Question

the angle of elevation of an aeroplane from a point on the ground is 60 degree after a flight of 30 seconds the angle of elevation changes to 30 degree if the plane is flying at a constant height of 3600 under root 3 M find the speed of the plane in kilometre per hour

Answer 1

In triangle ACE,
$\tan(30) = \frac{ac}{ce} \\ \frac{1}{ \sqrt{3} } = \frac{3600 \sqrt{3} }{ce} \\ ce = 3600 \sqrt{3} \times \sqrt{3} \\ ce = 10800 \: m$
CE = 10800 m
AC = BD =
$3600 \sqrt{3} \: m$
In triangle BED,
$\tan(60) = \frac{bd}{de} \\ \sqrt{3} = \frac{3600 \sqrt{3} }{de} \\ de = \frac{3600 \sqrt{3} }{ \sqrt{3} } = 3600 \: m$
CD + DE = CE
CD + 3600 = 10800
CD = 10800 - 3600 = 7200 m
Distance travelled = 7200 m
Time taken = 30 seconds
$speed = \frac{distance}{time} = \frac{7200}{30} = 240 \: m {s}^{ - 1}$
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

Answer:

Let P and Q be the two positions of the plane and A be the point of observation. Let ABC be the horizontal

line through A. It is given that angles of elevation of the plane in two positions P and Q from a point A are

60° and 30° respectively.

So, ∠PAB = 60°, ∠QAC = 30°

Again given that PB = 3600√3 m

In ΔABP, we have

tan 60° = BP/AB

=> √3 = 3600√3/AB

=> AB = 3600 m

In ΔACQ, we have

tan 30° = CQ/AC

=> 1/√3 = 3600√3/AC

=> AC = 3600√3 * √3  = 3600 * 3 = 10800 m

Now, Distance = BC = AC – AB = 10800 m – 3600 m = 7200 m

Thus, the plane travels 3000 m in 30 seconds

Hence, speed of plane = 7200/30 m/sec = 240 m/sec

                                                          = (240/1000) * 60 * 60 km/hr  

[Since 1 km = 1000 m, 1 hr = 60 minutes = 60 * 60 seconds]

                                                          = 864000/1000 km/hr

                                                          = 864 km/hr