$$\begin{aligned} A C^{2} & =x^{2}+4 \\ A D^{2} & =A C^{2}-3^{2} \\ & =x^{2}-5 \end{aligned}$$
(Pythagoras' Theorem on $\triangle A B C$ )
(Pythagoras' Theorem on $\triangle A C D$ )
$B D^{2}=x^{2}+\left(x^{2}-5\right)-2 x \sqrt{x^{2}-5} \cos 60^{\circ}$ (cosine rule on $\triangle A B D$ )
$B D^{2}=2^{2}+3^{2}-2 \cdot 2 \cdot 3 \cos 120^{\circ}$
(cosine rule on $\triangle B C D$ )
$\therefore 2 x^{2}-5-x \sqrt{x^{2}-5}=13+6$
$x \sqrt{x^{2}-5}=2 x^{2}-24$
$$x^{2}\left(x^{2}-5\right)=4 x^{4}-96 x^{2}+576$$
$3 x^{4}-91 x^{2}+576=0$
$$\begin{array}{l} \left(x^{2}-9\right)\left(3 x^{2}-64\right)=0 \\ x=3 \text { or } \frac{8}{\sqrt{3}} \end{array}$$

When $x=3, A D=\sqrt{x^{2}-5}=2$
$\tan \angle B A C=\frac{2}{3}, \tan \angle C A D=\frac{3}{2}=\tan \left(90^{\circ}-\angle B A C\right)$
$\angle B A D=90^{\circ} \neq 60^{\circ} \therefore$ reject $x=3$
When $x=\frac{8}{\sqrt{3}}, A D=\sqrt{x^{2}-5}=\frac{7}{\sqrt{3}}$
$\tan \angle B A C=\frac{\sqrt{3}}{4}, \tan \angle C A D=\frac{3 \sqrt{3}}{7}$
Method 2
$B D^{2}=2^{2}+3^{2}-2 \cdot 2 \cdot 3 \cos 120^{\circ}$
(cosine rule on $\triangle B C D$ )
$B D=\sqrt{19}$
$\angle A B C+\angle A D C=180^{\circ}$
$A, B, C, D$ are concyclic
(opp. $\angle$ s supp.)
$A C=\sqrt{x^{2}+4}=$ diameter $=2 R$
(converse, $\angle$ in semi-circle, $R=$ radius)
$\frac{B D}{\sin 60^{\circ}}=2 R$ (Sine rule on $\triangle A B D$ ) $\tan (\angle B A C+\angle C A D)=\frac{\frac{\sqrt{3}}{4}+\frac{3 \sqrt{3}}{7}}{1-\frac{\sqrt{3}}{4} \cdot \frac{3 \sqrt{3}}{7}}=\frac{19 \sqrt{3}}{19}=\sqrt{3}=\tan 60^{\circ} \frac{\sqrt{19}}{\frac{\sqrt{3}}{2}}=\sqrt{x^{2}+4}$
$\therefore x=\frac{8}{\sqrt{3}}=\frac{8}{3} \sqrt{3}$
$$\begin{array}{l} 76=3 x^{2}+12 \\ x=\frac{8}{\sqrt{3}}=\frac{8}{3} \sqrt{3} \end{array}$$