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In the given figure, PQ and PR are tange...

In the given figure, `PQ` and `PR` are tangents to a circle with centre `A`. If `/_QPA=27^(@)` then `/_QAR` equals

A

`63^(@)`

B

`117^(@)`

C

`126^(@)`

D

`153^(@)`

Text Solution

Verified by Experts

`/_QPR=2/_QPA =2xx27^(@)=54^(@)`[`:' AP` bisect `/_QPR`]
Now,`QPRA` is a cyclic quadrilateral
`implies /_QAR+/_QPR=180^(@)`
` implies /_ QAR=180^(@)-54^(@)=126^(@)`.
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