Question

From a point on a bridge across a river, the angle of depression of the banks on the opposite side of the river are $30{}^\circ $ and $45{}^\circ $ respectively. If the bridge is at the height of 30 m from the banks, find the width of the river.

Answer 1

Hint: While solving this question, we will first visualize the given situation and draw a satisfying figure for the same. After that we will use the concept of the trigonometric ratios, that is, $\sin \theta =\dfrac{Perpendicular}{Hypotenuse},\cos \theta =\dfrac{Base}{Hypotenuse}$ and $\tan \theta =\dfrac{Perpendicular}{Base}$ to get the desired value.

Complete step-by-step answer:
In this question, we are asked to find the width of the river, which has a bridge on it at a height of 30 m and that bridge forms angles of depression of the banks on the opposite side of the river as $30{}^\circ $ and $45{}^\circ $. Let us draw the figure of the same of the given situation. The diagram is given below.

In the figure, we have considered the bridge at point D, so the height of the bridge is BD = 30 m. And we have considered A and C as the two banks of the river. So, the width of the river can be represented as AC. According to the question, we can say that angle EBA and angle FBC are $45{}^\circ $ and $30{}^\circ $ respectively as the angles of depression. And we know that the angle of depression is formed with the line parallel to the ground. So, EF is parallel to AC. Therefore by opposite interior angles property, we can say, $\angle EBA=\angle BAD=45{}^\circ $ and $\angle FBC=\angle BCD=30{}^\circ $.
We know that the trigonometric ratios are defined by the ratios of the sides and tangent ratio is defined as the ratio of perpendicular to base, that is, $\tan \theta =\dfrac{Perpendicular}{Base}$. So, we can say,
In $\Delta ABD,\tan \angle BAD=\dfrac{BD}{AD}\ldots \ldots \ldots \left( i \right)$
In $\Delta BCD,\tan \angle BCD=\dfrac{BD}{DC}\ldots \ldots \ldots \left( ii \right)$
Now, we will consider equation (i) and will put the values of angle BAD and BD, that is,$\angle BAD=45{}^\circ $ and BD = 30 m. So, we get,
$\tan 45{}^\circ =\dfrac{30}{AD}$
We know that $\tan 45{}^\circ =1$, so applying that, we can write,
$\begin{align}
  & 1=\dfrac{30}{AD} \\
 & \Rightarrow AD=30\ldots \ldots \ldots \left( iii \right) \\
\end{align}$
Now, if we consider equation (ii), and put the values of angle BCD and BD, that is, $\angle BCD=30{}^\circ $ and BD = 30 m, then we get,
$\tan 30{}^\circ =\dfrac{30}{DC}$
We know that $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$, so applying that, we can write,
$\begin{align}
  & \dfrac{1}{\sqrt{3}}=\dfrac{30}{DC} \\
 & \Rightarrow DC=30\sqrt{3}\ldots \ldots \ldots \left( iv \right) \\
\end{align}$
Now, we have been asked to find the value of AC, which is nothing but the sum of AD and DC. So, we can write it as,
$AC=AD+DC$
From equation (iii) and (iv), we will put the values of AD and DC in the above equality, so we get,
$\begin{align}
  & AC=30+30\sqrt{3} \\
 & \Rightarrow AC=30\left( 1+\sqrt{3} \right) \\
\end{align}$
Hence, we can say that the width of the river is $30\left( 1+\sqrt{3} \right)$ metres.

Note: In this question, there are high possibilities that we make a mistake by writing the wrong value of $\tan 30{}^\circ $ or $\tan 45{}^\circ $. Also, we can make a mistake while writing the tangent ratios. So we should remember that $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$ , $\tan 45{}^\circ =1$ and the tangent ratio is, $\tan \theta =\dfrac{Perpendicular}{Base}$.