The angle of depression of two ships from the top of a lighthouse and on the same side of it are found to be $45{}^\circ \text{ and 30}{}^\circ $ , respectively. If the ships are 200 m apart, find the height of the light house.
VerifiedHint: Assume that the height of the light house from the ground is ‘h’. First, draw a rough diagram of the given conditions and then use the formula $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$ in the different right angle triangles and substitute the given values to get the height.
Complete step-by-step answer:
Let us start with the question by drawing a representative diagram of the situation given in the question.
According to the above figure:
We have assumed the height of the lighthouse from the ground as ‘h’. Therefore, AB = h. Also, assume that the distance BD is ‘x’ meters and distance BE is ‘y’ meters.
Also, we can say that $\angle AEB=30{}^\circ $ and $\angle ADB=45{}^\circ $ using the property of alternate interior angles.
Now, in right angle triangle ADB,
$\angle ADB=45{}^\circ $
We know that, $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$. Therefore,
\[ \tan 45{}^\circ =\dfrac{AB}{BD} \]
\[ \Rightarrow \tan 45{}^\circ =\dfrac{h}{x} \]
\[ \Rightarrow x=\dfrac{h}{\tan 45{}^\circ } \]
And we know that the value of $\tan 45{}^\circ $ is equal to 1.
$\therefore x=h.............(i)$
Now, in right angle triangle AEB,
\[\angle AEB=30{}^\circ \]
We know that, $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$. Therefore,
\[ \tan 30{}^\circ =\dfrac{AB}{BE} \]
\[ \Rightarrow \tan 30{}^\circ =\dfrac{h}{y} \]
\[ \Rightarrow y=\dfrac{h}{\tan 30{}^\circ } \]
And we know that the value of $\tan 30{}^\circ $ is equal to $\dfrac{1}{\sqrt{3}}$ .
$\therefore y=\sqrt{3}h.............(ii)$
Now according to the data given in the question, ships are 200 meters apart.:
BE-BD=200
$\Rightarrow y-x=200$
Now we will substitute x and y from equation (i) and equation (ii), respectively. On doing so, we get
$\sqrt{3}h-h=200$
$\Rightarrow h=\dfrac{200}{\sqrt{3}-1}$
Now we will multiply and divide the right-hand side of the equation with $\sqrt{3}+1$ . On doing so, we get
$h=\dfrac{200\left( \sqrt{3}+1 \right)}{\left( \sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)}$
$\Rightarrow h=\dfrac{200\left( \sqrt{3}+1 \right)}{3-1}=\dfrac{\times 100\left( \sqrt{3}+1 \right)}{{}}=100\left( \sqrt{3}+1 \right)$
So, the height of the light house is $100\left( \sqrt{3}+1 \right)$ meters.
Note: Do not use any other trigonometric function like sine or cosine of the given angle because the information which is provided to us is related to the base of the triangle and we have to find the height. So, the length of the hypotenuse is of no use. Therefore, the formula of the tangent of the angle is used. We can use sine or cosine of the given angles but then the process of finding the height will be lengthy.
