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In the adjoining figure , two circles in...

In the adjoining figure , two circles intersect each other at the points P and Q. If `angleQAD=80^(@)andanglePDA=84^(@)`, then find the value of `angleQBCandangleBCP`

Text Solution

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In the circle with center at O, ADPQ is a cyclic quadrilateral.
`thereforeangleADP+angleAQP=180^(@)" or "84^(@)+angleAQP=180^(@)[becauseangleADP=84^(@)]`
or, `angleAQP=180^(@)-84^(@)=96^(@)`
Now , `angleBQP+angleAQP=1 "straight angle "`
or, `angleBQP+96^(@)=180^(@)[becauseangleAQP=96^(@)]"or " angleBQP=180^(@)-96^(@)=84^(@)`
Again , BQPC is a cyclic quadrilateral.
`thereforeangleBCP+angleBQP=180^(@)" or " ,angleBCP+84^(@)=180^(@)`
or, `angleBCP=180^(@)-84^(@)=96^(@)`
Now , `angleDPQ+angleDAQ=180^(@) " or ",angleDPQ+80^(@)=180^(@)" or " , angleDPQ=180^(@)-80^(@)=100^(@)`
`thereforeangleDPQ+angleQPC=1" straight angle "=180^(@)`
`" or " 100^(@)+angleQPC=180^(@)" or " ,angleQPC=180^(@)-100^(@)=80^(@)`
Again , `angleQPC+angleQBC=180^(@)[because" BQPC is a cyclic quadrilateral"]`
or, `80^(@)+angleQBC=180^(@)" or" ,angleQBC=180^(@)-80^(@)=100^(@).`
Hence `angleQBC=100^(@) and angleBCP=96^(@)`
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