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.
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.
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,
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.
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.
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°.
Solution: According to question, OP is bisector of ∠BOC
Example 4: In figure OA and OB are opposite rays :
(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.
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.
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.
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.
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)
But ∠AOC = ∠COB (Given)
∴ ∠AOC + ∠ OC = 180º
⇒ 2∠AOC = 180º
⇒ ∠AOC = 90º
Example 10: In fig if ∠AOC + ∠BOD = 70º, find ∠COD.
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.
Solution: 2y + 3y + 5y = 180º
⇒ 10y = 180º
⇒ y = 180°/10º = 18º
