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Suppose Q is a point on the circle with ...

Suppose Q is a point on the circle with center P and radius 1, as shown in the figure , R is a point outside the circle such that `QR=1andangleQRP=2^(@)` . Let S be the point where the segment RP intersects the given circle. Then measure of `angleRQS` equals-

A

`86^(@)`

B

`87^(@)`

C

`88^(@)`

D

`89^(@)`

Text Solution

Verified by Experts

The correct Answer is:
D

By Cosine rule
`(QS)^(2)=2-2cos2^(@)`
`QS=2sin1^(@)`
Now by sine rule in `DeltaRQS`
`sintheta=(sin2^(@))/(2sin1^(@))`
`theta=89^(@)`
`angleRQS=180^(@)-(2+89^(@))=180^(@)-91^(@)=89^(@)`
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