## College Physics (4th Edition)

They need to walk at an angle of $20.2^{\circ}$ east of south. They need to walk a distance of $2.0~km$
We can find the total west component of the sum of the two parts of the walk: $(2.7~km)~cos~45^{\circ} -1.2~km = 0.7~km$ We can find the north component of the sum of the two parts of the walk: $(2.7~km)~cos~45^{\circ} = 1.9~km$ To return to the starting point, they need to walk 0.7 km east and 1.9 km south. We can find the direction east of south: $tan~\theta = \frac{0.7~km}{1.9~km}$ $\theta = tan^{-1}(\frac{0.7~km}{1.9~km})$ $\theta = 20.2^{\circ}$ We can find the distance $d$: $d = \sqrt{(0.7~km)^2+(1.9~km)^2} = 2.0~km$