Question

the pilot of an aeroplane observed that the angle of elevation of a kilometre stone is 30° when his aeroplane is at aparticuoar altittude.When he increased his altittude 300m the angle of elevation of a next kilometer stone is 60°.Find the height of the aeroplane when the first observation is made.

Answer 1

In triangle ACE,

\begin{lgathered}\tan(30) = \frac{ac}{ce} \\ \frac{1}{ \sqrt{3} } = \frac{3600 \sqrt{3} }{ce} \\ ce = 3600 \sqrt{3} \times \sqrt{3} \\ ce = 10800 \: m\end{lgathered}tan(30)=ceac31=ce36003ce=36003×3ce=10800m 

CE = 10800 m

AC = BD = 

3600 \sqrt{3} \: m36003m 

In triangle BED,

\begin{lgathered}\tan(60) = \frac{bd}{de} \\ \sqrt{3} = \frac{3600 \sqrt{3} }{de} \\ de = \frac{3600 \sqrt{3} }{ \sqrt{3} } = 3600 \: m\end{lgathered}tan(60)=debd3=de36003de=336003=3600m 

CD + DE = CE

CD + 3600 = 10800

CD = 10800 - 3600 = 7200 m

Distance travelled = 7200 m

Time taken = 30 seconds

speed = \frac{distance}{time} = \frac{7200}{30} = 240 \: m {s}^{ - 1}speed=timedistance=307200=240ms−1 

In 1 second = 240 m

In 3600 seconds (1 hour) = 240 × 3600 = 864000 m = 864 km

In hour = 864 km

Speed = 864 km/h