A man 1.68m tall is on top of a building that is 2000m in height looks at the person at the ground forming an angle of depression of 36 deg 45 min. How far is the building from the man? Show your solution. *
Answers
Step-by-step explanation:
Given :-
A man 1.68m tall is on top of a building that is 2000m in height looks at the person at the ground forming an angle of depression of 36 deg 45 min.
To find :-
How far is the building from the man?
Solution :-
Given that
The height of the man = 1.68 m
The height of the building = 2000 m
Angle of depression = 36° 45'
On converting the data into pictorial representation
∆ACD is a right angled triangle
<ACD = 90°
AB is the height of the man
BC is the height of the building
Total length = AC = AB + BC
=> AC = 1.68 m + 2000 m
=> AC = 2001.68 m
D is the point of the person
CD is the distance between the base of the building and the person = d m
< CAE = 36°45'
=> <CDA = 36°45'
Since AE || CD and AC is a transversal
We know that
Alternative interior angles are equal.
We know that
Tan θ = Opposite side to θ/Adjacent side to θ
=> Tan 36°45' = AC / CD
=> Tan 36° 45' = 2001.68/d
=> 0.7468 = 2001.68/d
=> d × 0.7468 = 2001.68
=> d = 2001.68/0.7469
=> d = 2680.34 m ( approximately)
Therefore ,The required distance = 2680.34 m
Answer:-
The distance between the base of the building and the man is 2680.34 m
Used formulae:-
→ Tan θ = Opposite side to θ/Adjacent side to θ
→ Tan 36°45' = Tan 36°(42'+3')
=> 0.7454+14 = 0.7468
→ If two parallel lines Intersected by a transversal then alternative interior angles are equal.
