#### 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]

#### Solution

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

Is there an error in this question or solution?

Solution 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 Concept: Heights and Distances.