As observed from the top of a lighthouse, 60 m high above sea the angle of depression of a ship changes from 45° to 60°. The distance travelled by the ship during the period of observation is?

20(3- √3) m

AB represents lighthouse = 60 m

distance travelled by ship = CD

In ΔABD, 

tan 45°  =        1      :       1

                     (AB)          (BD)

                       ↓                ↓

                     60              60 m

In ΔABC, 

tan 45°  =        \(\sqrt {3}\)      :       1

                     (AB)          (BC)

                       ↓                ↓

                     60              \(\frac{60}{\sqrt {3}}\) = \(\frac{60 \sqrt {3}}{3}\) = 20 \(\sqrt {3}\)

Therefore, CD = BD - BC = 60 - 20 \(\sqrt {3}\) = 20(3 - \(\sqrt {3}\)) m