Find $\overline{CD}$

By the exterior angle theorem, $\angle ACB = \angle CBD - \angle CAB = 60°- 30° = 30°$ , so that $\triangle ABC$ is isosceles and $BC = AB = 10$ .

Then for $\triangle BCD$ , $CD = BC \sin \angle CBD = 10 \sin 60° = 5 \sqrt{3}$ .

Therefore, $a = 5$ , $b = 3$ , and $a + b = \boxed{8}$ .

Let $CD=x$ and $BD=y$ .

In right triangle $\triangle ADC$ , $\tan \angle CAD=\frac{CD}{AD}\implies \tan 30°=\frac{x}{y+10}\implies \sqrt{3}x=y+10$

In right triangle $\triangle BDC$ , $\tan \angle CBD=\frac{CD}{BD}\implies \tan 60°=\frac{x}{y}\implies x=\sqrt{3}y$

Solving the above two equations, $y+10=\sqrt{3}x=\sqrt{3}(\sqrt{3}y)=3y\implies y=5\implies x=5\sqrt{3}$

Therefore, $\boxed{CD=5\sqrt{3}}\implies \boxed{a+b=8}$