

Solution:    Since ray OE bisects angle AOB. Therefore,

∠EOB = ∠EOA ….(i)

Now, ray OB stands on the line EF.

∴ ∠EOB + ∠FOB = 180º     …(ii)       [linear pair]

Again, ray OA stands on the line EF.

∴ ∠EOA + ∠FOA = 180º    ….(iii)

Form (ii) and (iii), we get

∠EOB + ∠FOB = ∠EOA + ∠FOA

⇒ ∠EOA + ∠FOB = ∠EOA + ∠FOA

[∵ ∠EOB = ∠EOA (from (i)]

⇒ ∠FOB = ∠FOA.

Example 8:    In figure OE bisects ∠AOC, OF bisects ∠COB and OE ⊥OF. Show that A, O, B are collinear.

Linear-Pair-Of-Angles-Example-8

Solution:    Since OE and OF bisect angles AOC and COB respectively. Therefore,

∠AOC = 2∠EOC ….(i)

and ∠COB = 2∠COF ….(ii)

Adding (i) and (ii), we get

∠AOC + ∠COB = 2∠EOC + 2∠COF

⇒ ∠AOC + ∠COB = 2(∠EOC + ∠COF)

⇒ ∠AOC + ∠COB = 2(∠EOF)

⇒ ∠AOC + ∠COB = 2 × 90º

[∵ OE ⊥ OF ∴ ∠EOF = 90º]

⇒ ∠AOC + ∠COB = 180º

But ∠AOC and ∠COB are adjacent angles.

Thus, ∠AOC and ∠COB are adjacent supplementary angles. So, ∠AOC and ∠COB form a linear pair. Consequently OA and OB are two opposite rays. Hence, A, O, B are collinear.

Example 9:    If ray OC stands on line AB such that

∠AOC = ∠COB, then show that

∠AOC = 90º.

Solution:    Since ray OC stands on line AB. Therefore,

∠AOC + ∠COB = 180º [Linear pair] …(i)

Linear-Pair-Of-Angles-Example-9

But ∠AOC = ∠COB     (Given)

∴ ∠AOC + ∠ OC = 180º

⇒ 2∠AOC = 180º

⇒ ∠AOC = 90º

Example 10:    In fig if ∠AOC + ∠BOD = 70º, find ∠COD.

Linear-Pair-Of-Angles-Example-10

Solution:    ∠AOC + ∠COD + ∠BOD = 180º

or   (∠AOC + ∠BOD) + ∠COD = 180º

or   70º + ∠COD = 180º

or   ∠COD = 180º – 70º

or   ∠COD = 110º

Example 11:    In fig. find the value of y.

Linear-Pair-Of-Angles-Example-11

Solution:    2y + 3y + 5y = 180º

⇒ 10y = 180º

⇒ y = 180°/10º = 18º

 

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