Question

From the top of a tower, 100, high, a man observes two cars on the opposite sides of the tower and in same straight line with its base, with angles of depression 30° and 45°. Find the distance between the cars. [Take `sqrt3` = 1.732]

Answer 1

Let PQ be the tower and A and B are two cars.
We have,

PQ = 100 m, ∠PAQ = 30^° and ∠PBQ = 45^°

In ∆APQ

`tan 30^@ = (PQ)/(AP)`

`=> 1/sqrt3 = 100/(AP)`

`=> AP =100sqrt3` m

Also In ∆ BPQ

`tan 45^@ = (PQ)/(BP)`

`=> 1= 100/(BP)`

`=> BP = 100 m`

Now, AB = AP + BP`

`= 100sqrt3 + 100`

`=100(sqrt3 + 1)`

=100 x (1.732 + 1)

= 100 x 2.732

= 273.2 m

So the distance between the cars is 273.2 m