Question 40Given right triangle $ABC$ with legs $\overline{BC}=3,$ $\overline{AC}=4.$ Find the length of the shorter angle trisector from $C$ to the hypotenuse:$\dfrac{32\sqrt3-24}{13}$$\dfrac{12\sqrt3-9}{13}$$6\sqrt3-8$$\dfrac{5\sqrt{10}}6$$\dfrac{25}{12}$