Question

A jet fighter at a height of $ 3000\text{ m} $ from the ground, passes directly over another jet fighter at an instance when their angles of elevation from the same observation point are $ 60{}^\circ $ and \[45{}^\circ \] respectively. Find the distance of the first jet fighter from the second jet at that instance. $ \left( \sqrt{3}=1.732 \right) $

Answer 1

Hint: First we assume a point of observation on the ground and draw a diagram using the information given in the question that the height of a jet fighter is $ 3000\text{ m} $ from the ground passes directly over another jet fighter at an instance. The angles of elevation from the same observation point are $ 60{}^\circ $ and \[45{}^\circ \] respectively. We assume the distance of the first jet fighter from the second jet at that instance will be $ x $ . Then by using trigonometric properties we solve the question.

Complete step-by-step answer:
We have given that a jet fighter at a height of $ 3000\text{ m} $ from the ground, passes directly over another jet fighter at an instance when their angles of elevation from the same observation point are $ 60{}^\circ $ and \[45{}^\circ \] respectively.
We have to find the distance of the first jet fighter from the second jet at that instance. \[\]

Here, we draw a diagram assuming a point of observation $ C $ on the ground. $ A $ is the jet fighter and $ D $ is the another jet fighter passes above $ A $ at an instance.
We have given that the height of jet fighter $ D $ from the ground is $ 3000\text{ m} $ . Angles of elevation from the point $ C $ are $ 60{}^\circ $ and \[45{}^\circ \].
Let the angle $ \angle DBC=60{}^\circ $ and $ \angle ABC=45{}^\circ $
We have to find the distance of the first jet fighter from the second jet at that instance.
Let us assume the distance vertical between the two jet fighter is $ x $ .
First let us consider a right angle triangle \[\Delta CBD\],
We know that $ \tan \theta =\dfrac{\text{Perpendicular}}{\text{Base}} $
We have $ \theta =60{}^\circ $ as given in the question, angle of elevation.
When we substitute the values, we get
 $ \begin{align}
  & \tan 60{}^\circ =\dfrac{BD}{\text{BC}} \\
 & \tan 60{}^\circ =\dfrac{3000}{\text{BC}} \\
\end{align} $
We know that $ \tan 60{}^\circ =\sqrt{3} $ , also we have given that $ \left( \sqrt{3}=1.732 \right) $
We get
\[\begin{align}
  & \sqrt{3}=\dfrac{3000}{BC} \\
 & 1.732=\dfrac{3000}{BC} \\
 & BC=\dfrac{3000}{1.732} \\
 & BC=1732...............(i) \\
\end{align}\]
Now, let us consider $ \Delta ABC $ ,
 $ \tan \theta =\dfrac{\text{Perpendicular}}{\text{Base}} $
 $ \tan 45{}^\circ =\dfrac{AB}{BC} $
We know that $ \tan 45{}^\circ =1 $
 $ \begin{align}
  & 1=\dfrac{3000-x}{BC} \\
 & BC=3000-x \\
\end{align} $
Substitute the value of BC from equation (i), we get
 $ \begin{align}
  & x=3000-1732 \\
 & X=1268\text{ m} \\
\end{align} $
So, the distance of the first jet fighter from the second jet at that instance is $ 1268\text{ m} $ .

Note: The key concept to solve this type of questions is the use of trigonometric angle properties. Also, in this type of questions first draw a diagram using the information given in the question. Always assume the point of observation on the ground. These points help to solve the question easily and we will get the correct answer. If we do not draw the points DAB on the same vertical line, we might not get the answers. The fighter jet is directly over the other, so they must be drawn in the same vertical line.