## College Physics (4th Edition)

(a) The pilot must fly 54.2 miles at an angle of $26.4^{\circ}$ north of east. (b) The pilot must fly an extra distance of 134.2 miles.
(a) We can find the total west component of the two extra parts of the flight: $55~mi - (25~mi)~sin~15^{\circ} = 48.5~mi$ We can find the total south component of the two extra parts of the flight: $0 +(25~mi)~cos~15^{\circ} = 24.1~mi$ To fly back to the original destination, the pilot needs to fly 48.5 miles east and 24.1 miles north. We can find the distance $d$: $d = \sqrt{(48.5~mi)^2+(24.1~mi)^2} = 54.2~mi$ We can find the direction north of east: $tan~\theta = \frac{24.1~mi}{48.5~mi}$ $\theta = tan^{-1}(\frac{24.1~mi}{48.5~mi})$ $\theta = 26.4^{\circ}$ (b) We can find the extra distance the pilot must fly: $55~mi+25~mi+54.2~mi = 134.2~mi$