Question

Two ships are sailing in the sea on the opposite sides of a lighthouse. The angle of elevations of the top of the lighthouse as observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:

• 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

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