Answer to Question #207156 in Physics for Sherlock

Question

37 m, 12° W of N +  –65 m, –200°

Answer 1

$$(-37\sin{12},37\cos{12})-\\(-65\cos{-200},-65\sin{-200})=(-68.77,58.42)$$

$$S=\sqrt{68.77^2+58.42^2}=90.2\ m\\\theta=\arctan{\frac{58.42}{68.77}}=40.4\degree\text{ N of W}$$