:)
In that case let angle BAX=\alpha and we easily get the following angles.
XAC=60-\alpha
ACX=60+\alpha
Without loss of generality let AB=AC=BC=1.
Then, using the sine rule
BX=\frac{\sin \alpha}{\sin 60}
XC=\frac{\sin(60- \alpha)}{\sin 60}
AX=\frac{\sin(60+...