Q. 44.9( 10 Votes )

If tangents PA an

Answer :

Given:


APB = 80°



Property 1: 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 2: Sum of all angles of a quadrilateral = 360°.


Property 3: 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.


By property 1,


PAO = 90°


PBO = 90°


By property 2,


APB + PAO + PBO + AOB = 360°


AOB = 360° - APB + PAO + PBO


AOB = 360° - (80° + 90° + 90°)


AOB = 360° - 260°


AOB = 100°


Now, in ∆POA and ∆POB


OA = OB [ radius of circle]


PA = PB [By property 3 (tangent from P)]


OP = OP [ common]


By SSS congruency,


∆POA POB


Hence, by CPCTC


POA = POB


Now,


AOB = 100°


POA + POB = 100° [∵∠AOB = POA + POB]


POA + POA = 100° [∵∠POA = POB]


2POA = 100°



POA = 50°


Hence, POA = 50°

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