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From the point of a tower 100m high, a man observe two cars on the opposite sides to the tower with angles of depression 30° and 45 respectively. Find the distance between the cars

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#### Solution

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.

Concept: Heights and Distances

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