A man sitting at a height of 20m on a tall tree on a small island in the middle of a river observes two poles directly opposite each other on the two banks of the river & in a line with the foot of the tree. If the angles of depression of the feet of the poles from a point at which the man is ${{60}^{\circ }}$ and ${{30}^{\circ }}$ respectively. Find the width of the river?
VerifiedHint: We start solving the problem by drawing the figure representing the given information. We then recall the definition of tangent of an angle in a right-angled triangle is equal to the ratio of the opposite side to the adjacent side. We apply tangent to the angle at the foot of the poles at two banks of the river to find its width.
Complete step by step answer:
According to the problem, we are given that the man sitting at a height of 20m on a tall tree on a small island in the middle of the river observes two poles directly opposite each other on the two banks of river & in a line with the foot of the tree. We need to find the width of the river if the angles of depression of the feet of the poles from a point at which the man is ${{60}^{\circ }}$ and ${{30}^{\circ }}$ respectively.
Let us draw the figure representing the given figure.
Let BC be the width of the river, AD be the height of the tree at which the man is sitting, points B and D be the foot of the poles that were present at the opposite end of the banks.
From the figure, we can see that the $\Delta ADC$ is a right-angled triangle. We know that the tangent of an angle in a right-angled triangle is equal to the ratio of the opposite side to the adjacent side.
So, we get $\tan {{30}^{\circ }}=\dfrac{20}{{{x}_{1}}}$.
$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{20}{{{x}_{1}}}$.
$\Rightarrow {{x}_{1}}=20\sqrt{3}$m -(1).
From the figure, we can see that the $\Delta ADB$ is a right-angled triangle.
So, we get $\tan {{60}^{\circ }}=\dfrac{20}{{{x}_{2}}}$.
$\Rightarrow \sqrt{3}=\dfrac{20}{{{x}_{2}}}$.
$\Rightarrow {{x}_{2}}=\dfrac{20}{\sqrt{3}}$m -(2).
From the figure, we can see $BC={{x}_{1}}+{{x}_{2}}=20\sqrt{3}+\dfrac{20}{\sqrt{3}}$.
$\Rightarrow BC=\dfrac{60+20}{\sqrt{3}}$.
$\Rightarrow BC=\dfrac{80}{\sqrt{3}}$m.
∴ The width of the river is $\dfrac{80}{\sqrt{3}}m$.
Note: Whenever we get this type of problem, we first draw the figure representing the given information as we can see that the figure provided half the solution to us. We should not make calculation mistakes while solving this problem. We can also find the shortest distance between the foot of the poles and the point where the man is sitting. Similarly, we can expect problems to find the angle of elevation of the foot of the poles if they were moved towards the tree by 5m.
