In the given figure, AC = AB and ∠ABC = 72°. OA and OB are two tangents. Determine :

(i) ∠AOB

(ii) angle subtended by the chord AB at the center.

(i) Since,

⇒ AB = AC

⇒ ∠ACB = ∠ABC = 72° [∵ angles opposite to equal sides are equal]

From figure,

⇒ ∠BAO = ∠ACB = 72° [Angles in alternate segment are equal]

⇒ OA = OB [Tangents from a fixed point outside the circle are equal.]

⇒ ∠OBA = ∠BAO = 72° [Since angle opposite to equal sides are equal]

In △ABO,

⇒ ∠BAO + ∠OBA + ∠AOB = 180° [By angle sum property of triangle]

⇒ 72° + 72° + ∠AOB = 180°

⇒ ∠AOB = 180° - 144° = 36°.

Hence, ∠AOB = 36°.

(ii) We know that,

Angle subtended by a chord at the centre of the circle is twice the angle subtended by it at any point of the circumference.

∴ ∠ADB = 2∠ACB = 2 × 72° = 144°.

Hence, angle subtended by AB at the centre of the circle = 144°.