Question

Vector $\overrightarrow{A}$ having magnitude $5\text{N}$ makes an angle ${53}^{\circ }$ North of East, $\overrightarrow{B}$ makes ${37}^{\circ }$ North of West, magnitude of the resultant is $13\text{N}$. Find the magnitude of the vector $\overrightarrow{B}$.

• A. $15\text{N}$
• B. $14\text{N}$
• C. $13\text{N}$
• D. $12\text{N}$

Answer 1

The correct option is D $12\text{N}$

Let $\overrightarrow{R}$ be the resultant and $|\overrightarrow{B}|=x$
We have
$|\overrightarrow{R}|=13\text{N}$,
$|\overrightarrow{A}|=5\text{N}$,
$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$,
From the diagram, resolving in x and y component.
$\overrightarrow{R}=(5\cos {53}^{\circ }-x\cos {37}^{\circ })\hat{i}+\left(5\sin {53}^{+}x\sin {37}^{\circ }\right)\hat{j}$
$\overrightarrow{R}=(5\times \frac{3}{5}-x\times \frac{4}{5})\hat{i}+(5\times \frac{4}{5}+x\times \frac{3}{5})\hat{j}$
$\overrightarrow{R}=(3-\frac{4x}{5})\hat{i}+(4+\frac{3x}{5})\hat{j}$
$|\overrightarrow{R}{|}^2=(3-\frac{4x}{5}{)}^2+(4+\frac{3x}{5}{)}^2$
$169=9+\frac{16x^2}{25}-\frac{24x}{5}+16+\frac{9x^2}{25}+\frac{24x}{5}$
$x^2=144$
$x=12\text{N}$