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In figure, if PA and PB are tangents to the circle with centre 0 such that `angleAPB=50^(@)`, then `angleOAB` is equal to A. `25^(@)`B. `30^(@)`C. `40^(@)`D. `50^(@)` |
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Answer» Correct Answer - A 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^(@)` |
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