Question

An airplane when flying at a height of $5000$ m from the ground passes vertically above another airplane at an instant when the angles of the elevation of the two planes from the same point on the ground are $60{}^\circ $ and $45{}^\circ $ respectively. Find the vertical distance between the airplanes at the instant.
A) $5000\left( \sqrt{3}-1 \right)m$
B) $5000\left( 3-\sqrt{3} \right)m$
C) $5000\left( 1-\dfrac{1}{\sqrt{3}} \right)m$
D) $4500m$

Answer 1

Hint: We will represent the given data in a diagram and list all the parameters given. We will assume some necessary parameters from the diagram. By using the trigonometric ratios $\left( \tan \theta \right)$ we will find the values of vertical distances corresponding to the angle. After finding those values from the diagram we will calculate the required value.

Complete step by step answer:
Given that an airplane is flying at a height of $5000 m$ from the ground passes vertically above another airplane at an instant when the angles of the elevation of the two planes from the same point on the ground are $60{}^\circ $ and $45{}^\circ $ respectively as shown in the below figure

In the above figure
$BC$ represents the path of the vertical plane which is at a height of $5000 m$ so $BC=5000 m$
$D$ is the point where another airplane is flying in the air.
The angle $\alpha $ represents elevation made by the plane at point $D$ from a point on the ground and from given data that is equal to $45{}^\circ $.
The angle $\beta $ represents the elevation made by the plane at point $C$ from a point on the ground and from given data that is equal to $60{}^\circ $.
Up to now we have
$\begin{align}
  & BC=5000m \\
 & \alpha =45{}^\circ \\
 & \beta =60{}^\circ \\
\end{align}$
$\Rightarrow$ Assume that the length of $BD=y$ and the length of the $AB=x$
Now use the trigonometric ratio $\tan $ for $\alpha ,\beta $ separately.
We know that $\tan \theta =\dfrac{\text{Opposite side to }\theta }{\text{Adjacent side to }\theta }$, hence
$\begin{align}
  & \tan \alpha =\dfrac{\text{Opposite side to }\alpha }{\text{Adjacent side to }\alpha } \\
 & =\dfrac{BD}{AB} \\
 & \tan \alpha =\dfrac{y}{x}
\end{align}$
$\Rightarrow$ Substituting the value of $\alpha $in the above equation then
$\Rightarrow$ $\tan 45{}^\circ =\dfrac{y}{x}$
$\Rightarrow$ We know that $\tan 45{}^\circ =1$ then
$\begin{align}
  & \dfrac{y}{x}=1 \\
 & y=x.....\left( \text{i} \right)
\end{align}$
$\Rightarrow$ Now finding the value of $\tan \beta $, so
$\begin{align}
  & \tan \beta =\dfrac{\text{Opposite side to }\beta }{\text{Adjacent side to }\beta } \\
 & =\dfrac{BC}{AB} \\
 & =\dfrac{5000}{x}
\end{align}$
$\Rightarrow$ Substituting the value of $\beta $ in the above equation then
$\tan 60{}^\circ =\dfrac{5000}{x}$
From equation $\left( \text{i} \right)$we have $y=x$ and we know $\tan 60{}^\circ =\sqrt{3}$ substituting these value in the above equation then
$\begin{align}
  & \sqrt{3}=\dfrac{5000}{y} \\
 & y=\dfrac{5000}{\sqrt{3}}
\end{align}$
$\Rightarrow$ Now the vertical distance between the planes is $DC$ and that is given by
$\begin{align}
  & DC=BC-BD \\
 & =5000-y \\
 & =5000-\dfrac{5000}{\sqrt{3}} \\
 & =5000\left( 1-\dfrac{1}{\sqrt{3}} \right)
\end{align}$
$\Rightarrow$ Hence the vertical distance between the two planes is $5000\left( 1-\dfrac{1}{\sqrt{3}} \right)$

Note:
For this type of problem, the diagrammatical representation gives half of the answer. While converting the given data into the diagrammatical form you need to know what was the relationship between the given data and required data. From that relation make an equation to solve the problem. Some of the relations are given below
$\begin{align}
  & \sin \theta =\dfrac{\text{Opposite side to }\theta }{\text{Hypotenuse}} \\
 & \cos \theta =\dfrac{\text{Adjacent side to }\theta }{\text{Hypotenuse}} \\
 & \tan \theta =\dfrac{\text{Opposite side to }\theta }{\text{Adjacent side to }\theta }=\dfrac{\sin \theta }{\cos \theta } \\
 & \cot \theta =\dfrac{1}{\tan \theta }=\dfrac{\cos \theta }{\sin \theta } \\
& \sec \theta =\dfrac{1}{\cos \theta } \\
 & \csc \theta =\dfrac{1}{\sin \theta }
\end{align}$