a man standing at 72m distance observe the angle of elevation of the vertical tower.he know that the angle is cosine 0.53. find the height of the tower.
saurabh11102000:
and one more is
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3
We have, cos θ = 0.53, let distance of the man from the foot of the tower be x.
AB = 20m, BC = x, then AC = √[x2 + (20)2] = √[x2 + 400]
cos θ = BC/AC = x/√[x2 + 400]
Or, 0.53 = x/√[x2 + 400]
Or, (0.53)2 = x2/[x2 + 400] [Squaring both sides]
Or, 0.2809 = x2/[x2 + 400]
Or, x2 = 0.2809x2 + 112.36
Or, x2 – 0.2809x2 = 112.36
Or, 0.7191x2 = 112.36
Or, x2 = 112.36/0.7191 = 12.5 m [Ans.]
AB = 20m, BC = x, then AC = √[x2 + (20)2] = √[x2 + 400]
cos θ = BC/AC = x/√[x2 + 400]
Or, 0.53 = x/√[x2 + 400]
Or, (0.53)2 = x2/[x2 + 400] [Squaring both sides]
Or, 0.2809 = x2/[x2 + 400]
Or, x2 = 0.2809x2 + 112.36
Or, x2 – 0.2809x2 = 112.36
Or, 0.7191x2 = 112.36
Or, x2 = 112.36/0.7191 = 12.5 m [Ans.]
Answered by
2
1.distance = 72m
let height = h
cos α = 0.53
sin α = √1-0.53² = 0.848
tan α = sinα/cosα = 1.6
tanα = h/72
⇒h = 72×1.6=115.2m
2.h = 20
cos α = 0.53
distance from tower = d
tan α = 20/d
⇒d = 20/tan α = 20/1.6 = 12.5m
let height = h
cos α = 0.53
sin α = √1-0.53² = 0.848
tan α = sinα/cosα = 1.6
tanα = h/72
⇒h = 72×1.6=115.2m
2.h = 20
cos α = 0.53
distance from tower = d
tan α = 20/d
⇒d = 20/tan α = 20/1.6 = 12.5m
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