Apply Morley's Theorem?

Let $r$ be circumradius of $\triangle ABC$ ,

$\alpha=\angle A/3, \beta=\angle B/3, \gamma=\angle C/3$

Side of a Morley triangle is:

$s=8 r* \sin(\alpha) \sin(\beta)\sin(\gamma)=$ $8 * 5* \sin(30) \sin( \arctan(3 / 4)/ 3) \sin( \arctan(4 / 3)/ 3)\approx 1.295$

$Area=\frac{\sqrt{3}}{4} s^2\approx \boxed{0.726}$

I just used brute force after plotting the triangle on a Cartesian Plane. I just banged out Point-Slope equations to find X,Y and Z (with paper and pencil (and calculator for trig ratios and long division) - and it took about half an hour using basic algebra techniques). I was humbled to see that XYZ is an equilateral triangle. There's got to be a simpler way.

link text
$Let\ \alpha=\angle X/3, \beta=\angle Y/3, \gamma=\angle Z/3.\\ The \ circumradius \ R=\dfrac{xyz}{4*area\ XYZ}=\dfrac{6*8*10}{4*\frac 1 2 *6*8}=5.\\ In\ the\ above\ link\ the\ \Delta\ XYZ\ is\ named\ as\ \Delta\ PQR,\ and\ original\ triangle\ is inscribed \ in\ a\ unit\ circle.\\ The\ formula\ given\ is\ PQ\ =\ 8*Sin(\alpha)*Sin(\beta)*Sin(\gamma).\\ In\ our\ case\ this\ would\ translate\ into,\\ XY=8 R* Sin(\alpha) Sin(\beta)Sin(\gamma)\\ But\ Morley's\ Theorem\ states\ that\ this\ is\ an\ equilateral\ \Delta.\\ \therefore\ the\ area\ of\ XYZ=\frac{\sqrt3} 4 *\{R*Sin(\alpha)Sin(\beta)Sin(\gamma)\}^2\\ \implies\ area \ of\ XYZ\ =\ \frac{\sqrt3} 4 *\left\{5*Sin\left(\dfrac{Sin^{-1} .8} 3\right)*Sin\left(\dfrac{Sin^{-1} .6} 3\right)*Sin\left(\dfrac{Sin^{-1} 1} 3\right) \right \}^2\\ area \ of\ XYZ\ =\large \ \ \ \ \color{#D61F06}{0.7262}$