Question

As observed from the top of a light house, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from ${30}^{\circ }to{40}^{\circ }$. Determine the distance travelled by the ship during the period of observation.

Answer 1

Let A and B be the two positions of the ship. Let d be the distance travelled by the ship during the period of observation i.e. AB = d metres.
Let the observer be at O, the top of the lighthouse PO.
It is given that PO = 100 m and the angle of depression from O of A and B are ${30}^{\circ }$ and ${45}^{\circ }$ respectively.
$\therefore \angle OAP={30}^{\circ }and\angle OBP={45}^{\circ }$
In $\Delta$ OPB, we have
$tan{45}^{\circ }=\frac{OP}{BP}$
$\Rightarrow 1=\frac{100}{BP}$
$\Rightarrow$ BP = 100 m
In $\Delta$ OPA, we have
$\Rightarrow tan{30}^{\circ }=\frac{OP}{AP}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{d+BP}$
$\Rightarrow$ d + BP = 100$\sqrt{3}$
$\Rightarrow$ d + BP = 100$\sqrt{3}$
$\Rightarrow$ d = 100$\sqrt{3}$ - 100
$\Rightarrow$ d = 100($\sqrt{3}$ - 1) = 100(1.732 - 1) = 73.2 m.
Hence, the distance travelled by the ship from A to B is 73.2 m.