Chapter 3 - Vectors - Questions - Page 56: 12

Magnitude=81 km angle=$40^{\circ}$ north of east

Here x component is given as : $X=50+25 cos 60^{\circ}$ =62.5 km y component, $Y=30+25 sin 60^{\circ}$ =51.7 km So magnitude is given by: $|S|=\sqrt (x^2+y^2)$ =81 km $\theta=y/x=tan ^- 51.7/62.5$=$40^{\circ}$ north of east.