Question

In the given figure, PA and PB are two tangents drawn from an external point P. If $\angle APB={40}^{\circ }$, then $\angle PAB=$

Answer 1

Given: $\angle APB={40}^{\circ }$

By Theorem- Tangents drawn to a circle from an external point are equal.
So, PA = PB

$\therefore \angle PAB=\angle PBA$ (Angles opposite to equal sides are equal.)
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$\because \angle PAB+\angle PBA+\angle APB={180}^{\circ }$ (Angle sum property)

$\therefore 2\angle PAB+{40}^{\circ }={180}^{\circ }$
$\Rightarrow \angle PAB={70}^{\circ }$
(1 mark)