2006-I-17

$\begin{array}{rcl}
\cos \angle BAD & = & \dfrac{AB^2 + AC^2 -BC^2}{2(AB)(AC)} \\
& = & \dfrac{40^2 +90^2 – 60^2}{2 (40)(90)} \\
& = & \dfrac{61}{72}
\end{array}$

Consider $\Delta ABD$,

$\begin{array}{rcl}
\cos \angle BAD & = & \dfrac{AD}{AB} \\
AD & = & 40 \times \dfrac{61}{72} \\
& = & \dfrac{305}{9} \text{ cm}
\end{array}$

1. 1. By applying the cosine law on $\Delta ACD$ and the result of (a), we have

      $\begin{array}{rcl}
      AD^2 + AC^2 -2(AD)(AC) \cos \angle DAC & = & CD^2 \\
      AC^2 – 31.819~739~26 AC – 2000 & = & 0 \\
      \end{array}$

      Therefore, $AC=63.376~952~45 \text{ cm}$ or $AC=-31.55721319\text{ cm}$ (rejected).

   2. By applying the Heron’s formula on $\Delta ABC$, we have

      $\begin{array}{rcl}
      s & = & \dfrac{1}{2} ( AB + AC + BC) \\
      & = & 81.688~472~23 \text{ cm}
      \end{array}$

      Hence, the area of $\Delta ABC$

      $\begin{array}{cl}
      = & \sqrt{s(s-AB)(s-AC)(s-BC)} \\
      = & 1~162.961~054 \text{ cm}^2
      \end{array}$

   3. The volume of the tetrahedron $ABCD$ with base $\Delta ACD$

      $\begin{array}{cl}
      = & \dfrac{1}{3} \times BD \times \text{the area of $\Delta ACD$} \\
      = & \dfrac{1}{3} \times BD \times \dfrac{1}{2} \times AD \times AC \times \sin \angle DAC \\
      = & 6~716.175~187 \text{ cm}^3
      \end{array}$

      Let $h\text{ cm}$ be the height of the tetrahedron $ABCD$ from the vertex $D$ to the base $\Delta ABC$. The volume of the tetrahedron $ABCD$ with base $\Delta ABC$

      $\begin{array}{cl}
      = & \dfrac{1}{3} \times h \times \text{the area of $\Delta ABC$} \\
      = & 387.653~684~7 h \text{ cm}^3
      \end{array}$

      Hence, we have

      $\begin{array}{rcl}
      387.653~684~7 h & = & 6~716.175~187 \\
      h & = & 17.325~193~73 \text{ cm}
      \end{array}$

      Therefore, the required height is $17.325~193~73\text{ cm}$.

2. Refer to part (b)(i)(3), the volume of tetrahedron $ABCD$

   $\begin{array}{cl}
   = & \dfrac{(AD)(AC)(BD)\sin \angle DAC}{6}
   \end{array}$

   Since $AD$, $AC$, $BD$ and $\dfrac{1}{6}$ are constants, then the volume of tetrahedron $ABCD$ varies directly as $\sin \angle DAC$.

   Hence, we have the following conclusion:

   When $\angle ADC$ increases from $30^\circ$ to $90^\circ$, the value of $\sin \angle DAC$ increases and so the volume of tetrahedron $ABCD$ increases.

   When $\angle ADC$ increases from $90^\circ$ to $150^\circ$, the value of $\sin \angle DAC$ decreases and so the volume of tetrahedron $ABCD$ decreases.