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Two tangents PA and PB are drawn to a circle with Center O from an external point P prove that ?APB = 2 ?OAB.

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From the given figure AB=PB 
So, angle PAB=angle PBA=frac12[180^circ-angle APB]

     =90^circ-frac12angle APB
Similarly,
angle OAB=90^circ-angle PAB
=90^circ-[90^circ-frac12angle APB]=frac12angle APB
i.e., 2angle OAB=angle APB

Posted by

Deependra Verma

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