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Question
A person observed the angle of elevation of the top of a tower as 30°. He walked 50 m towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as 60°. Find the height of the tower.
Solution
Let AB be the tower of height h. And person makes an angle of elevation of the top of a tower is 30°, he walks 50 m towards the foot of tower then makes an angle of elevation of 60°
Let BC = x, CD = 50, and ∠ACB = 60°, ∠ADB = 30°
Now we have to find the height of the tower.
We have the corresponding figure as follows
So we use trigonometric ratios.
In a triangle ABC
`=> tan C = (AB)/(BC)`
`=> tan 60^@ = h/x`
`=> sqrt3 = h/x`
`=> x = h/sqrt3`
Again in a triangle ADB,
`=> tan D = (AB)/(BC + CD)`
`=> tan 30^@ = h/(x + 50)`
`=> 1/sqrt3 = h/(x + 50)`
`=> sqrt3h = x + 50`
`=> 3h = h + 50sqrt3`
`=> 2h = 50sqrt3`
`=> h = 25sqrt3`
=> h = 25 x 1.73
=> h = 43.25
Hence the height of tower is 43.25 m
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After hitting the tree, how far did the ball travel in the sky when Kaushik saw the ball? - What is the speed of the bird in m/min if it had flown `20(sqrt3 + 1) m`?
