A radar device detects a rocket approaching directly from east due west. At one instant the rocket is 10 km away at an elevation angle of 30° above the horizon. At another instant the rocket is 8 km away at an angle of 150° in the vertical east-west plane. What is the rocket’s displacement between these two observations?

What the radar data is telling you

Each observation gives the rocket’s distance from the radar (range) and a direction angle in the vertical east-west plane. So we can place each position as a point in a 2D coordinate system: horizontal (east-west) and vertical (up).

Convert each sighting to coordinates

Let $x$ be horizontal with east positive, and let $z$ be vertical upward. Using polar-to-Cartesian conversion, $$x=r\cos\theta, \quad z=r\sin\theta.$$

First sighting: $r_1=10$ km, $\theta_1=30^\circ$ $$x_1=10\cos 30^\circ=10\cdot \frac{\sqrt{3}}{2}=8.660\text{ km}$$ $$z_1=10\sin 30^\circ=10\cdot \frac{1}{2}=5.000\text{ km}$$ So $\vec r_1=(8.660,\,5.000)$ km.

Second sighting: $r_2=8$ km, $\theta_2=150^\circ$ $$x_2=8\cos 150^\circ=8\left(-\cos 30^\circ\right)=-6.928\text{ km}$$ $$z_2=8\sin 150^\circ=8\sin 30^\circ=4.000\text{ km}$$ So $\vec r_2=(-6.928,\,4.000)$ km.

Displacement (change in position)

Displacement is final position minus initial position: $$\Delta\vec r=\vec r_2-\vec r_1.$$ So, $$\Delta x=x_2-x_1=-6.928-8.660=-15.588\text{ km}$$ $$\Delta z=z_2-z_1=4.000-5.000=-1.000\text{ km}$$ Therefore, $$\Delta\vec r=(-15.588,\,-1.000)\text{ km}.$$ This means the rocket moved $15.588$ km toward the west and $1.000$ km downward between observations.

Size of the displacement

The magnitude is $$|\Delta\vec r|=\sqrt{(\Delta x)^2+(\Delta z)^2}=\sqrt{(-15.588)^2+(-1.000)^2}\approx 15.62\text{ km}.$$ (Exactly, $|\Delta\vec r|=\sqrt{244}=2\sqrt{61}$ km.)