Question

A building is 50 feet high. At a distance away from the building, an observer notices that the angle of elevation to the top of the building is 41º. How far is the observer from the base of the building?
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Answer 1

Answer:

tan41 = opposite/ adjacent

0.86929 = 50/ distance between the building and the observer

distance between the building and the observer= 50/0.86929 = 57.52 meters

Answer 2

Answer:  

The observer is 57.157 [approx] feet far from the base of the building

Step-by-step explanation:

Given data

A building is 50 feet high

at a distance away from the building, an observer notices that the angle of elevation to the top of the building is 41°

Here we need to find the distance between building base to the observer

let A be the top of the building, B be the base of the building and C be the point of the observation, therefore right angled triangle ABC is formed. Here BC is the distance between observer and building base

⇒ From triangle ABC,

             tan 41° = $\frac{opposite side }{hypotenuse }$  =  $\frac{AB}{BC}$  =  $\frac{50}{BC}$  

             tan 41° =  $\frac{50}{BC}$  

             BC =   50/ tan 41°    

             BC = 50/ 0.8693    [ tan 41° = 0.8693 approx ]

             BC = 57.517 approx