In the figure (ii) given below, A, B and C are three points on a circle. The tangent at C meets BA produced at T. Given that ∠ATC = 36° and ∠ACT = 48°, calculate the angle subtended by AB at the centre of the circle.

Join OA, OB and CB. In △ATC,

Ext. ∠CAB = ∠ATC + TCA (∵ external angle in a triangle is equal to the sum of opposite interior angles.)

Ext. ∠CAB = 36° + 48° = 84°.

From figure,

∠ABC = ∠TCA = 48° (∵ angles in alternate segment are equal.)

Since sum of angles in a triangle = 180°.

In △ABC,

⇒ ∠ABC + ∠BAC + ∠ACB = 180°
⇒ 48° + 84° + ∠ACB = 180°
⇒ 132° + ∠ACB = 180°
⇒ ∠ACB = 180° - 132°
⇒ ∠ACB = 48°.

Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∴ ∠AOB = 2∠ACB (∵ angle subtended at centre is double the angle subtended at remaining part of the circle.)

∠AOB = 2 × 48° = 96°.

Hence, the angle subtended by AB at the center of the circle is 96°.