Chapter 6 Lines and Angles
Exercise – 6.1
Question 1 – In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Solution –
Given – ∠AOC + ∠BOE = 70°
∠BOD = 40°
∠AOC and ∠BOD are vertically opposite angles
∴ ∠AOC = ∠BOD = 40°
AB is a straight line,
∴ ∠AOC + ∠COE + ∠EOB = 180°
or (∠AOC + ∠BOE) + ∠COE = 180° …………. ( 1 )
put ∠AOC + ∠BOE = 70° in equation (1)
70° + ∠COE = 180°
∠COE = 180° – 70° = 110°
∴ Reflex ∠COE = 360° – 110° = 250°
∠AOC + ∠BOE = 70° …………. ( 2 )
put ∠AOC = 40° in equation ( 2)
∴ 40° + ∠BOE = 70°
∠BOE = 70° – 40° = 30°
∠BOE = 30° and reflex ∠COE = 250°
Question 2. In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.
Solution –
Question 3. In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Solution –
Given – ∠PQR = ∠PRQ ………..(1 )
∠PQR + ∠PQS = 180° [Linear pair of angles ]
∠PQR = 180° – ∠PQS ………..(2 )
Similarly,
∠PRQ + ∠PRT = 180° [Linear Pair of angles ]
∠PRQ = 180° – ∠PRT ……… (3)
From (1) , (2) and (3) –
put value ∠PQR and ∠PRQ from equation (1) and (2) in equation (1)
∠PQR = ∠PRQ
180° – ∠PQS = 180° – ∠PRT
– ∠PQS = – ∠PRT
∴ ∠PQS = ∠PRT
Question 4.
In figure, if x + y = w + ⇒, then prove that AOB is a line.
Solution:
Given – (x + y) = (w + z)
Sum of all the angles at a point O = 360°
∴ x + y + w + z = 360° …………. ( 1 )
Put (x + y) = (w + z) in equation (1)
(x + y) + (w + z) = 360°
(x + y) + (x + y) = 360°
2(x + y) = 360°
or, (x + y) = 360 / 2 = 180°
∠x and ∠y are linear pair angles.
if x+y = 180° then uncommon arms of these linear pair angles form a line .
∴ AOB is a straight line.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles
