From the middle of a ground Habu observes a flying bird toward north at an angle of elevation 30°, after 2 minutes he observes the same bird at an angle of elevation 60° toward south. If the bird always flies in a straight line at height 50√3 meters from the ground level, find its velocity.
Let the initial position of the bird be A and the final position be B. Let O be the position of Habu on the ground.
In right-angled triangle AOX:
tan 30º = \frac{\text{Height of bird}}{\text{Horizontal distance}}
\frac{1}{\sqrt{3}} = \frac{50\sqrt{3}}{x}
or, x = 50√3 × √3
= 50 × 3
= 150 m
In right-angled triangle BOY:
tan 60º = \frac{\text{Height of bird}}{\text{Horizontal distance}}
\sqrt{3} = \frac{50\sqrt{3}}{y}
Cross multiplying:
y = \frac{50\sqrt{3}}{\sqrt{3}} = 50 meters
Total horizontal distance traveled by the bird:
Total distance = x + y
= 150 + 50
= 200 meters
Time taken = 2 minutes = 120 seconds
Velocity of the bird:
v = \frac{\text{Distance}}{\text{Time}} = \frac{200}{120}
v = \frac{5}{3} m/s
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