Question

If $\overrightarrow{R}=\overrightarrow{P}+\overrightarrow{Q}$, also $|\overrightarrow{R}|=25\text{N}$ and $|\overrightarrow{P}|=10\text{N}$. Find the angle $\theta$ made by $\overrightarrow{Q}$ with the positive x-axis.

• A. ${\tan }^{-1}\left(\frac{23}{14}\right)$
• B. ${\tan }^{-1}\left(\frac{14}{23}\right)$
• C. ${\tan }^{-1}\left(\frac{2}{5}\right)$
• D. ${\tan }^{-1}\left(\frac{5}{2}\right)$

Answer 1

The correct option is B ${\tan }^{-1}\left(\frac{14}{23}\right)$

Given, $\overrightarrow{R}=\overrightarrow{P}+\overrightarrow{Q}$
$\Rightarrow \overrightarrow{Q}=\overrightarrow{R}-\overrightarrow{P}$
So, the above relation can be shown in the figure as
Resolving the vectors along the axes we get
$|\overrightarrow{Q_x}|=25\cos {53}^{\circ }+10\cos {37}^{\circ }=15+8=23\text{N}$
and, $|\overrightarrow{Q_y}|=25\sin {53}^{\circ }-10\sin {37}^{\circ }=20-6=14\text{N}$
So, we have
$\tan \alpha =\frac{|\overrightarrow{Q_y}|}{|\overrightarrow{Q_x}|}=\frac{14}{23}$
$\Rightarrow \alpha ={\tan }^{-1}\left(\frac{14}{23}\right)$