Question

An aeroplane when flying at a height of 5000m from the ground passes vertically above another aero plane at an instant when the angles of elevation of the two planes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the aeroplanes at the instant.

Answer 1

Hint: To find the vertical distance between the planes we establish a relation between height of plane from the ground and angles of elevation using the property of tan trigonometric function.

Complete step-by-step answer:
Given Data, height of plane flying = 5000m.
Now we draw an appropriate figure w.r.t the data given in the question,

Let the horizontal distance between the point and the plane be ‘x’ and the vertical distance of the other plane from the ground be ‘y’.
Now according to the figure,
Tan 60° = $\dfrac{{5000}}{{\text{x}}}$ and Tan 45° = $\dfrac{{\text{y}}}{{\text{x}}}$
From the trigonometric table of tan function,
Tan 60° = $\sqrt 3 $ and Tan 45° = 1
$ \Rightarrow \sqrt 3 = \dfrac{{5000}}{{\text{x}}}{\text{ and x = y}}$
$ \Rightarrow {\text{x = y = }}\dfrac{{{\text{5000}}}}{{\sqrt 3 }}$
∴Vertical Distance between two planes = 5000 – y = 5000 - $\dfrac{{5000}}{{\sqrt 3 }}$
$ \Rightarrow 5000\left( {\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right) = 2113.249$.
Hence the distance between two planes at the instant = 2113.249 m

Note: In order to solve questions of this type the key is to draw an appropriate figure with respect to all the given data and then use the triangle formed to establish relations between the required variables. In a triangle, tan θ =$\dfrac{{{\text{opposite side}}}}{{{\text{adjacent side}}}}$, where θ is the angle.