Exercise 6.1 Solutions (Class 9)
Lines and Angles
1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
>∠AOC = ∠BOD (Vertically opposite angles). Since ∠BOD = 40°, then ∠AOC = 40°.
Given ∠AOC + ∠BOE = 70°.
Substitute ∠AOC = 40°: 40° + ∠BOE = 70° => ∠BOE = 30°.
Since AB is a straight line, the sum of angles on it is 180°.
∠AOC + ∠COE + ∠BOE = 180°.
We know (∠AOC + ∠BOE) = 70°. So, 70° + ∠COE = 180° => ∠COE = 110°.
Reflex ∠COE = 360° – ∠COE = 360° – 110° = 250°.
2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
>Since XY is a straight line, ∠XOM + ∠MOP + ∠POY = 180°.
b + a + 90° = 180° => a + b = 90°.
Given a : b = 2 : 3. Let a = 2x and b = 3x.
2x + 3x = 90° => 5x = 90° => x = 18°.
So, a = 2 × 18° = 36° and b = 3 × 18° = 54°.
Since MN is a straight line, b + c = 180° (Linear pair).
54° + c = 180° => c = 180° – 54° = 126°.
Answer: c = 126°.
3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
>Proof:
ST is a straight line. Angles on the line at point Q form a linear pair.
∠PQS + ∠PQR = 180° => ∠PQS = 180° – ∠PQR. —(1)
Similarly, at point R:
∠PRT + ∠PRQ = 180° => ∠PRT = 180° – ∠PRQ. —(2)
We are given that ∠PQR = ∠PRQ.
From equations (1) and (2), since the right-hand sides are equal, the left-hand sides must also be equal.
Therefore, ∠PQS = ∠PRT. (Proved)
4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.
>The sum of all angles around a point O is 360°.
x + y + w + z = 360°.
It is given that x + y = w + z.
Substitute (x+y) for (w+z) in the sum equation:
(x + y) + (x + y) = 360°
2(x + y) = 360° => x + y = 180°.
Since x and y are adjacent angles and their sum is 180°, they form a linear pair. A linear pair of angles can only be formed on a straight line.
Therefore, AOB is a line. (Proved)
5. In Fig. 6.17, 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).
>Proof:
Since OR ⊥ PQ, ∠ROP = 90° and ∠ROQ = 90°.
From the figure, we can see that ∠ROP = ∠ROS + ∠POS.
So, 90° = ∠ROS + ∠POS => ∠ROS = 90° – ∠POS. —(1)
Also from the figure, ∠QOS = ∠ROQ + ∠ROS.
∠QOS = 90° + ∠ROS => ∠ROS = ∠QOS – 90°. —(2)
Adding equations (1) and (2):
2∠ROS = (90° – ∠POS) + (∠QOS – 90°)
2∠ROS = ∠QOS – ∠POS.
∠ROS = 1/2(∠QOS – ∠POS). (Proved)
6. It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Since XYP is a straight line, ∠XYZ + ∠ZYP = 180°.
64° + ∠ZYP = 180° => ∠ZYP = 116°.
Ray YQ bisects ∠ZYP, so ∠ZYQ = ∠QYP = 116° / 2 = 58°.
We need to find ∠XYQ.
∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122°.
We also need to find reflex ∠QYP.
Reflex ∠QYP = 360° – ∠QYP = 360° – 58° = 302°.
Answer: ∠XYQ = 122° and reflex ∠QYP = 302°.
