Answer
S $84.26^{\circ} $ W.
Work Step by Step
Step 1. We are given the wind as $\vec w=0i+30j$, and the sum vector as $\vec s=-s_xi+0j$
Step 2. Assume the vector of the airplane is $\vec a=a_xi+a_yj$ with $a_x^2+a_y^2=300^2$
Step 3. Since $\vec s = \vec a + \vec w$, we have $a_x=-s_x$ and $a_y+30=0$ or $a_y=-30$ mi/h
Step 4. Combine the results from 2 and 3, we have $a_x^2=300^2-30^2=89100$ or $a_x=-90\sqrt {11}$ mi/h (we choose negative sign here as the result is due west.)
Step 5. The vector of the airplane is then $\vec a=-90\sqrt {11}i-30j$
Step 6. In terms of angles, $tan\theta=\frac{30}{90\sqrt {11}}$ which gives $\theta\approx5.74^{\circ}$ in Quadrant III.
Step 7. As $\theta$ is the angle with respect to the x-axis, we have $90^{\circ}-5.74^{\circ}=84.26^{\circ}$ and conclude that the direction of the airplane should be S $84.26^{\circ} $ W.
