2011SP-I-18

$\begin{array}{rcl}
\dfrac{20}{\sin 30^\circ} & = & \dfrac{BC}{\sin 45^\circ} \\
BC & = & \dfrac{20\sin 45^\circ}{\sin 30^\circ} \\
BC & = & 28.284~271~25\mbox{ cm}
\end{array}$

In $\Delta BCD$,

$\begin{array}{rcl}
\cos 30^\circ & = & \dfrac{BD}{BC} \\
BD & = & BC\cos 30^\circ \\
& = & 24.494~897~43\mbox{ cm}
\end{array}$

1. In $\Delta ACD$,

   $\begin{array}{rcl}
   \cos 45^\circ & = & \dfrac{AD}{AC} \\
   AD & = & 20\cos 45^\circ \\
   & = & 14.142~135~62\mbox{ cm}
   \end{array}$

   In $\Delta ABD$ and by cosine law, we have

   $\begin{array}{rcl}
   \cos \angle BDA & = & \dfrac{AD^2+BD^2-AB^2}{2(AD)(BD)} \\
   & = & 0.687~046~82 \\
   \angle BDA & = & 46.603~208~66^\circ
   \end{array}$

   Therefore, the required angle is $46.6^\circ$.

2. Note that the volume of $ABCD$

   $\begin{array}{cl}
   = & \dfrac{1}{3}(\text{area of }\Delta ADB)(DC)
   \end{array}$

   Note also that the area of $\Delta ADB$

   $\begin{array}{cl}
   = & \dfrac{1}{2} (AD)(BD)\sin \angle ADB
   \end{array}$

   Since the lengths of $AD$, $BD$ and $DC$ are constant, the volume of $ABCD$ varies directly as $\sin \angle ADB$.

   When $\angle ADB$ increases from $40^\circ$ to $90^\circ$, $\sin \angle ADB$ increases and then the volume of $ABCD$ increases.

   When $\angle ADB$ increases from $90^\circ$ to $140^\circ$, $\sin \angle ADB$ decreases and then the volume of $ABCD$ decreases.