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?
 A: Cut the iscoles triangle in half to get a two right triangles with opposite side $\frac 12 y$ and hypotenuse $x$.
$\frac 12 Y = \sin (\frac 12 a) x$  
So apparently this is claiming $ 2\sin(\frac 12 a) = \tan a$ which isn't true but is apparently an approximation.  $\tan a = \frac{\sin (\frac 12 a + \frac 12 a)}{\cos(\frac 12 a + \frac 12 a)} = \frac {2\sin \frac 12 a\cos \frac 12 a}{\cos^2 \frac 12 a - \sin^2 \frac 12 a}=2\sin\frac 12 a*\frac {\cos \frac 12a}{\cos^2 \frac 12 a - \sin^2 \frac 12 a}$
And $\frac {\cos \frac 12a}{\cos^2 \frac 12 a - \sin^2 \frac 12 a}=\frac {\cos \frac 12a}{1 - 2\sin^2 \frac 12 a}$ which, I guess for small values of $a$ is close to 1. (you said $x > y$ so $a < 60$ and $\frac 12 a < 30$ So for $a = 60$ then term is $\frac{4\sqrt{3}}6=1.1547$ and as $a$ decreases it gets closer to $1$... I guess.)
A: 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 $\tan a \approx \frac yx$
