From the top of tower, 100 m high, a man observes two cars on the opposite sides of the tower with the angles of depression 30o& 45 o respectively. Find the distance between the cars.
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Let the Tower be AC = 100m
Distance between cars is BD
BD=BC+CD
Let the BC be 'x' and CD be 'y'
In Traingle ACD
Cot45°= Base/Perpendicular
Cot45°=y/100
1=y/100
y=100m
In Traingle ACB
Cot30°=x/100
1.732=x/100
1.732×100=x
x=173.2m
Distance between cars = BC+CD
=x+y
=173.2+100
=273.2m.
I hope it helps.
Distance between cars is BD
BD=BC+CD
Let the BC be 'x' and CD be 'y'
In Traingle ACD
Cot45°= Base/Perpendicular
Cot45°=y/100
1=y/100
y=100m
In Traingle ACB
Cot30°=x/100
1.732=x/100
1.732×100=x
x=173.2m
Distance between cars = BC+CD
=x+y
=173.2+100
=273.2m.
I hope it helps.
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