# College Physics 3.5 – Resultant of two vectors

## Solution:

To solve for the magnitude and direction of the resultant R, we can refer to the triangle formed by the tree vectors as shown in the figure below.

From the figure above, we are given the two sides A and B, and the included angle measured to be 70°. To solve for the magnitude of R, we can use the cosine law.

$\text{R}^2=\text{A}^2+\text{B}^2-2\text{AB}\:\cos 70^{\circ}$

$\text{R}=\sqrt{\text{A}^2+\text{B}^2-2\text{AB}\:\text{cos}\:70^{\circ} }\:$

$\text{R}=\sqrt{\text{(12 m)}^2+\text{(20 m)}^2-2\text{(12 m)(20 m)}\:\text{cos}\:70^{\circ} }\:$

$\text{R}=19.4892\:\text{m}$

To solve for the direction of the resultant, we just need to solve for angle C, then subtract 70° to get the value of θ. We can solve for angle C using the sine law.

$\displaystyle \frac{20\:\text{m}}{\sin \text{C}}=\frac{19.4892\:\text{m}}{\sin 70^{\circ} }$

$\displaystyle \sin \:\text{C}=\frac{20\:\text{m}\:\left(\sin \:\:70^{\circ \:}\right)}{19.4892\:\text{m}}$

$\displaystyle \text{C}=\text{sin}^{-1}\left(\frac{20\:\text{m}\:\left(\sin \:\:70^{\circ \:}\right)}{19.4892\:\text{m}}\right)$

$\text{C}=74.65^{\circ}$

Finally, to solve for θ, we subtract 70° from 74.65°.

$\theta =74.65^{\circ} -70^{\circ} =4.65^{\circ}$

Therefore, the compass direction of the resultant displacement is 4.65° South of West.