In the figure (ii) given below, two circles intersect at points P and Q. If ∠A = 80° and ∠D = 84°, calculate

(i) ∠QBC

(ii) ∠BCP

Join PQ as shown in the figure below:

PQAD is a cyclic quadrilateral as all vertices lie on the circumference of the circle.

Sum of opposite angles of cyclic quadrilateral = 180°

⇒ ∠DAQ + ∠DPQ = 180°
⇒ 80° + ∠DPQ = 180°
⇒ ∠DPQ = 180° - 80°
⇒ ∠DPQ = 100°.

Also,

⇒ ∠PDA + ∠PQA = 180°
⇒ 84° + ∠PQA = 180°
⇒ ∠PQA = 180° - 84°
⇒ ∠PQA = 96°.

Since exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

∠QBC = ∠DPQ = 100°

∠BCP = ∠PQA = 96°.

Hence, the value of ∠QBC = 100° and ∠BCP = 96°.