10.1.23 ** Given $\triangle A B C$, point $P$ is an internal point, $\angle P B C=\angle P C B=24^{\circ}$, $\angle A B P=30^{\circ}, \angle A C P=54^{\circ}$, find the size of $\angle B A P$.
Parse as shown in the figure, extend $BP$ to intersect $AC$ at point $E$. Since $\angle BAC = 48^{\circ}$, it follows that $\angle BEC = 48^{\circ} + 30^{\circ} = 78^{\circ} = \angle BCE$. Therefore, $BE = BC$. Let's assume $BE = BC = 1$. By the Law of Sines, $\frac{PE}{CE} = \frac{\sin 54^{\circ}}{\sin 48^{\circ}}$, so $PE = \frac{2 \sin 54^{\circ} \sin 12^{\circ}}{\sin 48^{\circ}}$. Also, $AC = \frac{\sin 54^{\circ}}{\sin 48^{\circ}}$, thus $AE = AC - CE = \frac{\sin 54^{\circ}}{\sin 48^{\circ}} - 2 \sin 12^{\circ}$. Since $\sin 54^{\circ} - 2 \sin 12^{\circ} \sin 48^{\circ} = \sin 54^{\circ} - \cos 36^{\circ} + \cos 60^{\circ} = \frac{1}{2}$, it follows that $AE^2 = \frac{1}{4 \sin^2 48^{\circ}}$. Also, $PE \cdot BE = \frac{2 \sin 54^{\circ} \sin 12^{\circ}}{\sin 48^{\circ}}$. If these two are equal, then $\triangle APE \sim \triangle BAE$, so $\angle PAE = \angle ABE = 30^{\circ}$, and thus $\angle BAP = 48^{\circ} - 30^{\circ} = 18^{\circ}$.

Next, we prove this equality, i.e., $\sin 12^{\circ} \sin 48^{\circ} \sin 54^{\circ} = \frac{1}{8}$. This is because the left side $= \frac{1}{2} (\cos 36^{\circ} - \cos 60^{\circ}) \sin 54^{\circ} = \frac{1}{2} (\cos^2 36^{\circ} - \frac{1}{2} \cos 36^{\circ}) = \frac{1}{4} (1 + \cos 72^{\circ} - \cos 36^{\circ}) = \frac{1}{4} (1 - 2 \sin 54^{\circ} \sin 18^{\circ}) = \frac{1}{4} (1 - 2 \cos 36^{\circ} \cos 72^{\circ}) = \frac{1}{4} (1 - \frac{2 \sin 36^{\circ} \cos 36^{\circ} \cos 72^{\circ}}{\sin 36^{\circ}}) = \frac{1}{4} (1 - \frac{\sin 144^{\circ}}{2 \sin 36^{\circ}}) = \frac{1}{8}$.

$\left. \cos 60^{\circ} \right) \sin 54^{\circ} = \frac{1}{2} (\cos^2 36^{\circ} - \frac{1}{2} \cos 36^{\circ}) = \frac{1}{4} (1 + \cos 72^{\circ} - \cos 36^{\circ}) = \frac{1}{4} (1 - 2 \sin 54^{\circ} \sin 18^{\circ}) = \frac{1}{4} (1 - 2 \cos 36^{\circ} \cos 72^{\circ}) = \frac{1}{4} (1 - \frac{2 \sin 36^{\circ} \cos 36^{\circ} \cos 72^{\circ}}{\sin 36^{\circ}}) = \frac{1}{4} (1 - \frac{\sin 144^{\circ}}{2 \sin 36^{\circ}}) = \frac{1}{8}$.