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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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Answer» Correct Answer - 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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