Question

Question
In the given figure, two circle intersect at P and Q. If ZA= 80° and ZD= 84°
calculate (i) < QBC and (ii) Z BCP
P
C С
D
P840
80°
А
B В​

Answer 1

Answer:

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Step-by-step explanation:

Given−Two circles intersect at P&Q. The lines APB & DQC intersect the circles at A,P,B and D,Q,C respectively. AD & BC have been joined. ∠ADQ=84° and ∠DAP=80°

To find out−∠QBC=? ∠BCP=?

Solution−We join PQ. ADPQ is a cyclic quadrilateral.

∴∠PQA=180°

∠ADP=180°−84°=96° [since the sum of the opposite angles of a cyclic quadrilateral=180°.]

Also ∠PQB=180°

∠PQA=180°−96°=84° [linearpair]

Again QBCP is a cyclic quadrilateral.

∴∠BCP=180°

∠BQP=180°−84°=96°....[since the sum of the opposite angles of a cyclic quadrilateral=180°]

Similarly ADPQ is a cyclic quadrilateral.

∴∠QPD=180°

∠QAD=180°−80°=100° [since the sum of the opposite angles of a cyclic quadrilateral=180°]

Also∠QPC=180°

∠QPD=180°−100°=80° [linearpair]

Again QBCP is a cyclic quadrilateral.

∴∠QBC=180°

∠QPC=180°−80°=100°......[since the sum of the opposite angles of a cyclic quadrilateral=180°]

∴∠QBC=100° and ∠BCP=96°