Question 1193625
<font face="Times New Roman" size="+2">


The altitude of an equiangular triangle forms two 30-60-90 triangles where the altitude is the long leg.  The long leg of a 30-60-90 triangle is in *[tex \Large 1:\frac{\sqrt{3}}{3}:\frac{2\sqrt{3}}{3}] proportion to the short side and hypotenuse respectively.


																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>