Question

Find $\theta$, if $0\le \theta \le {90}^o$
$2\sin \theta =\sqrt{3}$

Answer 1

The given trignometric function is $2\sin \theta =\sqrt{3}$ that is

$\sin \theta =\frac{\sqrt{3}}{2}........\left(1\right)$

We know that $\sin {60}^0=\frac{\sqrt{3}}{2}........\left(2\right)$

From equations 1 and 2 we get,

$\sin \theta =\sin {60}^0\Rightarrow \theta ={60}^0$

Hence, $\theta ={60}^0$.