In the given figure, ABCD is a cyclic quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠DCQ = 40° and ∠ABD = 60°, find:

(i) ∠DBC

(ii) ∠BCP

(iii) ∠ADB

(i) We know that,

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment, we have :

From figure,

PQ is a tangent and CD is a chord.

⇒ ∠DBC = ∠DCQ [Angles in the alternate segment are equal]

⇒ ∠DBC = 40°.

Hence, ∠DBC = 40°.

(ii) In △DCB

⇒ ∠DBC + ∠DCB + ∠CDB = 180° [By angle sum property of triangle]

⇒ 40° + 90° + ∠CDB = 180° [∠DCB = 90°, as angle in a semi-circle is a right angle]

⇒ ∠CDB = 180° - 130° = 50°.

From figure,

⇒ ∠BCP = ∠CDB = 50°. [Angles in the alternate segment are equal]

Hence, ∠BCP = 50°.

(iii) In ∆ABD,

∠BAD = 90° [Angle in a semi-circle is a right angle]

∠ABD = 60° [Given]

⇒ ∠ADB + ∠BAD + ∠ABD = 180° [By angle sum property of triangle]

⇒ ∠ADB + 90° + 60° = 180°

⇒ ∠ADB = 180° - 150° = 30°

Hence, ∠ADB = 30°.