In the given figure, two circles intersect at `P` and `Q`.
If `/_ A = 80^(@)` and `/_ D = 84^(@)`
calculate (i) `/_ QBC` and (ii) `/_ BCP`.

Explanation

Join `PQ`
Now, quad . `AQPD` is cyclic.
`implies /_QAD + /_ QPD = 180^(@)`
`implies 80^(@) + /_ QPD = 180^(@)`
`implies /_QPD = 180^(@)-80^(@) = 100^(@)`
(i) The side `CP` of quad. `QBCP` is produced to `D`.
`:. /_ QBC = /_ QPD = 100^(@) ` [ ext. `/_ = `int. opp. `/_ ]`
(ii) Side `AQ` of cyclic quad. `PDAQ` is produced to `B`.
`:. /_ PQB = /_ ADP = 84^(@)` [ext. `/_ =` int. opp. `/_ ]`.
In the cyclic quad. QBCP, we have
`/_PQB+ /_ BCP = 180^(@)`
`implies 84^(@) + /_ BCP = 180^(@)`
`implies /_ BCP = 180^(@) -84^(@) = 96^(@)`