10.1.25 As shown in the figure, in quadrilateral $A B C D, \angle B A C=30^{\circ}, \angle A B D=26^{\circ}, \angle D B C=$ $51^{\circ}, \angle A C D=13^{\circ}$, find $\angle C A D$.
Let $\angle A D C=\theta$, then by the Law of Sines, we have $\frac{A C}{C D}=\frac{\sin \theta}{\sin \left(\theta+13^{\circ}\right)}$, and $\frac{A C}{B C}=\frac{\sin 77^{\circ}}{\sin 30^{\circ}}$, $\frac{B C}{C D}=\frac{\sin 43^{\circ}}{\sin 51^{\circ}}$, thus $\frac{\sin \theta}{\sin \left(\theta+13^{\circ}\right)}=\frac{2 \sin 77^{\circ} \sin 43^{\circ}}{\sin 51^{\circ}}=$ $\frac{2 \cos 13^{\circ} \cos 47^{\circ} \cos 73^{\circ}}{\sin 51^{\circ} \cos 73^{\circ}}=\frac{\cos 39^{\circ}}{2 \sin 51^{\circ} \cos 73^{\circ}}=\frac{1}{2 \cos 73^{\circ}}$, hence $\cos 13^{\circ}+\cot \theta \sin 13^{\circ}=2 \cos 73^{\circ}, \cot \theta=\frac{2 \cos 73^{\circ}-\cos 13^{\circ}}{\sin 13^{\circ}}=$ $\frac{2 \cos \left(60^{\circ}+13^{\circ}\right)-\cos 13^{\circ}}{\sin 13^{\circ}}=\frac{-2 \sin 60^{\circ} \sin 13^{\circ}}{\sin 13^{\circ}}=-\sqrt{3}$, thus $\theta=150^{\circ}$, so $\angle C A D=180^{\circ}-150^{\circ}-13^{\circ}=17^{\circ}$.