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?
8 min 17 second
10 min 57 second
14 min 34 second
12 min 23 second
Explanation:
Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.
Then, $\angle$ADC = 30° , $\angle$ACB = 45°
Let AB = h, BC = x, CD = y
tan45°=$\dfrac{AB}{BC}=\dfrac{h}{x}$
=>1=$\dfrac{h}{x}$
=>h=x ⋯(1)
tan30°=$\dfrac{AB}{BD}=\dfrac{AB}{(BC+CD)}=\dfrac{h}{x+y}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{h}{x+y}$
=>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)
$\dfrac{(A)}{(B)}=>\dfrac{h(\sqrt{3}−1)}{h}=\dfrac{8}{t}$
=>(√3−1)=$\dfrac{8}{t}$
=>t=$\dfrac{8}{(\sqrt{3}−1)}=\dfrac{8}{(1.73−1)}$
=$\dfrac{8}{.73}=\dfrac{800}{73}$minutes
=10$\dfrac{70}{73}$minutes
≈10 minutes 57 seconds
Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.
Then, $\angle$ADC = 30° , $\angle$ACB = 45°
Let AB = h, BC = x, CD = y
tan45°=$\dfrac{AB}{BC}=\dfrac{h}{x}$
=>1=$\dfrac{h}{x}$
=>h=x ⋯(1)
tan30°=$\dfrac{AB}{BD}=\dfrac{AB}{(BC+CD)}=\dfrac{h}{x+y}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{h}{x+y}$
=>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)
$\dfrac{(A)}{(B)}=>\dfrac{h(\sqrt{3}−1)}{h}=\dfrac{8}{t}$
=>(√3−1)=$\dfrac{8}{t}$
=>t=$\dfrac{8}{(\sqrt{3}−1)}=\dfrac{8}{(1.73−1)}$
=$\dfrac{8}{.73}=\dfrac{800}{73}$minutes
=10$\dfrac{70}{73}$minutes
≈10 minutes 57 seconds