Question

An aeroplane flying horizontally at a height of $1.5$ km above the ground is observed at a certain point on earth to subtend an angle of $60$. After $15$ seconds, its angle of elevation is observed to be $30$. Calculate the speed of the, aeroplane in km/h.

• A. $240\sqrt{3}$ km/h
• B. $230\sqrt{3}$ km/h
• C. $210$ km/h
• D. $220$ km/h

Answer 1

The correct option is A $240\sqrt{3}$ km/h

Let $P$ and $Q$ be the two position of the plane.

Given angles of elevation of the plane in two position $P$ and $Q$ as $\angle PAB={60}^0$ and $\angle QAC={30}^0$

$PB=QC=$ $15$ km

Now in right angled $△ABP$

$\tan {60}^0=\frac{BP}{AB}$

$\Rightarrow \sqrt{3}=\frac{1.5}{AB}$

$\Rightarrow AB=\frac{1.5}{\sqrt{3}}$

$\Rightarrow \left(0.5\right)\sqrt{3}$ km

Again in right angled triangle $△ACQ$,

$\tan {30}^0=\frac{QC}{AC}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1.5}{AB}$

$\Rightarrow AC=\left(1.5\right)\sqrt{3}$ km

$PQ=BC=AC-AB$

$=\left(1.5\right)\sqrt{3}-\left(0.5\right)\sqrt{3}$

$=(1.5-0.5)\sqrt{3}$

$=\sqrt{3}$ km

The plane travels $PQ=$ $\sqrt{3}$ km in $15$ secs

$\therefore$ Speed of the aeroplane $=$ $\frac{distance}{time}$

$\Rightarrow \frac{\sqrt{3}}{\frac{15}{3600}}$

$\Rightarrow 240\sqrt{3}$ km/h