图中显示了一个高度为 9 米的棱柱形阶梯教室。 阶梯教室的地板 ABCD 是一个等腰梯形。 AB = 45 米，CD = 17 米，AD = 21 米，且 AB 平行于 DC。 J 是 AB 上的一个点，使得 DJ 垂直于 AB。 该阶梯教室建在水平地面上，且墙面是竖直的。

1. Find the horizontal offset from $A$ to $J$ using the isosceles trapezium property.

Since $ABCD$ is an isosceles trapezium,
$AJ=\frac{AB-CD}{2}=\frac{45-17}{2}=14\text{ m}.$

2. Find the perpendicular distance between the parallel sides.

In right-angled triangle $ADJ$,
$DJ^2=AD^2-AJ^2$
$DJ^2=21^2-14^2=441-196=245$
$DJ=\sqrt{245}=7\sqrt5\text{ m}.$

3. Calculate the floor area of trapezium $ABCD$.

Area of trapezium $ABCD$:
$\frac12(AB+CD)\times DJ=\frac12(45+17)(7\sqrt5)$
$=31(7\sqrt5)=217\sqrt5$
$\approx 485\text{ m}^2.$

1. Use the floor area from part (a) as the base area of the prism.

Floor area of lecture theatre $=651\text{ m}^2$.

Height of prism $=9\text{ m}$.

2. Calculate the volume of the prism.

$\text{Volume}=\text{base area}\times\text{height}$
$=651\times 9$
$=5859\text{ m}^3$

1. Find the horizontal distance from $J$ to $C$ on the floor.

From part (a), $DJ=21\text{ m}$.

Since $DC=17\text{ m}$ and $DJ\perp DC$,
$JC^2=DJ^2+DC^2$
$JC^2=21^2+17^2=730$
$JC=\sqrt{730}\text{ m}$

2. Use the tangent ratio for the angle of elevation from $J$ to $G$.

The vertical height of the lecture theatre is
$CG=9\text{ m}.$

Let the angle of elevation of $G$ from $J$ be $\theta$.

In right-angled triangle $JCG$,
$\tan \theta=\frac{CG}{JC}$
$\tan \theta=\frac{9}{\sqrt{730}}$

3. Calculate the angle and round appropriately.

$\theta=\tan^{-1}\left(\frac{9}{\sqrt{730}}\right)$
$\theta=18.4^\circ\text{, to 1 decimal place}$