(In reply to not a proof but ...
It's easy to show the result for right triangles:
In a right triangle, the sine of the 90-degree angle is 1. The other two angles are acute and complementary. Each one is greater than zero degrees. If one is called x, the sum of the sines of these other two angles is sin x + sin(90-x). That this total always exceeds 1 is illustrated by the fact that if the portion of the sine curve between zero and 90 degrees were changed into a straight line connecting (0,0) to (90,1) -- considering the function in degree measure -- the straight line would represent the condition under which sin x + sin(90-x) would be exactly 1. But the true sine curve is always above this straight line, so sin x + sin(90-x) always exceeds 1, except at the end points when one is zero and the other 90.
I tried to use this by dividing the acute triangle into two right triangles. The sum, then greater than 4 would be diminished by the 2 for the right angles lost, but I couldn't account for the two acute angles that would merge into one, as the sine of the combined angle would be less than the total of the sines of the two angles when they were in separate right triangles.
But combined with the table in the previous post, the totals look higher for acute triangles than for right triangles. But how to prove it....?
Posted by Charlie
on 2007-10-13 16:47:30