In the given figure, QAP is the tangent at point A and PBD is a straight line.

If ∠ACB = 36° and ∠APB = 42°, find :

(i) ∠BAP

(ii) ∠ABD

(iii) ∠QAD

(iv) ∠BCD

(i) We know that,

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

∴ ∠BAP = ∠ACB = 36°.

Hence, ∠BAP = 36°.

(ii) We know that,

An exterior angle in a triangle is equal to the sum of two opposite interior angles.

In △APB,

∠ABD = ∠APB + ∠BAP = 42° + 36° = 78°.

Hence, ∠ABD = 78°.

(iii) From figure,

∠ADB = ∠ACB = 36° (Angles in same segment are equal)

In △PAD,

∠QAD = ∠APB + ∠ADB = 42° + 36° = 78°. [Exterior angle is equal to sum of two opposite interior angles.]

Hence, ∠QAD = 78°.

(iv) We know that,

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

∴ ∠ACD = ∠QAD = 78°.

From figure,

∠BCD = ∠ACB + ∠ACD = 36° + 78° = 114°.

Hence, ∠BCD = 114°.