A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the
wer with a uniform speed. Six seconds later, the angle of depression of the car is found
to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Answers
Answer:3 seconds
Step-by-step explanation:let CD=h
angle of depression=30
after 6 sec angle of depression=60
so let AB=y and BC=x
in triangle BCD
tan60=CD/BC
=h/x
h=x....(i)
in triangle ACD
tan30=CD/AC
1/=h/x+y
x+y=h
from eq(i)
x+y=()
x+y=3x....(ii)
It is given that a car moves from point A to B in 6 sec.
let speed =k km/s
time=distance/speed
6=y/k
y=6k
on putting y=6k in eq(ii)
x+6k=3x
6k=2x
x=3k
time=distance/speed
=x/k
=3k/k
=3 seconds
let CD=h
angle of depression=30
after 6 sec angle of depression=60
so let AB=y and BC=x
in triangle BCD
tan60=CD/BC
\sqrt{3}=h/x
h=\sqrt{3}x....(i)
in triangle ACD
tan30=CD/AC
1/\sqrt{3}=h/x+y
x+y=\sqrt{3}h
from eq(i)
x+y=(\sqrt{3})\sqrt{3}
x+y=3x....(ii)
It is given that a car moves from point A to B in 6 sec.
let speed =k km/s
time=distance/speed
6=y/k
y=6k
on putting y=6k in eq(ii)
x+6k=3x
6k=2x
x=3k
time=distance/speed
=x/k
=3k/k
=3 seconds