Geometry & Trigonometry — IB Math AI

1. 2D and 3D Measurement

Key Measurement Formulas

Shape	Area / Volume
Circle: area, circumference	$A = \pi r^2$, $C = 2\pi r$
Sector: area, arc length	$A = \frac{1}{2}r^2\theta$, $l = r\theta$ ($\theta$ in radians)
Sphere	$V = \frac{4}{3}\pi r^3$, $SA = 4\pi r^2$
Cylinder	$V = \pi r^2 h$, $SA = 2\pi r^2 + 2\pi rh$
Cone	$V = \frac{1}{3}\pi r^2 h$, $SA = \pi r^2 + \pi rl$ (slant $l=\sqrt{r^2+h^2}$)
Prism	$V = A_{\text{base}} \times h$
Pyramid	$V = \frac{1}{3}A_{\text{base}} \times h$

2. Trigonometry in Applied Contexts

SOH-CAH-TOA and Sine/Cosine Rules

Right-angled triangles: $\sin\theta = \dfrac{\text{opp}}{\text{hyp}}$, $\cos\theta = \dfrac{\text{adj}}{\text{hyp}}$, $\tan\theta = \dfrac{\text{opp}}{\text{adj}}$

Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ (use when angle-side pairs are known)

Cosine rule: $a^2 = b^2 + c^2 - 2bc\cos A$ (use when 3 sides or 2 sides + included angle known)

Area of triangle: $\text{Area} = \dfrac{1}{2}ab\sin C$

Bearings

Bearings are measured clockwise from North, given as 3-digit numbers (e.g., 047°, 270°). Navigation problems use bearings combined with the sine and cosine rules. Draw a diagram always — it is essential for identifying angles.

Worked Example 1

Navigation Problem Using Bearings

A ship leaves port $A$ and travels 50 km on a bearing of 060° to reach $B$. It then travels 80 km on a bearing of 150° to reach $C$. Find the distance $AC$ and the bearing from $A$ to $C$.

Find angle at $B$: The bearing from $A$ to $B$ is 060°. The bearing from $B$ to $C$ is 150°. The angle $ABC$ (interior angle at $B$) requires careful diagram analysis: $\angle ABC = 180° - 60° + 150° - 180° = 90°$. Actually, the bearing of $BA$ from $B$ is $060+180=240°$. Angle between $BA$ and $BC$: $|240-150|=90°$.

Since angle $B = 90°$: $AC^2 = AB^2 + BC^2 = 50^2 + 80^2 = 2500 + 6400 = 8900$. So $AC = \sqrt{8900} \approx 94.3$ km.

Bearing from $A$ to $C$: $\tan(\angle \text{from North}) = \dfrac{BC\cdot\sin\!(\ldots)}{AB+\ldots}$ — use coordinate geometry or the cosine rule for the angle, then convert to bearing.

Worked Example 2

Volume of a Composite Solid

A grain silo consists of a cylinder of radius 3 m and height 8 m, topped with a cone of the same radius and height 4 m. Find the total volume.

Cylinder volume: $V_{\text{cyl}} = \pi(3)^2(8) = 72\pi$ m³

Cone volume: $V_{\text{cone}} = \dfrac{1}{3}\pi(3)^2(4) = 12\pi$ m³

Total: $V = 72\pi + 12\pi = 84\pi \approx 264$ m³

3. Voronoi Diagrams (HL)

A Voronoi diagram divides a plane into regions based on distance to a set of sites. Each region contains all points closer to its site than to any other site. The boundaries between regions are perpendicular bisectors of segments connecting adjacent sites.

Voronoi Diagram Construction

To find the perpendicular bisector of segment $AB$:

Find the midpoint $M = \left(\dfrac{x_A+x_B}{2}, \dfrac{y_A+y_B}{2}\right)$
Find gradient of $AB$: $m_{AB} = \dfrac{y_B-y_A}{x_B-x_A}$
Perpendicular bisector has gradient $m_\perp = -\dfrac{1}{m_{AB}}$ and passes through $M$

The circumcentre of three sites is the point equidistant from all three and is found by intersecting two perpendicular bisectors.

Worked Example 3

Voronoi Diagram: Adding a New Site

Three emergency services are located at $A(0,0)$, $B(6,0)$, and $C(3,4)$. A new call comes from $P(2,3)$. Which service is closest?

Distance $PA$: $\sqrt{(2-0)^2+(3-0)^2} = \sqrt{4+9} = \sqrt{13} \approx 3.61$

Distance $PB$: $\sqrt{(2-6)^2+(3-0)^2} = \sqrt{16+9} = \sqrt{25} = 5$

Distance $PC$: $\sqrt{(2-3)^2+(3-4)^2} = \sqrt{1+1} = \sqrt{2} \approx 1.41$

Conclusion: Service $C$ is closest. Point $P$ lies in the Voronoi cell of $C$.

Practice Problems

Q1. A triangular plot has sides 70 m, 80 m, and 100 m. Find its area using the cosine rule and the area formula.

Show Solution

First find angle $C$ (opposite the 100 m side): $100^2 = 70^2+80^2-2(70)(80)\cos C \Rightarrow 10000=4900+6400-11200\cos C \Rightarrow \cos C = \dfrac{1300}{11200} \approx 0.1161 \Rightarrow C \approx 83.3°$

Area $= \dfrac{1}{2}(70)(80)\sin(83.3°) \approx 2800 \times 0.9932 \approx 2781$ m²

Q2. A hollow sphere has outer radius 10 cm and wall thickness 1.5 cm. Find the volume of material used.

Show Solution

Inner radius $= 10 - 1.5 = 8.5$ cm.

$V_{\text{material}} = \dfrac{4}{3}\pi(10^3 - 8.5^3) = \dfrac{4}{3}\pi(1000-614.125) = \dfrac{4}{3}\pi(385.875) \approx 1618$ cm³

Q3. Two towns $A$ and $B$ are 40 km apart. Town $C$ is 55 km from $A$ and 35 km from $B$. Find angle $CAB$ and the area of triangle $ABC$.

Show Solution

Cosine rule: $BC^2 = AC^2+AB^2-2(AC)(AB)\cos A$

$35^2 = 55^2+40^2-2(55)(40)\cos A \Rightarrow 1225 = 3025+1600-4400\cos A \Rightarrow \cos A = \dfrac{3400}{4400} \approx 0.7727 \Rightarrow A \approx 39.6°$

Area $= \dfrac{1}{2}(55)(40)\sin(39.6°) \approx 1100\times0.637 \approx 701$ km²

4. Exam Tips

Always draw a clearly labelled diagram for trigonometry problems — examiners award method marks for diagrams.
For bearings: measure clockwise from North. The bearing from $B$ back to $A$ differs from $A$ to $B$ by 180° (if the angle is less than 180°, add 180; otherwise subtract).
When using the sine rule, check for the ambiguous case (two possible triangles) when given two sides and a non-included angle.
Voronoi boundary = perpendicular bisector. Vertex of Voronoi cell = circumcentre of three nearby sites.