Question

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 ${30}^{\circ }$ and ${45}^{\circ }$ respectively. If the lighthouse is $100m$ high, the distance between the two ships is:

Take $\sqrt{3}=1.73$

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

Answer 1

The correct option is C $273m$

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

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

In $△ABD,$ we have

$\tan {30}^{\circ }=\frac{BD}{BA}$ $[\because \tan \theta =\frac{\text{opposite side}}{\text{Adjacent sides}}]$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{BA}$

$\Rightarrow BA=100\sqrt{3}$

In $△CBD,$ we have

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

$\Rightarrow 1=\frac{100}{BC}$

$\Rightarrow BC=100m$

Distance between the two ships $=AC=BA+BC$
$=100\sqrt{3}+100$

$=100(\sqrt{3}+1)$

$=100(1.73+1)$

$=100\times 2.73=273m$