Q. 114.2( 6 Votes )

In Fig. 10.77, PA

Answer :

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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