Distance between points (0, 0) and (36, 15).
\(=\sqrt{(36-0)^2+(15-0)^2}=\sqrt{36^2+15^2}\)
\(=\sqrt{1296+225}=\sqrt{1521}=39\)
Yes, we can find the distance between the given towns A and B.
Assume town A at the origin point (0, 0).
Therefore, town B will be at point (36, 15) with respect to town A.
Hence, as calculated above, the distance between towns A and B will be 39 km.
What is the angle between the hour and minute hands at 4:30?
In the adjoining figure, TP and TQ are tangents drawn to a circle with centre O. If $\angle OPQ = 15^\circ$ and $\angle PTQ = \theta$, then find the value of $\sin 2\theta$. 
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD. 