Question

As we observed from the top of a 75 m lighthouse from the sea level, the angle of depression of 2 ships $${30}^{\circ }$$ and $${45}^{\circ }$$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between two ships.

Answer 1

Hint: Draw a rough figure of the lighthouse and 2 boats and mark the angle of depression. Consider the triangle where $${30}^{\circ }$$ and $${45}^{\circ }$$ are formed. Use Pythagoras theorem to find the distance between the 2 ships.

Complete step-by-step answer:
Given is the height of the lighthouse = 75 m.

Hence from the figure, we can say that, AD = 75 m
Let the angle of depression of first ship be $${45}^{\circ },\mathrm{\angle }PAC={45}^{\circ }.$$
The angle of depression of second ship will be $${30}^{\circ },\mathrm{\angle }PAB={30}^{\circ }.$$
We need to find the distance between the 2 ships, i.e. we need to find the length of BC.
From the figure, we can understand that lines PA and BD are parallel, i.e. $$PA||BD$$.
Thus AB and AC are transversals.
We can say that,
$$\begin{array}{rl}& \mathrm{\angle }ABD=\mathrm{\angle }PAB={30}^{\circ }\\ & \mathrm{\angle }ACD=\mathrm{\angle }PAC={45}^{\circ }\end{array}$$
where both these angles are alternate angles as $$PA||BD$$.
$$\mathrm{\angle }ABD={30}^{\circ }$$ and $$\mathrm{\angle }ACD={45}^{\circ }$$.
The lighthouse is perpendicular to the ground, i.e. $$AD\mathrm{\perp }BD$$.
$$\therefore \mathrm{\angle }ADB={90}^{\circ }$$.
From the figure, let us consider $$\mathrm{\Delta }ACD.$$
$$\tan {45}^{\circ }={\displaystyle \frac{SideoppositetoangleC}{SideadjacenttoangleC}}={\displaystyle \frac{AD}{CD}}$$.
We know AD = 75 m, let’s find CD and by trigonometric table $$\tan {45}^{\circ }=1$$.
$$\begin{array}{rl}& \tan {45}^{\circ }={\displaystyle \frac{75}{CD}}\\ & \therefore CD={\displaystyle \frac{75}{\tan 45}}={\displaystyle \frac{75}{1}}\\ & \therefore CD=75cm.\end{array}$$
Now let us consider $$\mathrm{\Delta }ABD$$.
$$\tan {30}^{\circ }={\displaystyle \frac{SideoppositetoangleB}{SideadjacenttoangleB}}={\displaystyle \frac{AD}{BD}}={\displaystyle \frac{AD}{BC+CD}}.$$
We know AD = 75 m, CD = 75 cm, and from trigonometric table,
$$\begin{array}{rl}& \tan {30}^{\circ }={\displaystyle \frac{1}{\sqrt{3}}}\\ & {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{75}{BC+75}}\end{array}$$
Cross multiply and find the value of BC.
$$\begin{array}{rl}& BC+75=75\sqrt{3}\\ & BC=75\sqrt{3}-75\\ & BC=75(\sqrt{3}-1)m.\end{array}$$
Thus we got the distance between 2 ships as $$75(\sqrt{3}-1)m$$.
Hence, distance between 2 ships $$=75(\sqrt{3}-1)m.$$

Note:
We have been given the angle of depression which is the angle from the top of the lighthouse to the bottom 2 ships. But as the lines formed are parallel they become alternate angles. So the elevation from the 2 boats to the top of the lighthouse will become $${45}^{\circ }$$ and $${30}^{\circ }$$.