Peter was standing on the top of a rock cliff facing the sea. He saw a boat coming towards the shore. As he kept seeing time just flew. Ten minutes less than half of an hour, the angle of depression changed from 30 to 60. How much more time in minutes will the boat take to
reach the shore ?

Let AB represent the rock cliff with its height be h

 D represents the initial point and C is the point at which he observers the boat after Ten minutes less than half of an hour (20 minutes).

BD = $$\frac{h}{\tan\ \left(30\right)}\ =\ \sqrt{\ 3}h$$

BC = $$\frac{h}{\tan\ \left(60\right)}\ =\ \frac{h}{\sqrt{\ 3}}$$

In 20 minutes boat covered CD length which is = $$\sqrt{\ 3}h\ -\ \frac{h}{\sqrt{\ 3}}=\ \frac{2h}{\sqrt{\ 3}}$$

Hence remaining lenght $$\frac{h}{\sqrt{\ 3}}$$(BC) will be covered in 10 minutes