Question

From the top of a tower 100 m 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° respectively. Find the distance between the cars. [Take `sqrt(3) = 1.732`.]

Answer 1

Let PQ be the tower
We have,
PQ = 100m, ∠PQR = 30° and ∠PBQ = 45°
In ΔAPQ,

`tan 30° = (PQ)/(AP)`

`⇒ 1/ sqrt(3) = 100/(AP)`

`⇒AP = 100 sqrt(3) m`

Also, in ΔBPQ,

` tan 45° = (PQ)/(BP)`

`⇒ 1 = 100/(BP)`

⇒ BP = 100M

Now , AB = AP+ BP 

   `= 100 sqrt(3) + 100`

   `= 100( sqrt(3) +1)`

   `= 100 xx (1.73 +1 )`

   ` = 100 xx 2.73` 

    = 273 m 

So, the distance between the cars is 273m.