1. **State the problem:** An airplane's engine produces a speed of 510 km/h in calm wind. The wind velocity vector is given as $\langle 30, -35 \rangle$. We want to find the direction the plane should fly so that its resultant velocity heads due west. 2. **Set up the vectors:** Let the plane's velocity relative to the air be $\vec{v_p} = 510 \langle \cos \theta, \sin \theta \rangle$, where $\theta$ is the angle from due west (0 degrees means flying exactly west). The wind velocity is $\vec{v_w} = \langle 30, -35 \rangle$. The resultant velocity $\vec{v_r} = \vec{v_p} + \vec{v_w}$ must point due west, so its vertical component must be zero. 3. **Write the condition for heading west:** $$ v_{r_y} = 510 \sin \theta - 35 = 0 $$ 4. **Solve for $\sin \theta$:** $$ 510 \sin \theta = 35 $$ $$ \sin \theta = \frac{35}{510} $$ $$ \sin \theta = 0.06862745 $$ 5. **Find $\theta$:** $$ \theta = \arcsin(0.06862745) \approx 3.9351^\circ $$ 6. **Check the horizontal component to ensure the plane heads west:** $$ v_{r_x} = 510 \cos \theta + 30 $$ Since the plane must head west, the resultant velocity vector points left (negative x-direction). So, $$ v_{r_x} < 0 $$ Calculate $\cos \theta$: $$ \cos 3.9351^\circ \approx 0.9976 $$ Then, $$ v_{r_x} = 510 \times 0.9976 + 30 = 508.8 + 30 = 538.8 > 0 $$ This is positive, meaning the resultant velocity points east, not west. 7. **Adjust angle to point west:** Since $\sin \theta$ is positive, $\theta$ could also be $180^\circ - 3.9351^\circ = 176.0649^\circ$. Calculate $\cos 176.0649^\circ$: $$ \cos 176.0649^\circ = -\cos 3.9351^\circ = -0.9976 $$ Calculate $v_{r_x}$: $$ v_{r_x} = 510 \times (-0.9976) + 30 = -508.8 + 30 = -478.8 < 0 $$ This is negative, so the resultant velocity points west. 8. **Final answer:** The plane should fly at an angle of approximately $176.0649^\circ$ from due west to head west considering the wind. **Rounded to four decimal places:** $$ \boxed{176.0649^\circ} $$