In the given figure, ∠BAD = 70°, ∠ABD = 50° and ∠ADC = 80°. Calculate :

(i) ∠BDC

(ii) ∠BCD

(iii) ∠BCA.

(i) By angle sum property of a triangle we get,

In ΔABD,

⇒ ∠BAD + ∠ABD + ∠ADB = 180°

⇒ 70° + 50° + ∠ADB = 180°

⇒ 120° + ∠ADB = 180°

⇒ ∠ADB = 180° - 120°

⇒ ∠ADB = 60°.

From figure,

⇒ ∠BDC = ∠ADC - ∠ADB

⇒ ∠BDC = 80° - 60°

⇒ ∠BDC = 20°.

Hence, ∠BDC = 20°.

(ii) We know that,

Sum of opposite angles of a cyclic quadrilateral is 180°.

⇒ ∠BAD + ∠BCD = 180°

⇒ 70° + ∠BCD = 180°

⇒ ∠BCD = 180° - 70°

⇒ ∠BCD = 110°.

Hence, ∠BCD = 110°.

(iii) We know that,

Angles in the same segment of a circle are equal.

⇒ ∠BCA = ∠ADB

⇒ ∠BCA = 60°.

Hence, ∠BCA = 60°.