In the given figure, `/_ A= 60^(@)` and `/_ ABC = 80^(@)`
Then find `angleBQC`

Explanation

(i) Side AD of cyclic quad. ABCD has been produced to P.
`:. /_ PDC = /_ ABC = 80(@)`
`[ :' ` ext. `/_ PDC =` int. opp. ` /_ ABC `] .
Again, side of BC of quad . ABCD has been produced to P.
`:. /_ PCD = /_ BAD = 60^(@) [ :'` ext. `/_ PCD =` int. opp `/_ BAD `].
In `Delta PCD` , we have
`/_ PDC + /_ PCD +/_ DPC = 180^(@)`
`[:'` sum of the angles of a `Delta ` is `180^(@) ]`.
`implies 80^(@) +60^(@) + /_ DPC = 180^(@)`.
`implies /_ DPC = 180^(@) -140^(@) = 40^(@)`
Hence, `/_ DPC = 40^(@)`
(ii) Since ABCD is a cyclic quadrilateral , we have
`/_ ABC + /_ ADC = 180^(@)`
`implies 80^(@) + /_ ADC = 180^(@)`
`implies /_ ADC = 180^(@) -80^(@) =100^(@)`
Again ,side DC of cyclic quad. ABCDis produced to Q.
`:. /_ QCB = /_ BAD = 60^(@)`
`[ :' ` ext. `/_ QCB = `int. opp. `/_ BAD ]`.
Also, side AB of cyclic quad. ABCD has been produced to Q.
`:. /_ QBC = /_ ADC = 100^(@)`
`[ :' ` ext. `/_ QBC =` int. opp. `/_ ADC]`.
We know that the sum of the angles of a triangle is `180^(@)`.
In`Delta QCB`, we have
`/_ QCB + /_ QBC + /_ BQC = 180^(@)`
`implies 60^(@) + 100^(@) + /_ BQC = 180^(@)`
`implies /_ BQC = 180^(@) - 160^(@) = 20^(@)`
Hence, `/_ BQC = 20^(@)`