Class Nine Maths NCERT Solution Chapter 6
Chapter 6: Lines and Angles
TEXTBOOK’S EXERCISE – 6.1
- In the, figure, lines AB and CD intersect at O. If ∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°, find ∠ BOE and reflex ∠ COE.
Sol. Lines AB and CD intersect at O.
∠AOC + ∠BOE = 70° …(1) [Given]
∠BOD = 40° …(2) [Given]
Since, ∠AOC = ∠BOD = 40°
[Vertically opposite angles]
∴ 40° + ∠BOE = 70° [From (1)]
⇒ ∠BOE = 70° – 40° = 30°
Also, ∠AOC + ∠BOE + ∠COE = 180°
[Since AOB is a straight line]
⇒ 70° + ∠COE = 180° [From (1)]
⇒ ∠COE = 180° – 70° = 110°
Now, reflex ∠COE = 360° – 110° = 250° Hence, ∠BOE = 30° and reflex ∠COE = 250°
2. In the given figure, lines XY and MN intersect at O. If ∠POY = 90°
and a : b = 2 : 3, find c.
Sol. In the figure, lines XY and MN intersect at O and ∠ POY = 90°.
Also, given a : b = 2 : 3
Let a = 2x and b = 3x.
Since, ∠XOM + ∠POM + ∠POY = 180° [Linear pair axiom]
⇒ 3x + 2x + 90° = 180°
⇒ 5x = 180° – 90° ⇒ x = 90/5° = 18°
∴ ∠XOM = b = 3x = 3 × 18° = 54°
and ∠POM = a = 2x = 2 × 18° = 36°
Now, ∠XON = c = ∠MOY = ∠POM + ∠POY [Vertically opposite angles]
= 36° + 90° = 126°
Hence, c = 126°
3. In the given figure, ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT.
Sol. ∠PQS + ∠PQR = 180° …(1) [Linear pair axiom]
∠PRQ + ∠PRT = 180° …(2) [Linear pair axiom]
But, ∠PQR = ∠PRQ [Given]
∴ From (1) and (2),
∠PQS = ∠PRT Proved.
4. In the figure, if x + y = w + z, then prove that AOB is a line.
Sol. Assume AOB is a line.
Therefore, x + y = 180° …(1) [Linear pair axiom]
w + z = 180° …(2) [Linear pair axiom]
Now, from (1) and (2)
x + y = w + z
Hence, our assumption is correct, AOB is a line. Proved.
5. In the figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
∠ROS = 12 (∠QOS – ∠ POS)
Sol. ∠ROS = ∠ROP – ∠POS …(1)
and ∠ROS = ∠QOS – ∠QOR …(2)
Adding (1) and (2),
∠ROS + ∠ROS = ∠QOS – ∠QOR + ∠ROP – ∠POS
⇒ 2∠ROS = ∠QOS – ∠POS
(Since ∠QOR = ∠ROP = 90°)
⇒ ∠ROS = 12 (∠QOS – ∠POS) Proved.
6. It is given that ∠ XYZ = 64° and XY is produced to a point P. Draw a figure from the given information. If ray YQ bisects ∠ ZYP, find ∠ XYQ and reflex ∠ QYP.
Sol. From figure,
∠XYZ = 64° [Given]
Now, ∠ZYP + ∠XYZ = 180° [Linear pair axiom]
⇒ ∠ZYP + 64° = 180°
⇒ ∠ZYP = 180° – 64° = 116°
Also, given that ray YQ bisects ∠ZYP.
But, ∠ZYP = ∠QYP + ∠QYZ = 116°
Therefore, ∠QYP = 58° and ∠QYZ = 58°
Also, ∠XYQ = ∠XYZ + ∠QYZ
⇒ ∠XYQ = 64° + 58° = 122°
and reflex ∠QYP = 360° – ∠QYP = 360° – 58° = 302° [since ∠QYP = 58°]
Hence, ∠XYQ = 122° and reflex ∠QYP = 302°.
HOME PAGE
NEXT