Question

PA and PB are two tangents to a circle with center O from a point P. If angle APB=50° . find angle AOB class 10​

Answer 1

Given : A circle with center O and PA and PB are tangents to circle from a common external point P to point A and B respectively and ∠APB = 50° To find : ∠AOB OA ⏊ AP and OB ⏊ PB

[ As tangent to at any point on the circle is perpendicular to the radius through point of contact] ∠BOP = ∠OAP = 90°

In Quadrilateral AOBP [ By angle sum property of quadrilateral] ∠BOP + ∠OAP + ∠AOB + ∠APB = 360° 90° + 90° + ∠AOB + 50° = 360° ∠AOB + = 130°

Now in △AOB OA = OB [Radii of same circle] ∠OBA = ∠ AOB

Also, By angle sum property of triangle ∠OBA + ∠OAB + ∠AOB = 180° ∠AOB + ∠AOB + 130 = 180 [using 2 and 3] 2∠AOB = 50°

∠AOB = 25°

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Answer 2

Answer:

Given : A circle with center O and PA and PB are tangents to circle from a common external point P to point A and B respectively and ∠APB = 50° To find : ∠AOB OA ⏊ AP and OB ⏊ PB

[ As tangent to at any point on the circle is perpendicular to the radius through point of contact] ∠BOP = ∠OAP = 90°

In Quadrilateral AOBP [ By angle sum property of quadrilateral] ∠BOP + ∠OAP + ∠AOB + ∠APB = 360° 90° + 90° + ∠AOB + 50° = 360° ∠AOB + = 130°

Now in △AOB OA = OB [Radii of same circle] ∠OBA = ∠ AOB

Also, By angle sum property of triangle ∠OBA + ∠OAB + ∠AOB = 180° ∠AOB + ∠AOB + 130 = 180 [using 2 and 3] 2∠AOB = 50°

∠AOB = 25°

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