$(a)\;30^{\circ} \\ (b)\;45^{\circ} \\ (c)\;60^{\circ} \\ (d)\;90^{\circ} $

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$\large\frac{\sin \theta }{\sin r} $$=\mu$

Now we have $ \angle r +90^{\circ} + \angle NOB =180^{\circ}$

$\angle NOB =\angle \theta$

$\therefore \angle =90^{\circ}-\theta$

$\large\frac{\sin \theta}{\sin (90 -\theta)} $$= \sqrt 3$

$\large\frac{\sin \theta}{\cos \theta} $$=\sqrt 3$

$\tan \theta= \sqrt 3$

$\theta =60^{\circ}$

Hence c is the correct answer.

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