Question

A plane is flying horizontally at a height of $1Km$ from ground. Angle of elevation of the plane at a certain instant is $60^{\circ }$. After $20$ seconds angle of elevation is found $30^{\circ }$. The speed of plane is

• A. $\frac{100}{\sqrt{3}}m/s$
• B. $\frac{200}{\sqrt{3}}m/s$
• C. $100\sqrt{3}m/s$
• D. $200\sqrt{3}m/s$

Answer 1

Answer: (A)

Let $AD$ be the height at which the plane is flying i.e.
$1km=1000m$ and $C$ be the position of the plane after $20s$.
It is given that, $\angle AOD=60^{\circ }$ and $\angle BOC=30^{\circ }$

In right angled $\Delta OAD$ and $\Delta OBC$, we have
$\tan 60^{\circ }=\frac{AD}{OA}$ and $\tan 30^{\circ }=\frac{BC}{OB}$
$\Rightarrow OA=\frac{1000}{\sqrt{3}}$ and $\frac{1}{\sqrt{3}}=\frac{1000}{OB}$
$[\because 1km=1000m]$
$\Rightarrow OA=\frac{1000}{\sqrt{3}}m$ and $OB=1000\sqrt{3}m$
$[\because AD=BC=1000m]$
$\Rightarrow$ Now, $OA+AB=1000\sqrt{3}$
$\Rightarrow AB=1000\sqrt{3}-\frac{1000}{\sqrt{3}}$
$\Rightarrow AB=\frac{3000-1000}{\sqrt{3}}=\frac{2000}{\sqrt{3}}m$
$\therefore$ Speed $=\frac{\text{ Distance }}{\mathrm{Time}}=\frac{2000}{\sqrt{3}\times 20}=\frac{100}{\sqrt{3}}m/s$