Question

Given $A=10N$, $B=15N$ and angle between them, $\theta ={60}^{\circ }$. Find the value of $|\overrightarrow{A}+\overrightarrow{B}|$ and $|\overrightarrow{A}-\overrightarrow{B}|$ respectively.

• A. $\sqrt{475}N$ and $\sqrt{175}N$
• B. $\sqrt{275}N$ and $\sqrt{75}N$
• C. $\sqrt{475}N$ and $\sqrt{275}N$
• D. $\sqrt{275}N$ and $\sqrt{175}N$

Answer 1

The correct option is A $\sqrt{475}N$ and $\sqrt{175}N$

$|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{(10{)}^2+(15{)}^2+2\left(10\right)\left(15\right)\cos \left({60}^{\circ }\right)}$
$=\sqrt{100+225+2\times 150\times \frac{1}{2}}$
$=\sqrt{100+225+150}$
$|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{475}N$
$|\overrightarrow{A}-\overrightarrow{B}|=\sqrt{(10{)}^2+(15{)}^2-2\left(10\right)\left(15\right)\cos \left({60}^{\circ }\right)}$
$=\sqrt{100+225-2\times 150\times \frac{1}{2}}$
$=\sqrt{100+225-150}$
$|\overrightarrow{A}-\overrightarrow{B}|=\sqrt{175}N$