ANSWER
85.9° North of East
EXPLANATION
First, we have to make a sketch:
We have to find the resultant vector of the car's displacement.
The horizontal component of the displacement of the car:
[tex]\begin{gathered} d_x=-10\cos 45+15\cos 60 \\ d_x=(-10\cdot0.7071)+(15\cdot0.5)=-7.071+7.5 \\ d_x=0.429\operatorname{km} \end{gathered}[/tex]The vertical component of the displacement of the car is:
[tex]\begin{gathered} d_y=-10\sin 45+15\sin 60 \\ d_y=(-10\cdot0.7071)+_{}(15\cdot0.8660)=-7.071+12.990 \\ d_y=5.919\operatorname{km} \end{gathered}[/tex]To find the direction of the car's resultant vector, apply the formula:
[tex]\theta=\tan ^{-1}(\frac{d_y}{d_x})[/tex]Therefore, we have:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{5.919}{0.429}) \\ \theta=\tan ^{-1}(13.7972) \\ \theta=85.9\degree \end{gathered}[/tex]Hence, the car's resultant vector is 85.9° North of East.
