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 tower with a uniform speed. Two minutes later, the angle of depression of the car is found to be 60°. How much time will the car take to reach the tower?
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let the speed of the car be "a"km/h and time taken to cover BC be "b" min.
we know speed = distance / time
therefore, time = distance / speed
2 min= DC/ a
DC =2a........(1)
In ∆ABD,
tan30°=AB/BD
1/√3=AB/DC+BC
1/√3=AB/2a+BC
AB=(2a+BC)/√3............(2)
In∆ABC
tan 60°=AB/BC
√3=AB/BC
√3BC=AB
√3BC=(2a+BC)/√3......{from (2)}
3BC=2a + BC
2BC=2a
2BC=DC.........{from(1)}
BC=DC/2
now time taken to cover DC is 2min
therefore , time taken to cover BC=2/2 = 1 min
in total the car takes 3min to reach the tower
and to cover BC it takes 1 min.
this is a brainliest answer!!!!!!!
we know speed = distance / time
therefore, time = distance / speed
2 min= DC/ a
DC =2a........(1)
In ∆ABD,
tan30°=AB/BD
1/√3=AB/DC+BC
1/√3=AB/2a+BC
AB=(2a+BC)/√3............(2)
In∆ABC
tan 60°=AB/BC
√3=AB/BC
√3BC=AB
√3BC=(2a+BC)/√3......{from (2)}
3BC=2a + BC
2BC=2a
2BC=DC.........{from(1)}
BC=DC/2
now time taken to cover DC is 2min
therefore , time taken to cover BC=2/2 = 1 min
in total the car takes 3min to reach the tower
and to cover BC it takes 1 min.
this is a brainliest answer!!!!!!!
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