13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answers (1)
M manish

Given that,
Height of the lighthouse (AB) is 75 m from the sea level. And the angle of depression of two different ships are \angle ADB = 30^0 and \angle ACB = 45^0 respectively

Let the distance between both the ships be x m.
According to question,

In triangle \Delta ADB,

\tan 30^0 = \frac{AB}{BD}=\frac{75}{x+y} = \frac{1}{\sqrt{3}}
\therefore x+y = 75 \sqrt{3}.............(i)

In triangle \Delta ACB,

\tan 45^0 = 1 =\frac{75}{BC}=\frac{75}{y}
\therefore y =75\ m.............(ii)

From equation (i) and (ii) we get;
x = 75(\sqrt{3}-1)=75(0.732)
 x = 54.9\simeq 55\ m

Hence, the distance between the two ships is approx 55 m.