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

In figure, if `O` 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

A

`100^(@)`

B

`80^(@)`

C

`90^(@)`

D

`75^(@)`

Text Solution

Verified by Experts

The correct Answer is:
A
Given, `angleQPR=50^(@)`
We know that, the tangent at any point of a circle is perpendicular to the through the point of contact.
`:.angleOPR=90^(@)`
`rArrangleOPQ+angleQPR=90^(@)` [from figure]
`rArrangleOPQ=90^(@)-50^(@)=40^(@)`
`[:'angleQPR=50^(@)]`
Now, OP = OQ = Radius of circle
`:.angleOQP=angleOPQ=40^(@)`
[since, angle opposite to equal sides are equal]
In `DeltaOPQ`, `angleO+angleP+angleQ=180^(@)`
[since, sum of angles of a triangle `=180^(@)`]
`rArrangleO=180^(@)-(40^(@)+40^(@))` `[:'angleP=40^(@)=angleQ]`
`=180^(@)-80^(@)=100^(@)`
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