Question

An airplane flying horizontally $1\text{km}$ above the ground is observed at an elevation of ${60}^{\circ }$. After $10$ seconds, its elevation is observed to be ${30}^{\circ }$. The speed of the airplane is

• A. $240\text{km/hr}$
• B. $420\text{km/hr}$
• C. $240\sqrt{3}\text{km/hr}$
• D. $420\sqrt{3}\text{km/hr}$

Answer 1

The correct option is C $240\sqrt{3}\text{km/hr}$

In $△OCA$,

$\begin{array}{cc}\cot {60}^{\circ }& =\frac{x}{1}\\ x& =\frac{1}{\sqrt{3}}\dots \left(i\right)\end{array}$

In $△ODB$,

$\begin{array}{cc}\cot {30}^{\circ }& =\frac{x+y}{1}\\ x+y& =\sqrt{3}\dots \left(ii\right)\end{array}$

Subtract equation $\left(i\right)$ from $\left(ii\right),$

$\begin{array}{cc}x+y-x& =\sqrt{3}-\frac{1}{\sqrt{3}}\\ y& =\sqrt{3}-\frac{1}{\sqrt{3}}\\ & =\frac{3-1}{\sqrt{3}}\\ & =\frac{2}{\sqrt{3}}\end{array}$

The distance covered by airplane in $10$ seconds is equal to $\frac{2}{\sqrt{3}}\text{km}.$

The distance covered by airplane in $1$ seconds is equal to $\frac{2}{10\sqrt{3}}\text{km}.$

And, the distance covered in $3600$ seconds or $1$ hour will be equal to,

$3600\times \frac{2}{10\sqrt{3}}\text{km}=240\sqrt{3}\text{km}$

Hence, the distance covered by an airplane in $1$ hour is equal to $240\sqrt{3}\text{km}$ so, the speed of the airplane is equal to $240\sqrt{3}\text{km/hr}.$