Question
A monument is in the shape of a right cone with a vertical height of 20 metres. Oliver stands 5 metres from the base of the monument. His eye level is 1.8 metres above the ground and the angle of elevation from Oliver’s eye level to the vertex of the cone is 58°, as shown on the following diagram.
(a) Find the radius of the base of the cone.
(b) Find the volume of the monument.
▶️Answer/Explanation
Detailed Solution
(a) Finding the radius of the base of the cone
Let the radius of the base of the cone be \( r \) meters.
The cone’s vertex is at a height of 20 meters above the ground.
Oliver’s eye level is 1.8 meters above the ground as given in figure, so the vertical distance from his eye level to the vertex is \( 20 – 1.8 = 18.2 \) meters.
Horizontally, Oliver is 5 meters from the edge of the base. If we assume he’s standing outside the cone and the distance is measured from the edge of the circular base, then the horizontal distance from his position to the center of the base is \( 5 + r \) meters (since he’s 5 meters from the edge, and \( r \) is the radius to the center).
In the right triangle formed by Oliver’s eye level, the vertex of the cone, and the point directly below the vertex at his eye level:
The vertical leg (opposite the angle of elevation) is 18.2 meters (height from eye level to vertex).
The horizontal leg (adjacent to the angle of elevation) is \( 5 + r \) meters.
The angle of elevation is 58°.
The tangent of the angle of elevation is defined as:
\[
\tan(58°) = \frac{\text{opposite}}{\text{adjacent}} = \frac{18.2}{5 + r}
\]
Solve for \( 5 + r \):
\[
5 + r = \frac{18.2}{\tan(58°)}
\]
Using \(\tan(58°) \approx 1.6003\)
\[
5 + r = \frac{18.2}{1.6003} \approx 11.373
\]
Solve for \( r \):
\[
r = 11.373 – 5 \approx 6.373
\]
So, the radius of the base is approximately \( r \approx 6.373 \) meters.
(b) Finding the volume of the monument
The volume \( V \) of a right cone is given by:
\[
V = \frac{1}{3} \pi r^2 h
\]
Where \( r \) is the radius of the base, and \( h \) is the height of the cone.
\( r \approx 6.373 \) meters (from part (a)).
\( h = 20 \) meters (given height of the cone).
\( r^2 \approx (6.373)^2 \approx 40.615 \).
So:
\[
V = \frac{1}{3} \pi (40.615)(20)
\]
\[
V = \frac{1}{3} \pi \cdot 812.3 \approx 270.767 \pi
\]
\[
V \approx 270.767 \times 3.1416 \approx 850.58
\]
The volume is approximately 850.58 cubic meters.
………………………Markscheme……………………
Solution: –
(a) attempt to use trigonometry to find the radius of the cone OR Oliver’s distance from centre (r+5)
$tan 58^\circ = \frac{18.2}{r+5}$ OR $\frac{r+5}{\sin 32^\circ} = \frac{18.2}{\sin 58^\circ}$ OR (r+5)=11.3726…
r = 6.37262… (m)
(r =) 6.37 (m)
(b) attempt to substitute h=20 and their radius into the correct volume of cone formula
$V = \frac{\pi(6.37262…)^2(20)}{3}$
= 850.540…
= 851 (m³)