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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)


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.

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