A radar observes an airplane moving horizontally with velocity v at height h above the ground, as shown in the figure

The ratio of angular acceleration to angular velocity of the airplane with respect to the radar is: 

At any instance, let x be the horizontal distance of the plane from the radar. Then

$x\tan \theta =h$

Differentiating throughout with respect to time, we have

$\begin{array}{l}\frac{d\left(x\tan \theta \right)}{dt}=\frac{dh}{dt}\\ \frac{\tan \theta dx}{dt}+x\frac{d\tan \theta }{dt}=0\\ v\tan \theta +x{\sec }^2\theta \frac{d\theta }{dt}=0\\ \Rightarrow \stackrel{ }{\frac{d\theta }{dt}}=-\frac{v\tan \theta }{x{\sec }^2\theta }\end{array}$

Putting $\stackrel{ }{\frac{d\theta }{dt}}=-\omega \text{ and }x=\frac{h}{\tan \theta }$, we have 

$\omega =\frac{v}{h}{\sin }^2\theta$

We know  $\alpha =\omega \frac{d\omega }{d\theta }$. Hence 

$\begin{array}{l}\frac{\alpha }{\omega }=\frac{d\omega }{d\theta }\\ =\frac{v}{h}\sin \left(2\theta \right)\end{array}$

Alternatively

$\begin{array}{l}\omega =\frac{v\sin \theta }{OP}\\ =\frac{v\sin \theta }{h\mathrm{cosec}\theta }\\ =\frac{v}{h}{\sin }^2\theta \end{array}$

$$\begin{gathered} \alpha =\frac{d\omega }{dt}=\frac{d\left({\sin }^2\theta \right)}{dt} \\ \Rightarrow \alpha =2\sin \theta .\cos \theta .\frac{d\theta }{dt} \\ \Rightarrow \frac{\alpha }{\omega }=\sin 2\theta \end{gathered}$$