Question

A man in a boat rowing away a lighthouse $100\text{ m}$ height takes $2\text{ m}$ to change the angle of elevation of the top at the lighthouse from ${60}^{\circ }$ to ${30}^{\circ }$. Find the speed of the boat in metres per minute.$(\sqrt{3}=1.732)$.

Answer 1

Hint: We have to find the speed of the boat, to calculate speed we need to find the distance that boat travelled from the lighthouse. We find the distance by using the angle elevation and tangent formula. After finding the distance we can find the speed of the boat in metres per minute.

Formula used: $\text{Distance = Speed times Time}$
$\text{tantheta = }{\displaystyle \frac{\text{Opposite}}{\text{Adjacent}}}$
$\tan {60}^{\circ }=\sqrt{3}$
$$\tan {30}^{\circ }={\displaystyle \frac{1}{\sqrt{3}}}$$

Complete step-by-step answer:

We construct the question as a diagram to understand easily.
Consider $\mathrm{\Delta }ABD$,
Here $AB$ is a height of the lighthouse is $100\text{ m}$. $A$ be the top of the lighthouse, $B$ be the bottom of the light house. $CD$ be the distance which man travelled to change the angle of elevation.
Let $x$ be the speed of the boat metres per minute.
$\Rightarrow CD=y=2x(\text{Distance = speed times time})$
$\Rightarrow AB=100\text{m}$
Now we are going to find the distance between $CB$ using the tangent formula,
From $\mathrm{\angle }ACB$, opposite side is $AB$ and adjacent side is $CB$.
Consider $\mathrm{\angle }ACB={60}^{\circ }$,
By using the tangent formula in $\mathrm{\angle }ACB$ we get,
$\Rightarrow \tan {60}^{\circ }={\displaystyle \frac{AB}{CB}}$
Substituting the value of $\tan {60}^{\circ }=\sqrt{3}$ and $AB=100\text{m}$ we get,
$\Rightarrow \sqrt{3}={\displaystyle \frac{100}{CB}}$
Let us solve this for $CB$,
$\Rightarrow CB={\displaystyle \frac{100}{\sqrt{3}}}$
$\Rightarrow CB={\displaystyle \frac{100}{1.732}}$ (For $\sqrt{3}=1.732$)
$\Rightarrow CB=57.73$
 Now we are going to find the distance between $DB$ using the tangent formula,
From $\mathrm{\angle }ADB$, opposite side is $AB$ and adjacent side is $DB$.
Consider $\mathrm{\angle }ADB={30}^{\circ }$,
By using the tangent formula in $\mathrm{\angle }ADB$ we get,
$\Rightarrow \tan {30}^{\circ }={\displaystyle \frac{AB}{DB}}$
Substituting the value of $$\tan {30}^{\circ }={\displaystyle \frac{1}{\sqrt{3}}}$$ and $AB=100\text{m}$ we get,
$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{100}{DB}}$
Let us solve this for $DB$,
$\Rightarrow DB=100\times \sqrt{3}$
$\Rightarrow DB=100\times 1.732$ (For $\sqrt{3}=1.732$)
$\Rightarrow DB=173.2$
Consider,
$CD=DB-CB$
Substituting the values of $DB=173.2$ and $CB=57.73$ we get,
 $\Rightarrow CD=172.3-57.73$
$\Rightarrow CD=115.47$
Since, $\Rightarrow CD=y=2x(\text{Distance = speed times time})$
$\Rightarrow CD=y=2x=115.47$
Let us solve this for $x$ we get,
$\Rightarrow x={\displaystyle \frac{115.47}{2}}$
$\Rightarrow x=57.73\text{ m/min}$

$\therefore$ The speed of the boat in metres per minute is $57.73\text{ m/min}$

Note: The term angle of elevation denotes the angle from the horizontal upward to an object. An observer’s line of sight would be above the horizontal. In this question we may go wrong on finding the speed of the boat and we will be careful on substituting the values. Given is already in required unit values. So we don’t change the unit values for speed and distance.