triangles and oberservation?

Question

If an observer is 200 ft. from the building and the angle of elevation to the top of a building is 60*, what is the height of the building?

Answer 1

traw a trianlge

the distance from the base is the short leg of the special triangle 30-60-90.

short leg = x
long leg = x sqrt(3)
hypot = 2x

the long leg is the height of the building

so the answer is 200 sqrt(3) ft

Answer 2

Let
h = height building

tan 60° = h/200

h = 200 tan 60° = 200√3 ft

Answer 3

tan(60º)=h/200 --> h=200·tan60º= 200· sq(3) = 346'41... ft.

Saludos.