knowing the angle alpha and the height inside an isosceles triangle, how can I get the length of the side y?
I did search in search engines for the formula for too long but could not find it and I cant remember a lot from trigonoetry from school and how to derive this. So I thought I ask here ... thanks for understanding and any help
On
Construct the perpendicular bisector of the base side. This cuts the triangle into two right triangles. Take one of them.
This triangle has sides: $h$ = height of the triangle; $\frac y2$ = half of the side opposite $\alpha$ (the side that isn't equal in the isoceles triangle); and $x$ = the side that is equal in the isoceles triangle.
The angles of the right triangle are: $\frac {\alpha }2$; $90$, $90-\frac {\alpha}2$.
$\sin \frac \alpha 2= \frac {\frac y2}s$
$\cos \frac \alpha 2 = \frac hs$
$\tan \frac \alpha 2 = \frac {\frac y2}h= \frac y{2h}$.
And so $y = 2h*\tan \frac \alpha 2$.
(You can also find out the other two sides of the isoceles triangle this way: $s = \frac h{\cos \frac\alpha 2}=h*\sec \frac \alpha 2$)
use the relation $$\tan(\alpha/2)=\frac{\frac{y}{2}}{h}$$