Question

Two ships are on either side of a 100 m lighthouse. The angles of elevation from the two ships to the top of the lighthouse are 30° and 45°. Find the distance between the ships.

• A. 300 m
• B. 173 m
• C. 273 m
• D. 200 m

Answer 1

The correct option is C

273 m

Let, BD be the lighthouse and A and C be the positions of the ships.

Then, BD = 100 m, $\angle BAD={30}^{\circ }$ , $\angle BCD={45}^{\circ }$

$tan{30}^{\circ }=\frac{BD}{BA}\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{BA}\Rightarrow BA=100\sqrt{3}$

$tan{45}^{\circ }=\frac{BD}{BC}\Rightarrow 1=\frac{100}{BC}\Rightarrow BC=100$

Distance between the two ships

$=AC=BA+BC=100\sqrt{3}+100=100(\sqrt{3}+1)=100(1.73+1)=100\times 2.73=273m$