(a) The radius of the base of the cone is approximately 6.37 m (M1A1A1A1A1A1, N6).

Working:
Vertical distance from eye level to vertex = 20 – 1.8 = 18.2 m (M1).
Horizontal distance from Oliver to center = 5 + r (A1).
Tangent relation: \( \tan 58^\circ = \frac{18.2}{5 + r} \) (A1).
\( \tan 58^\circ \approx 1.6003 \) (A1).
Solve: \( 5 + r = \frac{18.2}{1.6003} \approx 11.3726 \), \( r \approx 11.3726 – 5 \approx 6.3726 \) (A1).
Rounded to two decimal places: \( r \approx 6.37 \text{ m} \) (A1).

[6 marks]

(b) The volume of the monument is approximately 851 m³ (M1A1A1A1A1A1, N6).

Working:
Volume formula: \( V = \frac{1}{3} \times \pi \times r^2 \times h \) (M1).
Height \( h = 20 \text{ m} \), radius \( r \approx 6.3726 \text{ m} \) (A1).
\( r^2 \approx (6.3726)^2 \approx 40.6096 \) (A1).
\( V \approx \frac{1}{3} \times 40.6096 \times 20 \times \pi \), \( \frac{20}{3} \approx 6.6667 \) (A1).
\( V \approx 6.6667 \times 40.6096 \times 3.1416 \approx 850.540 \) (A1).
Rounded to nearest cubic meter: \( 851 \text{ m}^3 \) (A1).

[6 marks]

Total [12 marks]