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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
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