Question

please answer to my question 1,6,7,8,9​

Answer 1

Answer:

10 CIRCLES

Sol. We have, OA = OB = AB

Therefore, ∆OAB is a equilateral triangle.

⇒ ∠AOB = 60°

We know that angle subtended by an arc at the centre of a circle is double

the angle subtended by the same arc on the remaining part of the circle.

∴ ∠AOB = 2∠ACB

⇒ ∠ACB = 1

2 ∠AOB = 1

2 × 60°

⇒ ∠ACB = 30°

Also, ∠ADB = 1

2 reflex ∠AOB

=

1

2 (360° – 60°) = 1

2 × 300° = 150°

Hence, angle subtended by the chord at a point on the minor arc is 150°

and at a point on the major arc is 30° Ans.

Q.3. In the figure, ∠PQR = 100°, where P, Q and R are points on a circle with

centre O. Find ∠OPR.

Sol. Reflex angle POR= 2∠PQR

= 2 × 100° = 200°

Now, angle POR = 360° – 200 = 160°

Also,

EXERCISE 10.5

Q.1. In the figure, A, B and C are three points on a circle with

centre O such that ∠ BOC = 30° and ∠ AOB = 60°. If D is

a point on the circle other than the arc ABC, find ∠ ADC.

Sol. We have, ∠BOC = 30° and ∠AOB = 60°

∠AOC = ∠AOB + ∠BOC = 60° + 30° = 90°

We know that angle subtended by an arc at the centre

of a circle is double the angle subtended by the same arc on the remaining

part of the circle.

∴ 2∠ADC = ∠AOC

⇒ ∠ADC = 1

2 ∠AOC = 1

2 × 90° ⇒ ∠ADC = 45° Ans.

Q.2. A chord of a circle is equal to the radius of the circle. Find the angle

subtended by the chord at a point on the minor arc and also at a point on

the major arc