# In Fig. 10.77, PA

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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