A multi-storey car park is a cuboid ABCDEFGH, as shown in the diagram. E is directly above A, F above B, G above C and H above D. AB = 45 m, BC = 28 m and t...

Three-dimensional problems are solved by picking out flat right-angled triangles. In a cuboid, a vertical edge is at right angles to every line drawn on the horizontal face, which is what makes each triangle right-angled.

(a) Length of AC. [2]
AC is a diagonal of the horizontal rectangular floor ABCD:

\[AC^2 = AB^2 + BC^2 = 45^2 + 28^2 = 2025 + 784 = 2809\] \[AC = \sqrt{2809} = 53\ \text{m}\]

This value is exact, which makes the later parts cleaner.

(b) Length of AG. [3]
G is directly above C, so CG is vertical and equal to the height, 21 m. Triangle ACG lies in a vertical plane and is right-angled at C:

\[AG^2 = AC^2 + CG^2 = 2809 + 21^2 = 2809 + 441 = 3250\] \[AG = \sqrt{3250} = 57.008\ldots = 57.0\ \text{m (3 s.f.)}\]

(c) Angle between AG and the floor. [3]
The angle a line makes with a plane is the angle between the line and its shadow on that plane. The point G projects straight down onto C, so the shadow of AG on the floor is AC, and the required angle is angle GAC:

\[\tan(\angle GAC) = \frac{CG}{AC} = \frac{21}{53} = 0.396226\] \[\angle GAC = \tan^{-1}(0.396226) = 21.61\ldots = 21.6^{\circ}\ \text{(3 s.f.)}\]

(d) Angle GAB. [3]
Work in triangle ABG. First find BG, which lies in the vertical face BCGF and is right-angled at C:

\[BG^2 = BC^2 + CG^2 = 28^2 + 21^2 = 784 + 441 = 1225 \quad\Rightarrow\quad BG = 35\ \text{m}\]

The edge AB is perpendicular to the whole face BCGF, so angle ABG is a right angle. In triangle ABG:

\[\tan(\angle GAB) = \frac{BG}{AB} = \frac{35}{45} = 0.77778\] \[\angle GAB = \tan^{-1}(0.77778) = 37.87\ldots = 37.9^{\circ}\ \text{(3 s.f.)}\]

(e) Show the cable is not longer than 60 m. [2]
From part (b), \(AG = 57.008\ldots\) m. Since \(57.0 < 60\), the cable is shorter than 60 m and the engineer is wrong, by about 3 m.

Examination point: the angle between a line and a plane is never measured to an edge of the solid but to the projection of the line onto the plane. Here A to C is that projection, so angle GAC is the answer and angle GAB, which is between two lines rather than a line and a plane, is a different and larger angle.