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

## Question

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

**Reasoning:**

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, \(CD = BD − BC \)

**Steps:**

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