Question

the angle of elevation of an aeroplane from a point A on the ground is 60°. after a flight of 30 sec ,the angle of elevation change to 30°. if the plane is flying at a constant height of 3600m, then find the speed in km/hr​

Answer 1

Answer:

In triangle ACE,

\begin{gathered} \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{gathered}

tan(30)=

ce

ac

3

1

=

ce

3600

3

ce=3600

3

×

3

ce=10800m

CE = 10800 m

AC = BD =

3600 \sqrt{3} \: m3600

3

m

In triangle BED,

\begin{gathered} \tan(60) = \frac{bd}{de} \\ \sqrt{3} = \frac{3600 \sqrt{3} }{de} \\ de = \frac{3600 \sqrt{3} }{ \sqrt{3} } = 3600 \: m\end{gathered}

tan(60)=

de

bd

3

=

de

3600

3

de=

3

3600

3

=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=

time

distance

=

30

7200

=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

Answer 2

Step-by-step explanation:

In triangle ACE,

$\begin{gathered} \tan(30) = \frac{ac}{ce} \\ \frac{1}{ \sqrt{3} } = \frac{3600 \sqrt{3} }{ce} \\ ce = 3600 \sqrt{3} \times \sqrt{3} \\ ce = 10800 \: m\end{gathered}$

tan(30)= ca aa 31 = ce 3600 3

ce=3600 3 × 3

ce=10800m

CE = 10800 m

AC = BD =

3600 $\sqrt{3} \: m 3600 3$

In triangle BED,

$\begin{gathered} \tan(60) = \frac{bd}{de} \\ \sqrt{3} = \frac{3600 \sqrt{3} }{de} \\ de = \frac{3600 \sqrt{3} }{ \sqrt{3} } = 3600 \: m\end{gathered}$

tan(60)= db bd

3 = de 3600 3

de= 3 3600 3

=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= time distance

= 30 7200

=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