I have a non-right-angled isosceles triangle with two longer sides, X, and a short base Y.
I know the length of the long sides, X.
I also know the acute, vertex angle opposite the base Y, let's call it angle 'a'
I have been told I can calculate the length of the base Y by:
Y = tan(a) x X
I've sketched this out with a few hand drawn triangles and it does seem to work...... But why?
I can't derive that formula from any of the trigonometry I know. What am I missing?
Y = tan(a) x X
At best, that's an approximation only valid for very small angles $a\,$. Draw the altitude on the base, and you should figure out what the correct formula is, instead. $\endgroup$ – dxiv Mar 24 '17 at 23:15