Find the angle of elevation from the point on the ground 90 feet from the base of a building that is 200 feet tall

Sagot :

[tex]Look\ at\ the\ picture.\\\\tg\alpha=\frac{200}{90}\\\\tg\alpha=\frac{20}{9}\\\\tg\alpha=2.222...\\\\\alpha\approx66^o[/tex]
View image Аноним
View image Аноним
You didn't say so, but I'm assuming that what you want is the angle of elevation
when you're looking at the top of the building.

Also, since you didn't specify it, we have to assume that you and the building
are on level ground, and we also have to assume that your eye is located at
ground level. In other words, your eye is level with the base of the building.

Now ... The building is perpendicular to the ground.  So the building, the line
on the ground from the building to you, and the line of sight from your eye to
the top of the building, form a right triangle.

The angle we need is the angle located at your eye.  The line along the ground
from the building to you is the side of the triangle that's adjacent to that angle. 
And the building is the side of the triangle that's opposite that angle.

The tangent of the angle is (opposite) / (adjacent) = 200/90 = 2.2222...

Put that number into your calculator, and find the angle whose tangent it is.
(That's called the 'arctangent' of 2.2222, or tan⁻¹(2.2222) .)

The angle is about 65.77 degrees.  (rounded)