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प्रश्न
A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such away that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Determine the speed at which the bird flies `(sqrt(3) = 1.732)`
उत्तर
A is the initial position of the bird B is the final position of the bird Let the speed of the bird be s
Distance = speed × time
∴ AB = 2x
Let CD be x
∴ CE = x + 2s
In the ∆CDA, tan 45° = `"AD"/"CD"`
1 = `80/x`
x = 80 ...(1)
In the ∆BCE
tan 30° = `"BE"/"CE"`
`1/sqrt(3) = 80/(x + 2"s")`
x + 2s = `80sqrt(3)`
x = `80sqrt(3) - 2"s"` ...(2)
From (1) and (2) we get
`80 sqrt(3) - 2"s"` = 80
`80 sqrt(3) - 80` = 2s
⇒ `80(sqrt(3) - 1)` = 2s
s = `(80(sqrt(3) - 1))/2`
= 40(1.732 – 1)
= 40 × 0.732
= 29.28
Speed of the flying bird = 29.28 m/sec
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