Answer
$380\ mph, 64^\circ$
Work Step by Step
1. See diagram, the vector of the plane without wind is $355cos62^\circ i +355sin62^\circ j$
2. The wind vector is $28.5 i$
3. The resulting vector is $(355sins62^\circ+28.5) i +355cos62^\circ j\approx341.95i+166.66j$
4. The ground speed is given by $v=\sqrt {(341.95)^2+(166.66)^2}\approx380\ mph$
5. The bearing angle $\theta$ can be found as $theta=arctan(\frac{341.95}{166.66})\approx64^\circ$
