Ex.9.1 Q13 Some Applications of Trigonometry Solution - NCERT Maths Class 10

Go back to  'Ex.9.1'


As observed from the top of a \(75\,\rm{m}\) high lighthouse from the sea-level, the angles of depression of two ships are \(30^\circ\) and \(45^\circ.\) If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.


Text Solution


What is Known?

(i) Height of the lighthouse \(=75\,\rm{m} \) 

(ii) Angles of depression of two ships from the top of the lighthouse are \(30^\circ\) and \(45^\circ.\)

What is Unknown?

Distance between the two ships


Let the height of the lighthouse from the sea-level is \(AB\) and the ships are \(C\) and \(D\). The angles of depression of the ships \(C\) and \(D\) from the top \(A\) of the lighthouse, are \(45^\circ\) and \(60^\circ\) respectively.

Trigonometric ratio involving \(AB, BC, BD\) and angles is tan\(\theta \).

Distance between the ships, \(CD = BD − BC \)


In \(\Delta ABC\),

\[\begin{align} \tan {{45}^{0}}&=\frac{AB}{BC} \\  1&=\frac{75}{BC} \\  BC&=75\,  \end{align}\]

In \(\Delta ABD\),

\[\begin{align} \tan {{30}^{0}}&=\frac{AB}{BD} \\ \frac{1}{\sqrt{3}}&=\frac{75}{BD} \\  BD&=75\sqrt{3}  \end{align}\]

Distance between two ships \(CD=BD-BC\)

\[\begin{align}CD&=75\sqrt{3}-75 \\ &=75\left( \sqrt{3}-1 \right)  \end{align}\]

Distance between two ships \(C D = 75 ( \sqrt { 3 } - 1 ) \, \rm{m} \)