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Answer:
1.72m is height of man
Step-by-step explanation:
since we know tan60°=[tex]\sqrt{3}[/tex];
then here tan60°=height/base;
height =64-height of man (let be x)
base=36;
now
[tex]\sqrt{3}[/tex]=[tex]\frac{64-x}{36\\}[/tex]
x=64-36[tex]\sqrt{3}[/tex]
x=1.72 m
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Answer: Answer is 4(16 - 9[tex]\sqrt{3}[/tex])
Step-by-step explanation: Concept:
The term angle of elevation denotes the angle from the horizontal upward to an object. An observer’s line of sight would be above the horizontal.
tan θ = [tex]\frac{opposite}{adjacent}[/tex]
tan 60 =[tex]\sqrt{3}[/tex]
Given:
It is given that AB is the height of the man, the angle of elevation is 60°, the distance from the man to the tower BC is 36 m and the height of the tower is 64 m.
The height of the man AB.
In ΔADE
tan 60 = [tex]\frac{ED}{AD}[/tex]
[tex]\sqrt{3}[/tex] = [tex]\frac{ED}{36}[/tex]
36 [tex]\sqrt{3}[/tex] = ED
Now,
DC=CE-CD
CD = 64 -36 [tex]\sqrt{3}[/tex]
CD = 4(16 - 9 [tex]\sqrt{3}[/tex])
Now, given that AB=CD
Therefore, The height of the man is 4(16 - 9[tex]\sqrt{3}[/tex])
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