Question

from the top of a 60m tall lighthouse, the angle of depression of a ship moving east away from the lighthouse at 8kmph is 15 degrees. Calculate the angle of depression of the ship from the top of the lighthouse two minutes later, given that the ship stays on the same course at the same speed.

Answer 1

$tan 15^0 = \frac{60}{AB}\\ AB= \frac{60}{tan 15^0}= 223.92 m\\ 8 kmphr= 2.22 m/s\\$

Distance after 2 minutes

$BC= 2.22 * 2*60 = 266.4 m\\ AC= 223.92+266.4 = 490.32 m\\$

Let $\theta$ be the angle of depression

$tan \theta = \frac{60}{490.32}\\ \theta = tan^{-1}(\frac{60}{490.32})\\ \theta = 6.98^0$