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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 a way 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°. Find the speed of flying of the bird.
`("Take"sqrt3=1.732)`
उत्तर
`tan 45° = (AC)/(BC)`
`1 = 80/(BC)`
BC = 80 cm
`tan 30° = (DE)/(BC+CE)`
`1/sqrt3 = 80/(80+x)`
80 + x = 80 √3
x = 80 √3 − 80
x = 80 (√3 − 1) m
The bird travelled 80 (√3 − 1) m
Speed of bird = `"Distance"/"Time"`
`= 40 (sqrt3-1)/2`
= 40 (√3 − 1) m/s
∴ Speed = 40 (√3 − 1) m/s,
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