Question

From the top of the tower,100 m high a man observes two cars on the opposite side of the tower and in same straight line with its base with angles of depression 30 and 45 degrees find the distance between the cars

Answer 1

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

Answer 2

Plz find the attachment for answers