Answer :


fig.11


Let APB = θ


In Δ APB,


PA = PB


[ Tangents drawn from an exterior point to the circle are equal in length]


Δ APB is an isoceles triangle.


PAB = PBA = α [LET]


Now,


APB + PAB + PBA = 180° [ sum of all the angles of Δ=180°]


θ + α + α = 180°


2α = 180° – θ


α = PAB = 90° – (θ/2)


Also, OAAP


[ radius of a circle is always to the tangent at the point of contact.]


PAB + OAB = 90°


90° – (θ/2) + OAB = 90°


OAB = (θ/2) = (1/2) APB


APB = 2 OAB


Hence, Proved.


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