Exercise 6.1
Q1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠ BOE = 70° and ∠BOD = 40° find ∠BOE and reflex ∠COE.
Solution:
∠BOD = 40°
∠AOC = ∠BOD (Vertically opposite Angle)
∠AOC = 40°
∠AOC + ∠ BOE = 70° (Given)
∠BOE = 70°
∠BOE = 70° - 40°
∠BOE = 30°
AOB is straight line
∠AOC + ∠COE +∠BOE = 180° (linear pair)
⇒ 70° + ∠COE = 180°
⇒ ∠COE = 180° - 70°
⇒ ∠COE = 110°
Reflex ∠COE = 360 - 110°
= 250°
Q2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Solution:
∠POY=90° (given)
Let ∠a and ∠b = 2x and 3x
XOY is a straight line
∠a + ∠b + ∠POY = 180°
⇒2x + 3x + 90°= 180°
⇒5x = 180° - 90°
⇒5x = 90°
⇒x = 90°/5
⇒x = 18°
Now ∠a = 2 x 18°
= 36°
∠b =3 x 18°
= 54°
MON is a straight line
∠b + ∠c = 180°(linear pair)
∠54° + ∠c = 180°
⇒∠c = 180°- 54°
=126°
Q3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT
Solution :
Given : ∠PQR = ∠PRQ
To prove : ∠PQS = ∠PRT
Proof :
∠PQS + ∠PQR = 180° .................. (1) Linear pair
∠PRT + ∠PRQ = 180° .................. (2) Linear pair
From equation (1) and (2)
∠PQS + ∠PQR = ∠PRT + ∠PRQ
Or, ∠PQS + ∠PQR = ∠PRT + ∠PQR (∠PQR = ∠PRQ given)
Or, ∠PQS = ∠PRT Proved
Q4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.
Solution:
Given : x + y = w + z
To prove : AOB is a line.
Proof :
We know that;
x + y + w + z = 360०
(Angle Sustained on centre)
x + y + x + y = 360० (x + y = w + z given)
2x + 2y = 360०
2 (x + y) = 360०
x + y = 180० (linear pair)
Therefore, AOB is a line
Hence, Proved
Q5. 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
Solution:
Given:
POQ is a straight line. OR ⊥ PQ and OS is
another ray lying between rays OP and OR.
To prove:
Proof: OR ⊥ PQ (given)
∴ ∠QOR = 90० …………… (1)
POQ is straight line
∴ ∠POR + ∠QOR = 180० (linear pair)
⇒ ∠POR + 90० = 180०
⇒ ∠POR = 180०– 90०
⇒ ∠POR = 90०…………… (2)
Now, ∠ROS + ∠QOR = ∠QOS
Or, ∠ROS = ∠QOS – ∠QOR ……………. (3)
Again, ∠ROS + ∠POS = ∠POR
Or, ∠ROS = ∠POR – ∠POS ……………. (4)
Adding equation (1) and (2)
∠ROS + ∠ROS = ∠QOS – ∠QOR + ∠POR – ∠POS
2 ∠ROS = ∠QOS – 90०+ 90०– ∠POS
2 ∠ROS = (∠QOS – ∠POS)
Hence Proved
Q6. 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.
Solution:
Given: ∠ XYZ = 64°and XY is produced to point P. YQ bisects ∠ ZYP.
To Find: ∠XYQ and reflex ∠QYP.
YQ bisects ∠ZYP
∴ ∠ZYQ = ∠QYP ................. (1)
∵ XY is produced to point P.
∴ PX is a straight line.
Now, ∠ XYZ + ∠ZYQ + ∠QYP = 180° (linear pair)
Or, 64° + ∠ZYQ + ∠QYP = 180°
⇒ ∠ZYQ + ∠QYP = 180° - 64°
⇒ ∠ZYQ + ∠ZYQ = 116° [Using equation (1) ]
⇒ 2∠ZYQ = 116°
⇒ ∠ZYQ = 116°/2
⇒ ∠ZYQ = 58°
∠ZYQ = ∠QYP = 58°
∠XYQ = ∠XYZ + ∠ZYQ
= 64° + 58°
= 122°
∵ ∠QYP = 58°
∴ Reflex ∠QYP = 360° - 58°
= 302°
∠XYQ = 122°, Reflex ∠QYP = 302°
