Trisector?

$\angle C=90^{\circ}$ . As $\angle A<\angle B$ , $AC>AB$ . Therefore the shorter trisection is the one closer to $B$ . Denote this by $D$ .

We have $\angle CBA=60^{\circ},\angle BCA=30^{\circ},\angle BDC=90^{\circ},\angle CDA=90^{\circ},\angle DCA=60^{\circ},\angle DAC=30^{\circ}$ .

Let $BD=x$ and $BC=y$ . So $DA=10-x$ and $CA=\sqrt{100-y^2}$ (By pythagorean theorem ).

Again by Pythagorean theorem, we have $CD=\sqrt{y^2-x^2}=\sqrt{(100-y^2)-(10-x)^2}$ . Simplifying we get $y=\sqrt{10x}$ .

Now consider $\Delta BCD$ . By sine rule (law of sines) we have $\dfrac{CD}{\sin(60^{\circ})}=\dfrac x {\sin(30^{\circ})}=\dfrac{\sqrt{10x}}{\sin(90^{\circ})}$

Simplifying we get $x=\dfrac 5 2$ .

Hence $CD=\dfrac{2 \times \dfrac 5 2 \times \sqrt{3}}{2}=\dfrac{5 \sqrt3}{2} \approx \boxed{4.33}$ .