2012-I-18

In $\Delta ABP$, by sine law, we have

$\begin{array}{rcl}
\dfrac{AP}{\sin \angle ABP} & = & \dfrac{AB}{\sin \angle APB} \\
\dfrac{AP}{\sin 60^\circ} & = & \dfrac{20}{\sin 48^\circ} \\
AP & = & 23.307~042~56 \text{ cm}
\end{array}$

1. In $\Delta PAB$, by sine law, we have

   $\begin{array}{rcl}
   \dfrac{PB}{\sin \angle PAB} & = & \dfrac{AB}{\sin \angle APB} \\
   \dfrac{PB}{\sin 72^\circ} & = & \dfrac{20}{\sin 48^\circ} \\
   PB & = & 25.595~455~52\text{ cm}
   \end{array}$

   Add a point $X$ on $AD$ such that $PX \perp AD$.

   In $\Delta PAX$,

   $\begin{array}{rcl}
   \sin \angle PAX & = & \dfrac{PX}{PA} \\
   PX & = & PA \times \sin 72^\circ \\
   & = & 22.166~314~7\text{ cm}
   \end{array}$

   Also,

   $\begin{array}{rcl}
   \cos \angle PAX & = & \dfrac{AX}{PA} \\
   AX & = & PA \times \cos 72^\circ \\
   & = & 7.202~272~24\text{ cm}
   \end{array}$

   Add a point $Y$ on $BC$ such that $PY\perp BC$.

   Note that $BY=AX$. In $\Delta PBY$,

   $\begin{array}{rcl}
   PY & = & \sqrt{PB^2 – BY^2} \\
   & = & 24.561~242~19\text{ cm}
   \end{array}$

   In $\Delta PXY$, by cosine law, we have

   $\begin{array}{rcl}
   \cos \angle PYX & = & \dfrac{PY^2+XY^2-PX^2}{2(PY)(XY)} \\
   & = & 0.521~053~766 \\
   \angle PYX & = & 58.597~037~33^\circ \\
   \end{array}$

   Therefore, $\alpha=58.6^\circ$.

2. Let $P’$ be the projection of $P$ on the base $ABCD$.

   In $\Delta PYP’$, $\sin \alpha = \dfrac{PP’}{PY}$.

   In $\Delta PBP’$, $\sin \beta = \dfrac{PP’}{PB}$.

   Hence, we have

   $\begin{array}{rcl}
   PY \sin \alpha & = & PB \sin \beta
   \end{array}$

   Since $PY < PB$, then we have

   $\begin{array}{rcl}
   \sin \alpha & > & \sin \beta \\
   \alpha & > & \beta
   \end{array}$