Question

An airplane is flying horizontally at a height of $3150\text{m}$ above a horizontal plane ground. At a particular instant, it passes another plane vertically below it. At this instant, the angles of elevation of the planes from a point on the ground are ${30}^{\circ }$ and ${60}^{\circ }.$ Hence, the distance between the two planes at that instant is:

• A. $1050\text{m}$
• B. $2100\text{m}$
• C. $4200\text{m}$
• D. $5250\text{m}$

Answer 1

The correct option is B $2100\text{m}$

Let the distance between the planes be $x$ and horizontal distance from the point of observation be $y.$

In $△ABC,$

$\begin{array}{cc}\tan {30}^{\circ }& =\frac{3150-x}{y}\\ \frac{1}{\sqrt{3}}& =\frac{3150-x}{y}\dots \left(i\right)\end{array}$

In $△ABD,$

$\begin{array}{cc}\tan {60}^{\circ }& =\frac{3150}{y}\\ \sqrt{3}& =\frac{3150}{y}\dots \left(ii\right)\end{array}$

Divide equation $\left(i\right)$ by $\left(ii\right),$

$\begin{array}{cc}\frac{3150-x}{3150}& =\frac{1}{3}\\ 9450-3x& =3150\\ 3x& =6300\\ x& =2100\text{m}\end{array}$

Hence, the distance between the planes will be $2100\text{m}.$