Recent content by asymptote

Wow, that's the very long way around :) Though elegant, you can avoid all of this by simply looking at a 30-60-90 triangle and choosing the angle whose \tan^{-1} = \frac{1}{sqrt(3)} Find the angle where \frac{opposite}{adjacent} =\frac{1}{sqrt(3)}

This is one of the 'special' triangles, for which we know exact trigonometric ratios. Look at the 60°-30°-90° triangle for the answer. The point of the problem is to solve the question without using a calculator, only the special triangle.