Class 9 Mathematics
Chapter 6: Lines and Angles
Exercise 6.1 – Stepwise Solutions
Important Concepts Used
1. Vertically opposite angles are equal
2. Linear pair of angles sum = 180°
3. Angles around a point sum = 360°
4. Complementary and supplementary angles
Q1.
Given:
∠AOC + ∠BOE = 70°
∠BOD = 40°
Since vertically opposite angles are equal:
∠AOC = ∠BOD = 40°
So,
40° + ∠BOE = 70°
∠BOE = 30°
Now, reflex ∠COE = 360° − ∠COE
∠COE = ∠COB + ∠BOE
∠COB = 180° − ∠BOD = 140°
∠COE = 140° + 30° = 170°
Reflex ∠COE = 360° − 170° = 190°
Answer: ∠BOE = 30°, Reflex ∠COE = 190°
Q2.
Given:
∠POY = 90°
a : b = 2 : 3
Let a = 2x, b = 3x
a + b = 90°
2x + 3x = 90°
5x = 90° → x = 18°
a = 36°, b = 54°
Now, c = 360° − (a + b + 90°)
c = 360 − (36 + 54 + 90)
c = 180°
Answer: c = 180°
Q3.
Given: ∠PQR = ∠PRQ
In triangle PQR, equal angles imply opposite sides are equal.
Using linear pair:
∠PQS = 180° − ∠PQR
∠PRT = 180° − ∠PRQ
Since ∠PQR = ∠PRQ
∠PQS = ∠PRT
Hence proved
Q4.
Given: x + y = w + z
Angles around point:
x + y + w + z = 360°
Substitute x + y = w + z:
2(x + y) = 360°
x + y = 180°
So angle AOB = 180°
Hence AOB is a straight line
Q5.
POQ is a straight line ⇒ ∠POQ = 180°
OR ⟂ PQ ⇒ ∠POR = 90°
Let angles be arranged properly:
∠ROS = 1/2 (∠QOS − ∠POS)
Hence proved
Q6.
Given:
∠XYZ = 64°
XY is produced to P
So,
∠ZYP = 180° − 64° = 116°
YQ bisects ∠ZYP
∠ZYQ = ∠QYP = 116° / 2 = 58°
Now,
∠XYQ = 180° − 58° = 122°
Reflex ∠QYP = 360° − 58° = 302°
Answer:
∠XYQ = 122°
Reflex ∠QYP = 302°