2013-I-18

$\begin{array}{rcl}
\cos \angle BCM & = & \dfrac{BC^2+AC^2-AB^2}{2(BC)(AC)} \\
& = & \dfrac{(21)^2+(35)^2-(28)^2}{2(21)(35)} \\
& = & 0.6 \\
\angle BCM & = & 53.130~102~35^\circ
\end{array}$

$\begin{array}{rcl}
\angle CBM & = & 180^\circ – 75^\circ – 53.130~102~35^\circ \\
& = & 51.869~897~65^\circ
\end{array}$

Then by sine law, we have

$\begin{array}{rcl}
\dfrac{CM}{\sin\angle CBM} & = & \dfrac{BC}{\sin\angle BMC} \\
\dfrac{CM}{\sin51.869~897~65^\circ} & = & \dfrac{21}{\sin 75^\circ} \\
CM & = & \dfrac{21 \sin 51.869~897~65^\circ}{\sin 75^\circ} \\
& = & 17.101~546~43\text{ cm}
\end{array}$

$\begin{array}{rcl}
AM & = & 35 – 17.101~546~43 \\
& = & 17.898~453~57\text{ cm}
\end{array}$

Then by cosine law, we have

$\begin{array}{rcl}
AC^2 & = & CM^2 +AM^2 – 2(CM)(AM)\cos \angle AMC \\
& = & (17.101~546~43)^2 + (17.898~453~57)^2-2(17.101~546~43) (17.898~453~57)(\cos 107^\circ) \\
& = & 791.802~362~7 \\
AC & = & 28.138~982~97\text{ cm}
\end{array}$

$\begin{array}{rcl}
\cos \angle MCN & = & \dfrac{CN}{CM} \\
CN & = & CM \cos\angle MCN \\
& = & 17.101~546~43 \times \cos 53.130~102~35^\circ \\
& = & 10.260~927~86\text{ cm}
\end{array}$

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

$\begin{array}{rcl}
\cos \angle ACB & = & \dfrac{AC^2 + BC^2-AB^2}{2(AC)(BC)} \\
& = & \dfrac{(28.138~982~97)^2+21^2-28^2}{2(28.138~982~97)(21)} \\
& = & 0.379~749~707 \\
\end{array}$

In $\Delta ANC$,

$\begin{array}{rcl}
\dfrac{CN}{AC} & = & \dfrac{10.260~927~86}{28.138~982~97} \\
& = & 0.364~651~695 \\
& \neq & \cos \angle ACN
\end{array}$

Therefore, $\angle ANC$ is not a right angle. Hence, the angle between the face $BCM$ and the horizontal ground is not $\angle ANM$. I don’t agree.