Question

A man standing on top of a tower of 100mts high observed two cars on both sides of the tower with angle of depression 30° and 45° .what is the distance between the two cars​

Answer 1

Answer:

Step-by-step explanation: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

Answer:

Step-by-step explanation:

100mts

i think

the answer is 274.5