In the given figure, O is the centre of a circle and $∠ A C B = 30^{\circ} . Then, ∠ A O B = ?$

(a) $30^{\circ}$

(b) $15^{\circ}$

(c) $60^{\circ}$

(d) $90^{\circ}$

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Solution

(c) 60°
We know that the angle at the centre of a circle is twice the angle at any point on the remaining part of the circumference.
Thus, ∠AOB = (2 × ∠ACB) = (2 × 30°) = 60°

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Q. In the given figure, O is the centre of a circle and ∠ACB = 30°. Then, ∠AOB = ?
(a) 30°
(b) 15°
(c) 60°
(d) 90°
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Q. In the given figure, O is the centre of a circle, ∠AOB = 90° and ∠ABC = 30°. Then, ∠CAO = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

In figure, A, B and C are three points on a circle with centre O such that $∠$ BOC = $30^{\circ}$ and $∠$ AOB = $60^{\circ}$ . If D is a point on the circle other than the arc ABC, find $∠$ ADC.

$In the given figure,O is the centre of the circle. If ∠ ACB = 60^{\circ} then find ∠ O A B .$

In the given figure, $∠ A O B = 90^{\circ} a n d ∠ A B C = 30^{\circ} . T h e n, ∠ C A O = ?$

(a) $30^{\circ}$

(b) $45^{\circ}$

(d) $90^{\circ}$

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In the given figure, O is the centre of a circle and ∠ACB=30∘. Then, ∠AOB=? (a) 30∘ (b) 15∘ (c) 60∘ (d) 90∘

(c) 60° We know that the angle at the centre of a circle is twice the angle at any point on the remaining part of the circumference. Thus, ∠AOB = (2 × ∠ACB) = (2 × 30°) = 60°

In the given figure, O is the centre of a circle and $∠ A C B = 30^{\circ} . Then, ∠ A O B = ?$

(a) $30^{\circ}$

(b) $15^{\circ}$

(c) $60^{\circ}$

(d) $90^{\circ}$

(c) 60°
We know that the angle at the centre of a circle is twice the angle at any point on the remaining part of the circumference.
Thus, ∠AOB = (2 × ∠ACB) = (2 × 30°) = 60°