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

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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.


 Video Solution
Some Applications Of Trigonometry
Ex 9.1 | Question 13

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


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

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