Answer
$357^\circ$, $220\ mph$
Work Step by Step
1. See figure, based on the given conditions, we have $AB=630/3=210\ mi, AC=15\ m$ and the wind vector $AC$ forms an angle of $360-318=42^\circ$ with the downward vertical as shown.
2. Assume the airplane vector $AD$ has angle $\theta$ ($t$ in the figure) from vertical and amplitude $v$, we have $v\ sin\theta=15sin42^\circ\approx10.0$ and $v\ cos\theta-15cos42^\circ=210$ or $v\ cos\theta\approx221.1$
3. Solve $\begin{cases} v\ sin\theta=10.0\\ v\ cos\theta=221.1 \end{cases}$ we have $tan\theta=0.045$ and $\theta=atan(0.045)\approx3^\circ$ which gives a bearing of $360-3=357^\circ$
4. The amplitude can be found as $v=\sqrt {10.0^2+221.1^2}\approx220\ mph$
