Calculate length of AC

Draw a perpendicular from $A$ to the chord $BD$ which will bisect the chord, let the point of intersection be $F$ . Let angle $ABC = \theta$

In triangle $AFB$ , $cos(\theta)=\frac{10}{AB}$

In triangle $ABC$ , $cos(\theta)=\frac{AB}{36}$ .

That gives us: $AB= 6\sqrt{10}$

Using the pythagorean theorem, we get that $AC=6\sqrt{26}$ .

Using the same perpendicular as David Nasr, but without using any trigonometry and just Pythagoras. Since $DBA$ is isosceles, $BF=FD=10$ . Now from Pythagoras, we know that $AB^2=BF^2+FA^2$ and, again by using Pythagoras, $AC^2=BC^2-AB^2=BC^2-BF^2-FA^2$ . Since $\angle AFC=90^\circ$ , there's one more triangle we can use Pythagoras on, which gives us $AC^2=AF^2+FC^2$ .

Adding these two equations, we arrive at $2AC^2 = BC^2+FC^2-BF^2 = 36^2+26^2-10^2$ and a quick calculation later, it is revealed to us that $\boxed{AC=6\sqrt{26}}$ .