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Question
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
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.
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