Question

Let $\overrightarrow{A}$ and $\overrightarrow{B}$ be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angles ${30}^0$ and ${60}^0$ respectively, find the resultant.

Answer 1

Angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $\theta$
= ${60}^0-{30}^0={30}^0$

$\overrightarrow{A}$ and $\overrightarrow{B}$ = 10 unit

R= $\sqrt{{10}^2+{10}^2+\mathrm{2.10.10}cos{30}^0}$

= $\sqrt{200+200cos{30}^0}$

= $\sqrt{200+100}$

= $\sqrt{300}=19.3$

Let $\beta$ = the angle between $\overrightarrow{R}$ and $\overrightarrow{A}$

$\therefore \beta =ta{n}^{-1}\left(\frac{10sin{30}^0}{10+10cos{30}^0}\right)$

= $ta{n}^{-1}\left(\frac{1}{2+\sqrt{3}}\right)ta{n}^{-1}\left(\frac{1}{3.372}\right)$

= $ta{n}^{-1}\left(0.26795\right)={15}^0$

$\therefore$ Resultant makes angle

= ${15}^0+{30}^0$

= ${45}^0$ with x- axis.