Question

A plane is flying horizontally at a height of $1km$ from ground. Angle of elevation of the plane at a certain instant is ${60}^o$. After $20s$, angle of elevation is found ${30}^o$. The speed of plane is :

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

Answer 1

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, $\angle AOD={60}^o$ and $\angle BOC={30}^o$
In right angled $\Delta OAD$ and $\Delta OBC$, we have
$\tan {60}^o=\frac{AD}{OA}$ and $\tan {30}^o=\frac{BC}{OB}$
$\Rightarrow OA=\frac{1000}{\sqrt{3}}$ 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 $=\frac{Distance}{Time}=\frac{2000}{\sqrt{3}\times 20}=\frac{100}{\sqrt{3}}m/s$.