Question

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

Answer 1

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

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

We know that $\tan {30}^o=\frac{1}{\sqrt{3}}........\left(2\right)$

From equations 1 and 2 we get,

$\tan \theta =\sin {30}^o\Rightarrow \theta ={30}^o$

Hence, $\theta ={30}^o$.