Question

From a lighthouse 100 m high, it is observed that two ships are approaching it from west and south. If angles of depression of the two ships are ${30}^{\circ }$and${45}^{\circ }$ respectively, then the distance between the ships in metres is:

• A. $100(\sqrt{3}+1)$
• B. $100(\sqrt{3}-1)$
• C. $200$
• D. $400$

Answer 1

The correct option is C $200$

Height of the light house = $100m$
Angles of elevation of the ships = ${30}^{\circ }$ and ${45}^{\circ }$
When the angle of elevation is ${30}^{\circ }$, let the distance be $d_a$
Now, $\tan 30=\frac{100}{d_a}$
$d_a=100\sqrt{3}m$
When the angle of elevation of the ships = ${45}^{\circ }$, let the distance be $d_b$
Now, $\tan 45=\frac{100}{d_b}$
$d_b=100m$
Distance between the ships = $\sqrt{d_a^2+d_b^2}$
= $\sqrt{(100\sqrt{3}{)}^2+{100}^2}m$
= $100\left(\sqrt{4}\right)m$

=$200m$