Question

Find $\overrightarrow{A}+\overrightarrow{B}and\overrightarrow{A}-\overrightarrow{B}$ in the diagram shown in figure. Given A=4 units and B=3 units.

Answer 1

$R=\sqrt{A^2+B^2+2AB\cos \theta }$

$=\sqrt{16+9+2\times 4\times 3\cos {60}^{\circ }}=\sqrt{37}$ units

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

$=\frac{3\sin {60}^{\circ }}{4+3\cos {60}^{\circ }}=0.472$

$\alpha ={\tan }^{-1}\left(0.472\right)={25.3}^{\circ }$

Thus, resultant of $\overrightarrow{A}and\overrightarrow{B}is\sqrt{37}$ units at angle ${25.3}^{\circ }from\overrightarrow{A}$.

Subtraction, $S=\sqrt{A^2+B^2-2AB\cos \theta }$

$=\sqrt{16+9-2\times 4\times 3\cos {60}^{\circ }}$

$=\sqrt{13}$ units

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

$=\frac{3\sin {60}^{\circ }}{4-3\cos {60}^{\circ }}=1.04$

$\therefore \alpha ={\tan }^{-1}\left(1.04\right)={46.1}^{\circ }$

Thus, $\overrightarrow{A}-\overrightarrow{B}is\sqrt{13}$ units at ${46.1}^{\circ }from\overrightarrow{A}$.