Question

The angle of elevation of a jet plane from a point on the ground is$60°$. After a flight of $20sec$ at the speed of $432km/hr$, the angle of elevation changes to $30°$. If the jet plane is flying at a constant height, then its height is:

• A. $1200\sqrt{3}m$
• B. $1800\sqrt{3}m$
• C. $3600\sqrt{3}m$
• D. $2400\sqrt{3}m$

Answer 1

The correct option is A

$1200\sqrt{3}m$

Given

The initial angle of elevation $=60°$

The changed angle of elevation $=30°$

Step 1 - Distance traveled $\left(PQ\right)$

Speed $\left(v\right)=432km/hr$

$v=432\times \frac{5}{18}m/sec$

$v=120m/sec$

we know that,

distance traveled $\left(PQ\right)=$speed $\times$time

$$\begin{gathered} =120\times 20 \\ =2400m \end{gathered}$$

Step 2- the jet plane is flying at a constant height

$$\begin{gathered} In△PAC \\ \tan 60°=\frac{h}{AD} \\ \Rightarrow \sqrt{3}=\frac{h}{AD} \\ \Rightarrow AD=\frac{h}{\sqrt{3}}m \end{gathered}$$

$$\begin{gathered} In△AQD \\ \tan 30°=\frac{h}{AC} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{AC} \\ \Rightarrow AC=\sqrt{3}hm \end{gathered}$$

Now,

$$\begin{gathered} \because AD=AC+CD \\ \Rightarrow \sqrt{3}h=\frac{h}{\sqrt{3}}+2400(\because CD=PQ=2400m) \\ \Rightarrow \sqrt{3}h-\frac{h}{\sqrt{3}}=2400 \\ \Rightarrow \frac{2h}{\sqrt{3}}=2400 \\ \Rightarrow h=\left(2400\times \frac{\sqrt{3}}{2}\right)m \\ \Rightarrow h=1200\sqrt{3}m \end{gathered}$$

Therefore, the jet plane is flying at a constant height of $1200\sqrt{3}m$

Hence, Option A is correct.