Question

And aeroplane is flying horizontally 1 km above the ground is observed at an angle elevation of 60° after 10 seconds it’s elevation is object to be 30° find the uniform speed of the aeroplane in kilometre per second

Answer 1

time (t)=10/60*60=1/360
in ∆ADE,
tan 60°=1/DE,
DE=1/√3
in ∆BCE,
tan 30°=1/DE+DC,
DE=√3-DC
now see the soln..further in the given pic

Answer 2

When the elevation was 60, distance of the point of observation to the point on the ground

below the plane = 1xcot60 = 1/
$\sqrt{3}$
km

 

Similarly when the angle becomes 30, distance becomes = 1xcot30 =
$\sqrt{3}$
km

 

So, distance travelled =
$\sqrt{3}$
-1/
$\sqrt{3}$
km = 2/
$\sqrt{3}$
km

 

time taken = 10sec = 10/3600 hr

 

hence speed = distance / time = 2x3600/10x
$\sqrt{3}$
= 720/
$\sqrt{3}$
km/hr = 415.7 km/hr