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

Options

A. `100^(@)`
B. `80^(@)`
C. `90^(@)`
D. `75^(@)`

Answer

A

Explanation

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^(@)`