Question

Two boats approach a lighthouse in mid-sea from the opposite directions. The angles of elevation of the top of the lighthouse from the two boats are ${{30}^{0}}$ and ${{45}^{0}}$ respectively. If the distance between the two boats is 100 m, find the height of the lighthouse.

Answer 1

Hint: For solving this problem first we will draw the geometrical figure as per the given data. After that, we will use the basic formula of trigonometry $\tan \theta =\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)}$ . Then, we will solve correctly to get the height of the lighthouse.
Complete step-by-step answer:
Given:
It is given that there are two boats approaching a lighthouse in mid-sea from the opposite directions. The angles of elevation of the top of the lighthouse from the two boats are ${{30}^{0}}$ and ${{45}^{0}}$ respectively, and distance between two boats is 100 m. We have to find the height of the lighthouse.
Now, first, we will draw a geometrical figure as per the given data. For more clarity look at the figure given below:

In the above figure, BA represents the height of the lighthouse. Position of the first boat is represented by the point C and angle of elevation of the top of the lighthouse from point C is equal to $\angle ACB={{30}^{0}}$ . Moreover, the position of the second is at point D and the angle of elevation of the top of the lighthouse from point D is equal to $\angle ADB={{45}^{0}}$ .
Now, as point C, D and A lies on the same line segment and it is given that, the distance between two boats is 100 m. Then,
$\begin{align}
  & CD=CA+DA=100 \\
 & \Rightarrow CA+DA=100...........\left( 1 \right) \\
\end{align}$
Now, we consider $\Delta DAB$ in which $\angle DAB={{90}^{0}}$ , BA is the length of the perpendicular, DA is the length of the base and $\angle ADB={{45}^{0}}$ . Then,
$\begin{align}
  & \tan \left( \angle ADB \right)=\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)} \\
 & \Rightarrow \tan {{45}^{0}}=\dfrac{BA}{DA} \\
 & \Rightarrow 1=\dfrac{BA}{DA} \\
 & \Rightarrow DA=BA.........................\left( 2 \right) \\
\end{align}$
Now, we consider $\Delta CAB$ in which $\angle CAB={{90}^{0}}$ , BA is the length of the perpendicular, CA is the length of the base and $\angle ACB={{30}^{0}}$ . Then,
$\begin{align}
  & \tan \left( \angle ACB \right)=\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)} \\
 & \Rightarrow \tan {{30}^{0}}=\dfrac{BA}{CA} \\
 & \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{BA}{CA} \\
 & \Rightarrow CA=BA\sqrt{3}.........................\left( 3 \right) \\
\end{align}$
Now, put $CA=BA\sqrt{3}$ from equation (3) and $DA=BA$ from equation (2) into equation (1). Then,
$\begin{align}
  & CA+DA=100 \\
 & \Rightarrow BA\sqrt{3}+BA=100 \\
 & \Rightarrow BA\left( \sqrt{3}+1 \right)=100 \\
 & \Rightarrow BA\left( \sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)=100\left( \sqrt{3}-1 \right) \\
\end{align}$
Now, as we know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ so, we can write $\left( \sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)=3-1=2$ in the above equation. Then,
$\begin{align}
  & BA\left( \sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)=100\left( \sqrt{3}-1 \right) \\
 & \Rightarrow BA\times 2=100\left( \sqrt{3}-1 \right) \\
 & \Rightarrow BA=50\left( \sqrt{3}-1 \right) \\
\end{align}$
Now, from the above result, we conclude that the height of the lighthouse will be $BA=50\left( \sqrt{3}-1 \right)\approx 36.6025$ metres.

Note: Here, the student should first try to understand what is asked in the problem. After that, we should try to draw the geometrical figure as per the given data. Moreover, we should apply the basic formula of trigonometry properly without any error and avoid calculation mistakes while solving to get the correct answer. We can use cot as well instead of tan.