Question

an aeroplane flying horizontally at a height of 1.5 km above the ground is observed at a certain point on earth to subtend an angle of 60 degree . after 15 seconds ,its angle of elevation is observed to be 30 degree . calculate the speed of aeroplane in km/h (√3=1.73)

Answer 1

[IN DIAGRAM, BC = ED = 1500 m. It is not 2500m. It is an error. I regret for the inconvenience]

In ∆ABC,
cot 60° = AC/BC
=> 1/√3 = AC/1500
=> AC = 1500/√3 ....(i)

In ∆AED,
cot 30° = AE/ED
=> √3 = AE/1500
=> AE = 1500√3 ....(ii)

From figure,
BD = CE 
and, CE = AE - AC
so, BD = AE - AC

Now, substituting value of AE and AC from equations (i) and (ii),
BD = 1500√3 - 1500/√3
= 1500 (√3 - 1/√3)
= 1500 × 2/√3
BD = 3000/√3

BD is covered by the aeroplane in 15 seconds. So,
Speed of Aeroplane = Distance (BD) ÷ Time (t)
= 3000/√3 ÷ 15 m/sec
= 3000/ (15 × 1.732) m/sec
= 115.473 m/sec

Hence, speed of the aeroplane is 115.473 m/sec

Answer 2

Answer:

Step-by-step explanation:

Please refer to the photo for the answer.

I HAVE NOT CONVERTED THE UNIT TO METRE BECAUSE THE WE NEED THE ANSWER IN KM/HR