Question

An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angle of elevation of the two planes from the same point on the ground at 30 degree and 60 degree respectively. find the distance between the two planes at that instant..

Answer 1

Let A and B be the positions of the two planes at the given.

Let O, be the point of observation.

Draw ABC perpendicular to the horizontal through O.

Then,

$\sf\blue{\angle{AOC}=60°}$

$\sf\blue{\angle{BOC}=30°}$

$\sf\blue{AC=3125m}$

From right ∆OCA, we have

$\longrightarrow$$\sf\green{ \tan60°= \frac{AC}{OC}}$

$\longrightarrow$ $\sf\green{\sqrt{3} = \frac{3125}{OC} }$

$\therefore$ $\sf\green{OC = \frac{3125}{ \sqrt{ 3} } m}$

From right ∆OCB, we have

$\longrightarrow$$\sf\red{ \tan30°= \frac{BC}{OC}}$

$\longrightarrow$ $\sf\red{\frac{1}{√3} = \frac{BC}{3125 / √3} }$

$\therefore$ $\sf\red{BC=\frac{3125}{√3×√3}=\frac{3125}{3}m}$

Hence, the vertical distance between the two planes is

AB= AC - BC

$\longrightarrow$$\sf{3125-\frac{3125}{3}}$

$\longrightarrow$$\sf{\frac{6250}{3}}$

$\longrightarrow$$\sf{2083.33m}$