Question

In the following figure, O is the centre of the circle, ∠BCO = 30°. Find x and y.

Answer 1

Given, O is the centre of the circle and ∠BCO = 30°. In the given figure join OB and AC.

In ΔBOC, CO = BO   ...[Both are the radius of circle]

∴ ∠OBC = ∠OCB = 30°   ...[Angles opposite to equal sides are equal]

∴ ∠BOC = 180° – (∠OBC + ∠OCE)   ...[By angle sum property of a triangle]

= 180° – (30° + 30°)

= 120°

∠BOC = 2∠BAC

We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

∴ `∠BAC = 120^circ/2 = 60^circ`

Also, ∠BAE = ∠CAE = 30°  ...[AE is an angle bisector of angle A]

⇒ ∠BAE = x = 30°

In ΔABE, ∠BAE + ∠EBA + ∠AEB = 180°  ...[By angle sum property of a triangle]

⇒ 30° + ∠EBA + 90° = 180°

∴ ∠EBA = 180° – (90° + 30°)

= 180° – 120°

= 60°

Now, ∠EBA = 60°

⇒ ∠ABD + y = 60°

⇒ `1/2 xx ∠AOD + y = 60^circ`  ...[In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle]

⇒ `90^circ/2 + y = 60^circ`   ...[∵ ∠AOD = 90°, given]

⇒ 45° + y = 60°

⇒ y = 60° – 45°

∴ y = 15°