1. **Problem statement:** An airplane's engine thrust produces a speed of 590 km/h in calm wind. The wind velocity vector is given by $\langle 36, -43 \rangle$. We want to find the direction the plane should fly so that its resultant velocity is due west. 2. **Understanding the problem:** The plane's velocity relative to the air plus the wind velocity must result in a vector pointing due west. Due west means the resultant velocity vector points along the negative x-axis, so its y-component is zero. 3. **Define variables:** Let the plane's velocity vector relative to the air be $\vec{v_p} = 590 \langle \cos \theta, \sin \theta \rangle$, where $\theta$ is the angle measured from due west (negative x-axis). The wind velocity vector is $\vec{v_w} = \langle 36, -43 \rangle$. 4. **Resultant velocity:** The resultant velocity vector is $$\vec{v_r} = \vec{v_p} + \vec{v_w} = \langle 590 \cos \theta + 36, 590 \sin \theta - 43 \rangle.$$ 5. **Condition for heading due west:** The resultant velocity must have zero y-component: $$590 \sin \theta - 43 = 0.$$ 6. **Solve for $\sin \theta$:** $$590 \sin \theta = 43$$ $$\sin \theta = \frac{43}{590}.$$ 7. **Calculate $\theta$:** $$\theta = \arcsin\left(\frac{43}{590}\right) \approx \arcsin(0.0728814) \approx 4.1791^\circ.$$ 8. **Interpretation:** Since $\theta$ is measured from due west, the plane should fly approximately $4.1791^\circ$ north of west to counteract the wind and head due west. **Final answer:** $$\boxed{4.1791}$$ degrees from due west.