Question

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 lighthous, find the distance between the two ships. ​

Answer 1

Answer:

Step-by-step explanation:

$\bf\underline{Given:-}$

Height of the Lighthouse = 75 m

Angles of depression = 30° and 45°

$\bf\underline{Solution:-}$

Let AB be the lighthouse.

According to the Question,

$\sf\bigtriangleup In \: ABC$

$\sf\implies Tan \: 45\degree=\frac{AB}{BD}$

$\sf\implies 1=\frac{75}{BC}$

$\bf\implies AB=75 \: m$

$\sf Again \: In \: \bigtriangleup ABC,$

$\sf\implies Tan \: 30 = \frac{AB}{BD}$

$\sf\implies\frac{1}{\sqrt{3} } =\frac{75}{BD}$

$\bf\implies BD=75 \: \sqrt{3} \: m$

$\sf\implies AD+AB=75 \: \sqrt{3} \: m$

$\bf\implies DA=75 \: (\sqrt{3} -1) \: m$

Hence, the distance between the two ships is $\bf 75 \: (\sqrt{3} -1) \: m.$