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

Answer: Option C

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100m, ∠ACB = 30° and ∠ADB = 45°
$$\frac{AB}{AC}=\tan 30^{\circ }=\frac{1}{\sqrt{3}}$$
$$\Rightarrow AC=AB\times \sqrt{3}=100\sqrt{3}\text{m}$$
$$\frac{AB}{AD}=\tan 45^{\circ }=1$$
$$\Rightarrow AD=AB=100\text{m}$$
∴ CD = (AC+AD) = (100√3 +100)
= 100(√3 +1) = 100(1.73+1) =100 × 2.73 = 273m