Question

A man standing on the deck of a ship, which is 8m above water level. He observes the angle of elevation of the top of a hill as ${{60}^{0}}$ and the angle of depression of the base of the hill as ${{30}^{0}}$. Calculate the distance of the hill from the ship and the height of the hill.

Answer 1

Hint: First of all draw the figure for better visualization. Let us assume that the height of the hill is “h” and distance between the hill and the ship is “d”. When you draw the figure, you will find that the height of the ship, distance “d” and line of sight of the observer from the deck of the ship to the base of the hill forms a right angled triangle in which we will apply $\tan {{30}^{0}}$ to find the value of “d”. To find the height of the hill, take the other right triangle in which apply $\tan {{60}^{0}}$ which is equal to $\dfrac{h-8}{d}$. In this formula, the value of “d” we have already calculated and the value of $\tan {{60}^{0}}$ we already know so solving this expression will give the value of “h”.

Complete step-by-step solution -
In the below diagram, we have shown distance between the ship and the hill as “d” and the height of the hill (CD) as “h”.
 

In the above question, we have asked to find the value of “d” and “h”.
To find the value of “d”, take the right angled triangle ABD and apply the trigonometric ratio of tan in it.
We know that $\tan \theta =\dfrac{P}{B}$ where “P” stands for perpendicular with respect to angle $\theta $ and “B” stands for the base of the triangle with respect to angle $\theta $.
$\tan {{30}^{0}}=\dfrac{AB}{BD}$
The value of AB from the diagram is 8m and BD is “d” so substituting this value in the above equation we get,
$\tan {{30}^{0}}=\dfrac{8}{d}$
We know that the value of $\tan {{30}^{0}}=\dfrac{1}{\sqrt{3}}$ so substituting this value in the above equation we get,
$\dfrac{1}{\sqrt{3}}=\dfrac{8}{d}$
On cross – multiplication of the above equation we get,
$d=8\sqrt{3}m$
From the above, we have calculated the distance between the ship and the hill as $8\sqrt{3}m$.
Now, to find the value of the height of the hill, i.e. “h” we are considering the triangle CEA.
In $\Delta CEA$,
$\tan {{60}^{0}}=\dfrac{CE}{AE}$
From the figure that we have drawn above, $AE=BD=d$ and the length of CE is equal to h – 8 so substituting these values in the above equation we get,
$\tan {{60}^{0}}=\dfrac{h-8}{d}$
We have calculated the value of “d” as $8\sqrt{3}m$ and we know that the value of $\tan {{60}^{0}}=\sqrt{3}$ so substituting these values in the above equation we get,
$\sqrt{3}=\dfrac{h-8}{8\sqrt{3}}$
Cross – multiplying the above equation we get,
$\begin{align}
  & 8\left( 3 \right)=h-8 \\
 & \Rightarrow 24=h-8 \\
 & \Rightarrow h=32m \\
\end{align}$
From the above, we have calculated the height of the hill as 32m.
Hence, the distance between the ship and the hill is $8\sqrt{3}m$ and the height of the hill is 32m.

Note: Generally, it is seen that students get confused with the values of trigonometric ratios $\tan {{30}^{0}}\And \tan {{60}^{0}}$. They interchange the values of $\tan {{30}^{0}}\And \tan {{60}^{0}}$. They are confused that whether the value of $\tan {{30}^{0}}$ is $\sqrt{3}$ or $\dfrac{1}{\sqrt{3}}$ and same problem repeats with $\tan {{60}^{0}}$ also.
The remedy for this confusion is tan is an increasing function so on increasing the value of $\theta $, the value of tan also increases as $\dfrac{1}{\sqrt{3}}$ is less than $\sqrt{3}$ so $\tan {{30}^{0}}=\dfrac{1}{\sqrt{3}}$ and $\tan {{60}^{0}}=\sqrt{3}$.