IBDP Maths AA: Topic SL 3.3: Applications of right and non-right angled trigonometry: IB style Questions SL Paper 2

QUESTION

A monument is in the shape of a right cone with a vertical height of 20 meters. Oliver stands 5 meters 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

Detail 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 metres

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 centre of the base is 5+r metres(since he’s 5 metres from the edge and r is the radius to the centre)

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 metres

The angle of elevation is 58°.

The tangent of the angle of elevation is defined as:

$$ tan(58^{\circ})=\frac{perpendicular}{base}=\frac{18.2}{5+r}$$

solve for $$ 5+r=\frac{18.2}{tan58^{\circ}}$$

Using a calculator,

\tan(58°) \approx 1.6003

. So:

$1.6003 = \frac{18.2}{r + 5}$

Solve for

$r + 5$

$r + 5 = \frac{18.2}{1.6003} \approx 11.373$

$r = 11.373 – 5 \approx 6.373 \text{ meters}$

(b)

The volume of a cone is:

$V = \frac{1}{3} \pi r^2 h$

Here,

$h = 20$

meters, and

$r \approx 6.373$

meters. Compute:

$r^2 = (6.373)^2 \approx 40.615$

$V = \frac{1}{3} \pi \cdot 40.615 \cdot 20$

$\frac{1}{3} \cdot 20 = \frac{20}{3} \approx 6.6667$

$V \approx 6.6667 \cdot 40.615 \cdot \pi \approx 270.77 \cdot \pi$

Using

$\pi \approx 3.1416$

$V \approx 270.77 \cdot 3.1416 \approx 850.77 \text{ cubic meters}$

Let’s round to two decimal places, as is common: 850.77 m³.

————Markscheme—————–

(a) attempt to use trigonometry to find the radius of the cone OR Oliver’s distance from center (r + 5)

$tan58^{o}=\frac{18.2}{r+5} \textrm{OR} \frac{r+5} {sin32^{o}}=\frac{18.2}{sin58^{o}} \textrm{OR} (r+5=)11.3726…$

r=6.37262…(m)

(r=)6.37(m)

(b) attempt to substitute h = 20 and their radius into the correct volume of cone formula

$v=\frac{\pi \left ( 6.37262… \right )^{2}\left ( 20 \right )}{3}$

$=850.540…$

$=851(m^{3})$

IBDP Maths AA: Topic SL 3.3: Applications of right and non-right angled trigonometry: IB style Questions SL Paper 2

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