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
Answers
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
Step-by-step explanation:
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