In figure, if `PA` and `PB` are tangents to the circle with centre `O` such that `angleAPB=50^(@)`, then `angleOAB` is equal to

Options

A. `25^(@)`
B. `30^(@)`
C. `40^(@)`
D. `50^(@)`

Answer

A

Explanation

Given, PA and PB are tangent lines.
:. PA=PB
[since, the length of tangents drawn from an external point to a circle is equal]
`rArranglePBA=anglePAB=0` [say]
In `DeltaPAB,` `angleP+angleA+angleB=180^(@)`
[since, sum of angles of a triangle `=180^(@)`]
`rArr50^(@)+0+0=180^(@)`
`rArr20=180^(@)-50^(@)=130^(@)`
`rArr0=65^(@)`
Also, `OAbotPA`
[since, tangent at any point of a circle is perpendicular to the radius through the point of contact]
`:.anglePAO=90^(@)`
`rArranglePAB+angleBAO=90^(@)`
`rArr65^(@)+angleBAO=90^(@)`
`rArrangleBAO=90^(@)-65^(@)=25^(@)`