# Vector Addition and Subtraction|College Physics| Problem 3.5

## Solution:

Solve for the x and y components of the displacement.

The x-component of the displacement is: $\displaystyle R_x=A_x+B_x$ $\displaystyle R_x=A\:sin\left(20^{\circ} \right)+B\:cos\left(40^{\circ} \right)$ $\displaystyle R_x=-\left(12.0\:m\right)sin\left(20^{\circ} \right)-\left(20.0\:m\right)cos\left(40^{\circ} \right)$ $\displaystyle R_x=-19.425\:m$

The y-component of the displacement is: $\displaystyle R_y=A_y+B_y$ $\displaystyle y=Acos20^{\circ} -Bsin40^{\circ}$ $\displaystyle R_y=\left(12.0\:m\right)cos\left(20^{\circ} \right)-\left(20.0\:m\right)sin\left(40^{\circ} \right)$ $\displaystyle R_y=-1.579$

The resultant displacement is therefore, $\displaystyle R=\sqrt{R_x+R_y}$ $\displaystyle R=\sqrt{\left(-19.425\:m\right)^2+\left(-1.579\:m\right)^2}$ $\displaystyle R=19.5\:m$

The compass direction is: $\displaystyle \theta =tan^{-1}\left|\frac{R_y}{R_x}\right|$ $\displaystyle \theta =tan^{-1}\left|\frac{1.579\:m}{19.425\:m}\right|$ $\displaystyle \theta =4.65^{\circ} \:S\:of\:W$