Question

An aeroplane is flying at a height of 210 m. Flying at this height at some instant the angles of depression of two points in a line in opposite direction on both the banks of the river are ${{45}^{\circ }}$and ${{60}^{\circ }}$. Find the width of the river. (Use $\sqrt{3}=1.73$)

Answer 1

Hint: Here, we have to draw the figure for easy calculation. Given that the angles of depression are ${{45}^{\circ }}$and ${{60}^{\circ }}$. Therefore by the property of transversals we will get $\angle BAD$ and $\angle ABD$. Now, from the two right triangles, $\Delta DAC$ and $\Delta DBC$, we can find AC and BC and then calculate the width of the river AB.

Complete step-by-step answer:

So here, first we have to draw the figure for the given data. According to the given data we have an aeroplane flying at a height of 210 m with angles of depression ${{45}^{\circ }}$ and ${{60}^{\circ }}$with the river.

In the figure, let D be the aeroplane at a height of 210m from the ground. i.e. CD=210m, let AB be the width of the river.
The angles of depression are:
$\angle ADX={{60}^{\circ }},\angle BDY={{45}^{\circ }}$
Here, let AD be the transversal drawn between the two parallel lines XY and AB. So, by the properties of transversal we say that the interior opposite angles are equal. i.e.
$\angle ADX=\angle BAD={{60}^{\circ }}$
Similarly, we can say that,
$\angle BDA=\angle ABD={{45}^{\circ }}$
To find the width of the river AB, consider two triangles, $\Delta DAC$ and $\Delta DBC$.
First, let us consider $\Delta DAC$ where tangent of A is given by,
$\begin{align}
  & \tan {{60}^{\circ }}=\dfrac{opposite\text{ }side}{adjacent\text{ }side\text{ }} \\
 & \tan {{60}^{\circ }}=\dfrac{CD}{AC} \\
 & \tan {{60}^{\circ }}=\dfrac{210}{AC}\text{ }....\text{ (1)} \\
\end{align}$
We know that the value of $\tan {{60}^{\circ }}=\sqrt{3}$.
Therefore, our equation (1) becomes:
$\sqrt{3}=\dfrac{210}{AC}\text{ }$
Here, we have to calculate AC by cross multiplication. i.e.
$\begin{align}
  & AC=\dfrac{210}{\sqrt{3}}\text{ } \\
 & \text{AC=}\dfrac{70\times 3}{\sqrt{3}} \\
\end{align}$
By cancellation we get:
$\text{AC=70}\sqrt{3}$
Next, consider the $\Delta DBC$ where tangent of B is given by,
$\begin{align}
  & \tan {{45}^{\circ }}=\dfrac{opposite\text{ }side}{adjacent\text{ }side\text{ }} \\
 & \tan {{45}^{\circ }}=\dfrac{CD}{BC} \\
 & \tan {{45}^{\circ }}=\dfrac{210}{BC}\text{ }.....\text{ (2)} \\
\end{align}$
We know that the value of $\tan {{45}^{\circ }}=1$.
Therefore, our equation (2) becomes:
$1=\dfrac{210}{BC}$
Here, we have to calculate BC by cross multiplication. i.e.
$BC=210$
From the figure we can say that,
$\begin{align}
  & AB=AC+BC \\
 & AB=70\sqrt{3}+210 \\
 & AB=70\times 1.73+210\text{ }....\text{ (}\sqrt{3}=1.73) \\
 & AB=121.1+210 \\
 & AB=331.1 \\
\end{align}$
Hence, the width of the river, $AB=331.1$

Note: For these types of problems, figure is necessary for the calculations. Here, AD and BD are transversals, a transversal is a line that cuts two or more lines, often parallel lines. From the figure we can say that XY and AB are parallel lines. Therefore, we can say that the alternate interior angles are equal .