Tan function and isosceles triangles 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?

bisect angle $a$
$\sin \frac a2 = \frac {y}{2x}\\
\cos \frac a2 = \frac {\sqrt{4x^2-y^2}}{2x}\\
\tan \frac a2 = \frac {y}{\sqrt{4x^2-y^2}}\\
\tan a = \frac {2\tan\frac a2}{1-\tan^2 \frac a2}\\
\tan a = \frac {2y}{\sqrt{4x^2-y^2}(1-\frac {y^2}{4x^2-y^2})}\\
\tan a = \frac {y\sqrt{4x^2-y^2}}{2x^2-y^2}$
$\frac yx$ is small, $\frac{\sqrt{4x^2-y^2}}{2x^2-y^2} \approx \frac {1}{x}$
and $\tan a \approx \frac yx$
