2016-I-19

$\begin{array}{rcl}
\dfrac{AB}{\sin \angle ADB} & = & \dfrac{BD}{\sin \angle BAD} \\
\dfrac{10}{\sin \angle ADB} & = & \dfrac{15}{\sin 86^\circ} \\
\sin \angle ADB & = & \dfrac{10\sin 86^\circ}{15} \\
\sin \angle ADB & = & 0.665\ 042\ 7 \\
\angle ADB & = & 41.685\ 601\ 32^\circ
\end{array}$

所以，可得

$\begin{array}{rcl}
\angle ABD & = & 180^\circ – \angle BAD – \angle ADB \\
\angle ABD & = & 180^\circ – 86^\circ – 41.685\ 601\ 32^\circ \\
\angle ABD & = & 52.314\ 398\ 68^\circ
\end{array}$

考慮 $\Delta BCD$，

$\begin{array}{rcl}
CD^2 & = & BC^2 + BD^2 – 2(BC)(BD)\cos \angle CBD \\
CD^2 & = & 8^2 + 15^2 – 2(8)(15)\cos 43^\circ \\
CD & = & 10.652\ 469\ 74 \text{ cm}
\end{array}$

$\begin{array}{rcl}
\dfrac{AD}{\sin \angle ABD} & = & \dfrac{BD}{\sin \angle BAD} \\
\dfrac{AD}{\sin 52.314\ 398\ 68^\circ} & = & \dfrac{15}{\sin 86^\circ} \\
AD & = & \dfrac{15 \sin 52.314\ 398\ 68^\circ}{\sin 86^\circ} \\
AD & = & 11.899\ 644\ 75 \text{ cm}
\end{array}$

考慮 $\Delta ACD$，

$\begin{array}{rcl}
\cos \angle ACD & = & \dfrac{AC^2 + CD^2 – AD^2}{2 (AC)(CD)} \\
\cos \angle ACD & = & \dfrac{6^2 + 10.652\ 469\ 74^2 – 11.899\ 644\ 75^2}{2(6)(10.652\ 469\ 74)} \\
\cos \angle ACD & = & 0.055\ 138\ 666 \\
\angle ACD & = & 86.839\ 184\ 13^\circ
\end{array}$

由於 $\angle ACD \neq 90^\circ$，則 $C$ 並不是 $A$ 在平 $BCD$ 的垂足。所以 $CB$ 不是 $AB$ 在平面 $BCD$ 的投影。由此，$\angle ABC$ 並不是 $AB$ 與平面 $BCD$ 的交角。我不同意該工匠。