In the figure, given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75°; ∠ABD = 58° and ∠ADC = 77°. Find :

(i) ∠BDC,

(ii) ∠BCD,

(iii) ∠BCA.

(i) We know,

Sum of angles in a triangle = 180°.

In △ABD,

⇒ ∠ADB + ∠ABD + ∠DAB = 180°

⇒ ∠ADB + 58° + 75° = 180°

⇒ ∠ADB + 133° = 180°

⇒ ∠ADB = 180° - 133° = 47°

From figure,

∠BDC = ∠ADC - ∠ADB = 77° - 47° = 30°.

Hence, ∠BDC = 30°.

(ii) We know that,

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

⇒ ∠BAD + ∠BCD = 180°

⇒ ∠BCD = 180° - 75° = 105°.

Hence, ∠BCD = 105°.

(iii) Join AC.

We know that,

Angles subtended by the same chord on the circle are equal.

∠BCA = ∠ADB = 47°

Hence, ∠BCA = 47°.