Question

Two forces $\overrightarrow{A}=5\text{N}$ and $\overrightarrow{B}=2\text{N}$ make an angle of ${60}^{\circ }$. Find the angle made by resultant force with $\overrightarrow{A}$.

• A. ${45}^{\circ }$
• B. ${30}^{\circ }$
• C. ${\tan }^{-1}(\frac{\sqrt{3}}{6})$
• D. ${\tan }^{-1}(\frac{6}{\sqrt{3}})$

Answer 1

The correct option is C ${\tan }^{-1}(\frac{\sqrt{3}}{6})$

Let the angle made by the resultant force with $\overrightarrow{A}$ be $\alpha$.

We know that,

$\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }$

$\tan \alpha =\frac{2\sin \left({60}^{\circ }\right)}{5+2\cos \left({60}^{\circ }\right)}=\frac{2\times \frac{\sqrt{3}}{2}}{5+2\times \frac{1}{2}}=\frac{\sqrt{3}}{6}$

$\therefore \alpha ={\tan }^{-1}(\frac{\sqrt{3}}{6})$

Hence, option $\left(\text{A}\right)$ is correct.