# hyperbolic equilateral triangle : $\cosh \left(\frac{1}{2} a\right) \sin \left(\frac {1}{2} \alpha\right) = \frac{1}{2}$

In an equilateral triangle with side length $a$ and angle $\alpha$, $$\cosh \left(\frac{1}{2} a\right) \sin \left(\frac {1}{2} \alpha\right) = \frac{1}{2}.$$
This uses the identity $\sinh(2x)=2\sinh(x)\cosh(x)$ from the Wikipedia article on hyperbolic functions.