Question
Find angle $'x\ '$ if :
✓
Answer
The above figure we have
$\tan 60^{\circ}=\frac{30}{ AD } $
$\sqrt{3}=\frac{30}{ AD }$
$AD =\frac{30}{\sqrt{3}}$
Again
$\sin x=\frac{A D}{20}$
$AD = 20 \sin x$
Now
$20 \sin x=\frac{30}{\sqrt{3}}$
$\sin x=\frac{30}{20 \sqrt{3}}$
$\sin x=\frac{\sqrt{3}}{2}$
$\sin x = \sin 60^\circ$
$x = 60^\circ$
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