## University Physics with Modern Physics (14th Edition)

We should run 28.8 m at an angle of $11.4^{\circ}$ north of east.
We can find the east component $d_x$ of the direction $d$. $d_x = 38.0 - 18.0~sin(33.0^{\circ})$ $d_x = 28.2~m$ We can find the north component $d_y$ of the direction $d$. $d_y = 20.8 - 18.0~cos(33^{\circ})$ $d_y = 5.70~m$ We can use $d_x$ and $d_y$ to find the magnitude of the distance $d$. $d = \sqrt{(28.2~m)^2+(5.70~m)^2}$ $d = 28.8~m$ We can find the angle north of east. $tan(\theta) = \frac{5.70}{28.2}$ $\theta = tan^{-1}(\frac{5.70}{28.2}) = 11.4^{\circ}$ We should run 28.8 m at an angle of $11.4^{\circ}$ north of east.