Question

Two circles intersect at points P and Q. If angle A=80° and angle D=84°,calculate angle QBC and angle BCP.

Answer 1

join p and q it will make a cyclic quadrilateral
and sum of opposite angle of cyclic quardrilateral is 180
pda+pqa=180
84+pqa=180
pqa=180-84
pqa=96 is on straight line so qpc=180-96=84
like wise qpd=100 is on straight line pqb=180-100=80

now pqbc is a cyclic quadrilateral
qpc+qbc=180
84+qbc=180
qbc=180-84=96

pqb+bcp=180
bcp=180-80
bcp=100

Answer 2

Answer:

Hope it helps!! Mark this answer as brainliest if u found it useful and follow me for quick and accurate answers...

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°