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In the adjoining figure , if angleBAD=6...

In the adjoining figure , if `angleBAD=60^(@).angleABC=80^(@)` , then find the values of `angleDPCandangleBQC` .

Text Solution

Verified by Experts

`angleDPC=angleAPB` [by the figure] ltbrge `180^(@)-(anglePAB+anglePBA)`
= `180^(@)-(60^(@)+80^(@))=180^(@)-140^(@)=40^(@)`
Again, ABCD is a cyclic quaadrilateral.
`thereforeangleADC+angleABC=180^(@)`
or, `angleADC+80^(@)=180^(@)"or", angleADC=180^(@)-80^(@)=100^(@)`
`thereforeangleBQC=angleAQD`
`=180^(@)-(angleQAD+angleQDA)=180^(@)-60^(@)(60^(@)+100^(@))`
`[becauseangleQAD=60^(@)("given"),angleQDA=100^(@)]`
`=180^(@)-160^(@)=20^(@)`
`thereforeangleDPC=40^(@)andangleBQC=20^(@)`
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Knowledge Check

  • In the adjoining figure, O is the centre of circle. If angle BCD=28^(@), angle AEC= 38^(@) , then the value of angle AXB is-

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