In the given diagram, an isosceles ∆ABC is inscribed in a circle with centre O. PQ is a tangent to the circle at C. OM is perpendicular to chord AC and ∠COM = 65°. Find :

(a) ∠ABC

(b) ∠BAC

(c) ∠BCQ

(a) From figure,

∠AOC = ∠AOM + ∠COM = 65° + 65° = 130°.

We know that,

Angle at the center is twice the angle formed by the same arc at any other point of the circle.

⇒ ∠AOC = 2∠ABC

⇒ ∠ABC = $\dfrac{1}{2} \times ∠AOC = \dfrac{1}{2} \times 130°$ = 65°.

Hence, ∠ABC = 65°.

(b) In △ABC,

⇒ AB = AC (Given)

⇒ ∠ACB = ∠ABC = 65° (Opposite angles of equal sides are equal)

By angle sum property of triangle,

⇒ ∠ACB + ∠ABC + ∠BAC = 180°

⇒ 65° + 65° + ∠BAC = 180°

⇒ ∠BAC = 180° - 65° - 65° = 50°.

Hence, ∠BAC = 50°.

(c) We know that,

The angle formed between the tangent and the chord through the point of contact of the tangent is equal to the angle formed by the chord in the alternate segment.

∴ ∠BCQ = ∠BAC = 50°.

Hence, ∠BCQ = 50°.