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In the given figure, $PQ$ is tangent to a circle centred at $O$ and $\angle BAQ = 30^\circ$; show that $BP = BQ$.
._q_1.png)
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Join $OQ$
$OQ=OA$
$\Rightarrow \angle 2 = 30^\circ$
$\angle 3 = 90^\circ - 30^\circ = 60^\circ$
$\angle 4 = 90^\circ - 60^\circ = 30^\circ$
$\angle 6 = \angle 1 + \angle 2 = 60^\circ$
Hence $\angle 5 = 90^\circ - 60^\circ = 30^\circ = \angle 4$
$\therefore BP=BQ$._a_1.png)
$OQ=OA$
$\Rightarrow \angle 2 = 30^\circ$
$\angle 3 = 90^\circ - 30^\circ = 60^\circ$
$\angle 4 = 90^\circ - 60^\circ = 30^\circ$
$\angle 6 = \angle 1 + \angle 2 = 60^\circ$
Hence $\angle 5 = 90^\circ - 60^\circ = 30^\circ = \angle 4$
$\therefore BP=BQ$