Question

The angle of elevation at the top of a building is 34 degrees. At a point 80 feet closer , the angle of elevation to the top of the same building is 45 degrees. Approximate the height of the building.

Answer 1

According to the diagram, AB is the building.

Let distance BC= x CD= 80 ft

So distance BD= x + 80

In triangle ABD,

anlge ACB = 45 degree

So Triangle ACB is an isosceles triangle.

So AB = BC = x

Consider triangle ABD,

tan (34) = $\frac{AB}{BD}$

0.6745 = $\frac{x}{x+80}$

Cross multiply

0.6745 (x + 80 ) = x

0.6745 x + 53.46 = x

53.46 = x - 0.6745x

53.46 = 0.3255 x

Now divide both side by 0.3255

x = 164.24 ft

So height of the building is 164 ft approximately