Question

$\sf If \sqrt{3}Sin \theta - Cos \theta = 0 \ and \ 0^{\circ} \ \textless \ \theta \ \textless \ 90^{\circ}, \ find \ the \ value \ of \ \theta.$

Answer 1

Given:

$\sqrt{3}\:sin\theta-cos\theta=0,\:0^{\circ}<\theta<90^{\circ}$

To find: the value of $\theta$

Step-by-step explanation:

Given, $\sqrt{3}\:sin\theta-cos\theta=0$

$\Rightarrow \sqrt{3}\:sin\theta=cos\theta$

$\Rightarrow \dfrac{sin\theta}{cos\theta}=\dfrac{1}{\sqrt{3}}$

$\Rightarrow tan\theta=tan30^{\circ}$

$\Rightarrow \theta=30^{\circ}$

Answer: the value of $\theta$ is $30^{\circ}$

Note:

The value of $\theta$ can be found using one of $sin(A-B)$ or $cos(A+B)$ rule. You just need to take $\dfrac{\sqrt{3}}{2}=cos30^{\circ}$ or $\dfrac{1}{2}=cos60^{\circ}$.