A man on the top of a vertical observation tower observers a car moving at a uniform speed coming di......

A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.
Then, angleADC = 30° , angleACB = 45°
Let AB = h, BC = x, CD = y
tan45°=AB/BC=h/x
=>1=h/x
=>h=x?(1)
tan30°=AB/BD=AB/(BC + CD)=h/(x+y)
=>1/√3=h/(x+y)
=>x + y = √3h
=>y = √3h - x
=>y = √3h−h(? Substituted the value of x from equation 1 )
=>y = h(√3−1)
Given that distance y is covered in 8 minutes.
i.e, distance h(√3−1)is covered in 8 minutes.
Time to travel distance x
= Time to travel distance h (? Since x = h as per equation 1).
Let distance h is covered in t minutes.
since distance is proportional to the time when the speed is constant, we have
h(√3−1)∝8?(A)
h∝t?(B)
(A)/(B)=>h(√3−1)/h=8/t
=>(√3−1)=8/t
=>t=8/(√3−1)=8/(1.73−1)
=8/.73=800/73minutes
=10 70/73minutes
≈10minutes 57seconds

By: MIRZA SADDAM HUSSAIN