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In Fig. PA and PB are tangents to the circle with centre O such that ∠APB = 50°. Write the measure of ∠OAB

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Given:

∠APB = 50°

Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3: Sum of all angles of a triangle = 180°.

By property 1,

AP = BP (tangent from P)

Therefore, ∠PAB = ∠PBA

Now,

By property 3 in ∆PAB,

∠PAB + ∠PBA + ∠APB = 180° 

⇒ ∠PAB + ∠PBA = 180° – ∠APB 

⇒ ∠PAB + ∠PBA = 180° – 50° 

⇒ ∠PAB + ∠PBA = 130°

By property 2,

∠PAO = 90°

Now,

∠PAO = ∠PAB + ∠OAB

⇒ ∠OAB = ∠PAO – ∠PAB

⇒ ∠OAB = 90° – 65° = 25°

Hence, ∠OAB = 25°

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