The angle of elevation of a building 80 m high from a man on the ground is 50.41degrees. How far is the man from the building?

Question

1.   The angle of elevation of a building 80 m high from a man on the ground is 50.41 degrees. How far is the man from the building?

Answer 1

Given:

L = 80 ft

θ = 50.41°

Find:

B = ? ft

Solution:

SOH CAH TOA is your friend here

sin θ = opp / hypot

cos θ = adj / hypot

tan θ = opp / adj

What is given (or soon to be calculated) in this problem is the opposite (dimension L) and the adjacent (dimension B) sides of a right triangle. The angle θ is made with the hypotenuse of the right triangle and the adjacent side meaning the TOA part will be set up like:

tan θ = L / B

Solving for B:

B = L / tan θ = L * cot θ (this swap utilizes the trig fact that cot θ = 1 / tan θ)

Plugging the values in:

B = ( 80 ft ) * cot( 50.41° ) ∴ B ≈ 66.16 ft

Therefore, the man is about 66.16 ft away from the building.

I hope this helps! Message me in the comments if you have any questions, comments, or concerns!