Question

A man 2m tall is 50m away from a building 40m high.What is the angle of elevation of the top of the building from his eyes?

Answer 1

total height of tower =40m

height of tower when we subtract man's height=40-2=38m
height of man =2m
distance =50m
therefore..
tan theta =38/50=19/25

-----
ED =BC=50m
CD=BE=2m
AD=40m
AD=AC+CD
40=AC+2
therefore AC=40-2=38
now theta =opposite/adjecent=AC/BC
therefore tan theta= 38/50=19/25

Answer 2

The angle of elevation of the top of the building from the man's eyes is approximately 37.61 degrees

Let's first depict the problem in a diagram:

              C

            / |

           /  |h = 40m

          /   |

         /    |

        /θ    |

       A------B

          d = 50m

In this diagram, A represents the man's position, B represents the base of the building, and C represents the top of the building. We are looking for the angle θ.

Using trigonometry, we know that:

tan(θ) = opposite / adjacent

In this case, the opposite side is the height of the building, h, and the adjacent side is the distance from the man to the building, d + his height, which is 52m. So we have:

tan(θ) = h / (d + man's height)

tan(θ) = 40 / 52

θ =$tan^{-1} (40/52)$

θ ≈ 37.61°

Therefore, the angle of elevation of the top of the building from the man's eyes is approximately 37.61 degrees

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