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Question

Two tangents PA and PB are drawn to a circle with center O from an external point P. Prove that APB=2OAB [3 MARKS]

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Solution

Concept: 1 Mark
Application: 2 Mark

OAPA (tangent and the radius at the point of contact)

OAP=90

Let OAB=x

PAB=90x(1)

PA = PB (tangent from same external point)

APB is isosceles

PAB=PBA=90x

In APB, By angle sum property

APB + PAB + PBA=180

P+90x+90x=180

P+1802x=180

P=2x

APB=2x

i.e., APB=2OAB


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