Math, asked by saurabh11102000, 1 year ago

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
saurabh11102000: Vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower ?
Anonymous: u already said the distance as 72m
Anonymous: The question is itself a question

Answers

Answered by Anurup12
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.]



                 
     
Answered by TPS
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

TPS: Thank you Sourav.
saurabh11102000: ur is right but method is wrong if u want u can check on google book
TPS: i have used trigonometry. What is wrong in it?
saurabh11102000: it is not wrong but this method is also right
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