2009-I-17

1. By applying the cosine law to $\Delta BCD$, we have

   $\begin{array}{rcl}
   CD^2 & = & BD^2 + BC^2 -2 (BD)(BC)\cos \angle ABC \\
   & = & 6^2 + 25^2 -2(6)(25)\cos 57^\circ \\
   & = & 497.608~289~5 \\
   CD & = & 22.307~135~39 \text{ cm} \\
   & \approx & 22.3\text{ cm}
   \end{array}$

2. By applying the sine law to $\Delta ABC$, we have

   $\begin{array}{rcl}
   \dfrac{BC}{\sin \angle BAC} & = & \dfrac{AC}{\sin ABC} \\
   \dfrac{25}{\sin \angle BAC} & = & \dfrac{28}{\sin 57^\circ} \\
   \sin \angle BAC & = & \dfrac{25\sin57^\circ}{28} \\
   & = & 0.748~813~007 \\
   \angle BAC & = & 48.487~661~26^\circ \\
   & \approx & 48.5^\circ
   \end{array}$

3. $\begin{array}{cl}
   & \angle ACB \\
   = & 180^\circ – 57^\circ – 48.487~661~26^\circ \\
   = & 74.512~338~74^\circ
   \end{array}$

   Hence, the area of $\Delta ABC$

   $\begin{array}{cl}
   = & \dfrac{1}{2} \times AC \times BC \times \sin \angle ACB \\
   = & \dfrac{1}{2} \times 28 \times 25 \times \sin 74.512~338~74^\circ \\
   = & 337.290~793~4 \text{ cm}^2 \\
   \approx & 337 \text{ cm}^2
   \end{array}$

4. Let $h \text{ cm}$ be the shortest distance from $E$ to the horizontal ground. Note that the shortest distance from $E$ to the horizontal ground is the height of the tetrahedron $ABCE$ with base $\Delta ABC$. Then the volume of the tetrahedron $ABCE$

   $\begin{array}{cl}
   = & \dfrac{1}{3} \times h \times 337.290~793~4 ~\ldots \unicode{x2460}
   \end{array}$

   By applying the sine law to $\Delta ABC$, we have

   $\begin{array}{rcl}
   \dfrac{AB}{\sin \angle ACB} & = & \dfrac{AC}{\sin \angle ABC} \\
   \dfrac{AB}{\sin 74.512~338~74^\circ } & = & \dfrac{28}{\sin 57^\circ} \\
   AB & = & \dfrac{28 \sin 74.512~338~74^\circ }{\sin 57^\circ} \\
   & = & 32.173~852~88\text{ cm}
   \end{array}$

   Note that $CE$ is perpendicular to the thin metal plate $ABE$, therefore $\angle CEB = \angle CED = \angle CEA = 90^\circ$. By applying the Pythagoras Theorem to $\Delta BCE$, we have

   $\begin{array}{rcl}
   CE^2 & = & BC^2 – BE^2 \\
   CE & = & \sqrt{25^2 – 24^2} \\
   & = & 7 \text{ cm}
   \end{array}$

   Also by applying the Pythagoras Theorem to $\Delta ACE$, we have

   $\begin{array}{rcl}
   AE^2 & = & AC^2 – CE^2 \\
   AE & = & \sqrt{28^2 – 7^2} \\
   & = & \sqrt{735}\text{ cm}
   \end{array}$

   By applying Heron’s formula to $\Delta ABE$, we have

   $\begin{array}{rcl}
   s & = & \dfrac{AB + BE + AE}{2} \\
   & = & 41.642~368~15 \text{ cm}
   \end{array}$

   Therefore, the area of $\Delta ABE$

   $\begin{array}{cl}
   = & \sqrt{ s (s-AB)(s-BE)(s-AE)} \\
   = & 317.937~742~9 \text{ cm}^2
   \end{array}$

   Note that $CE$ is the height of the tetrahedron $ABCE$ with base $\Delta ABE$. Hence, the volume of the tetrahedron $ABCE$

   $\begin{array}{cl}
   = & \dfrac{1}{3} \times 317.937~742~9 \times 7 \\
   = & 741.854~733~5 \text{ cm}^3 ~\ldots \unicode{x2461}
   \end{array}$

   By comparing $\unicode{x2460}$ and $\unicode{x2461}$, we have

   $\begin{array}{rcl}
   \dfrac{1}{3} \times h \times 337.290~793~4 & = & 741.~854~733~5 \\
   h & = & 6.598~354~429\text{ cm} \\
   & \approx & 6.60\text{ cm}
   \end{array}$

   Therefore, the required distance is $6.60\text{ cm}$.

$\begin{array}{rcl}
DE^2 & = & CD^2 – CE^2 \\
DE & = & \sqrt{22.307~135~39^2 – 7^2} \\
& = & 21.180~375~1\text{ cm}
\end{array}$

Let $\theta$ be the angle between $DE$ and the horizontal ground. Hence, we have

$\begin{array}{rcl}
\sin \theta & = & \dfrac{h}{DE} \\
& = & \dfrac{6.598~354~429}{21.180~375~1} \\
\theta & = & 18.151~551~2^\circ
\end{array}$

Consider $\Delta CDE$,

$\begin{array}{rcl}
\tan \angle CDE & = & \dfrac{CE}{DE} \\
& = & \dfrac{7}{21.180~375~1} \\
\angle CDE & = & 18.288~442~66^\circ \\
& \neq & \theta
\end{array}$

Therefore, the claim is not agree.