A building is 50 feet high. At a distance away from the building, an observer notices that the angle of elevation from the ground to the top of the building is 41 degrees. How far is the observer from the base of the building to the nearest tenth of a foot?

Question

A building is 50 feet high.  At a distance away from the building, an observer notices that the angle of elevation from the ground to the top of the building is 41 degrees.  How far is the observer from the base of the building to the nearest tenth of a foot?

anonymous · Answer

Call x the distance from the observer to the building
tan 41 = 50'/x -> x = 50'/tan 41 = ...... in feet

eg145341 · Answer

So x would equal 57.5?