Question

Find $\theta$, if $0\le \theta \le {90}^o$
$\sqrt{3}\tan \theta =1$

Answer 1

The given trignometric function is $\sqrt{3}\tan \theta =1$ that is

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

We know that $\tan {30}^0=\frac{1}{\sqrt{3}}........\left(2\right)$

From equations 1 and 2 we get,

$\tan \theta =\tan {30}^0\Rightarrow \theta ={30}^0$

Hence, $\theta ={30}^0$.