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In figure, if 0 is the centre of a circl...

In figure, if 0 is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of `50^(@)` with PQ, then `anglePOQ` is equal to

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The correct Answer is:
A
`because` PR is a tangent.
`:." "angleOPR=90^(@)" "angleOPQ+angleQPR-90^(@)`
`implies" "angleOPQ=90^(@)-QPR=90^(@)-50^(@)=40^(@)`
In `triangleOPQ,`
`OP=OQ" "`(radii of a circle)
`implies" "angleOQP=angleOPQ=40^(@)" "`(angles opposite to equal sides are equal)
Now, in `triangleOPQ,`
`anglePOQ=180^(@)-angleOPQ-angleOQP=180^(@)-40^(@)-40^(@)=100^(@)`
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