IB Mathematics AA – Topic 3: Geometry & Trigonometry

Comprehensive Guide to Geometry & Shapes

Introduction to Geometry & Trigonometry

Geometry and trigonometry form the foundation of spatial reasoning and measurement in mathematics. From ancient architecture to modern engineering, GPS navigation to computer graphics, these concepts enable us to describe, measure, and analyze shapes, angles, and distances in both two and three dimensions.

Key concepts: Understanding areas, volumes, and surface areas of standard shapes provides tools for practical problem-solving. Circles introduce unique measurements like arc length and sector area. Trigonometry extends beyond right triangles to any triangle through the sine and cosine rules, enabling calculation of unknown sides and angles in real-world scenarios.

Why this matters: These topics connect abstract mathematics to tangible applications—calculating material quantities in construction, determining distances in navigation using bearings, analyzing forces in physics, and optimizing designs in engineering. Mastery of these concepts is essential for IB success and countless STEM fields.

In this guide: We'll master formulas for 2D and 3D shapes (areas, volumes, surface areas), explore circles with sectors and arcs, develop proficiency in right-angle trigonometry, apply the powerful sine and cosine rules for any triangle, calculate areas using multiple methods, and solve practical bearing problems essential for IB exams.

1. 2D and 3D Shapes

2D Shapes - Key Formulas

Essential 2D Formulas:

Rectangle

Area: \(A = l \times w\)

Perimeter: \(P = 2(l + w)\)

Triangle

Area: \(A = \frac{1}{2}bh\) (base × height)

Perimeter: \(P = a + b + c\) (sum of all sides)

Parallelogram

Area: \(A = b \times h\) (base × perpendicular height)

Trapezoid (Trapezium)

Area: \(A = \frac{1}{2}(a + b)h\) (average of parallel sides × height)

3D Shapes - Volume and Surface Area

Essential 3D Formulas:

Cube (side length \(a\))

Volume: \(V = a^3\)

Surface Area: \(SA = 6a^2\) (6 faces)

Rectangular Prism (Cuboid)

Volume: \(V = l \times w \times h\)

Surface Area: \(SA = 2(lw + lh + wh)\)

Cylinder (radius \(r\), height \(h\))

Volume: \(V = \pi r^2 h\)

Surface Area: \(SA = 2\pi r^2 + 2\pi rh\) (2 circles + curved surface)

Sphere (radius \(r\))

Volume: \(V = \frac{4}{3}\pi r^3\)

Surface Area: \(SA = 4\pi r^2\)

Cone (radius \(r\), height \(h\), slant height \(l\))

Volume: \(V = \frac{1}{3}\pi r^2 h\)

Surface Area: \(SA = \pi r^2 + \pi rl\) (base + curved surface)

Note: \(l = \sqrt{r^2 + h^2}\)

Pyramid (base area \(B\), height \(h\))

Volume: \(V = \frac{1}{3}Bh\)

⚠ Common Pitfalls:

Units confusion: Area uses square units (cm²), volume uses cubic units (cm³)
Radius vs diameter: Most formulas use radius, not diameter (divide by 2!)
Height confusion: For cones/pyramids, height is perpendicular from base, not slant height
Forgetting all faces: Surface area includes ALL surfaces, not just visible ones

2. Circles, Sectors, and Arcs

Circle Fundamentals

Basic Circle Formulas (radius \(r\)):

Circumference: \(C = 2\pi r\) or \(C = \pi d\)

Area: \(A = \pi r^2\)

Sectors and Arcs

Sector and Arc Formulas

For angle \(\theta\) in degrees at center of circle with radius \(r\):

Arc Length

\(l = \frac{\theta}{360°} \times 2\pi r\)

Arc is fraction of circumference

Sector Area

\(A = \frac{\theta}{360°} \times \pi r^2\)

Sector is fraction of circle area

If angle in radians:

Arc length: \(l = r\theta\)

Sector area: \(A = \frac{1}{2}r^2\theta\)

💡 Circle Tips:

Check if angle is in degrees or radians—formulas differ!
Sector area is to circle area as arc length is to circumference
A semicircle has angle 180° (or \(\pi\) radians)
Quarter circle has angle 90° (or \(\frac{\pi}{2}\) radians)

Example 1: Circles and Sectors

Problem: A circle has radius 8 cm. A sector of this circle has an angle of 120° at the center.

(a) Find the arc length of the sector

(b) Find the area of the sector

Solution:

(a) Arc length:

Using \(l = \frac{\theta}{360°} \times 2\pi r\)

\(l = \frac{120°}{360°} \times 2\pi(8)\)

\(= \frac{1}{3} \times 16\pi\)

\(= \frac{16\pi}{3}\) cm

Arc length: \(\frac{16\pi}{3}\) cm ≈ 16.8 cm

(b) Sector area:

Using \(A = \frac{\theta}{360°} \times \pi r^2\)

\(A = \frac{120°}{360°} \times \pi(8)^2\)

\(= \frac{1}{3} \times 64\pi\)

\(= \frac{64\pi}{3}\) cm²

Sector area: \(\frac{64\pi}{3}\) cm² ≈ 67.0 cm²

(c) Perimeter of sector:

Perimeter = arc length + 2 radii

\(P = \frac{16\pi}{3} + 8 + 8\)

\(= \frac{16\pi}{3} + 16\) cm

Perimeter: \(\frac{16\pi}{3} + 16\) cm ≈ 32.8 cm

3. Right-Angle Triangle Trigonometry

The Three Ratios

SOH-CAH-TOA

For a right triangle with angle \(\theta\):

Sine

\(\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}}\)

Cosine

\(\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)

Tangent

\(\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}}\)

Finding Sides and Angles

Solving Right Triangles:

To find a side: Identify which ratio to use, substitute, solve for unknown
To find an angle: Use inverse trig functions: \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\)
Pythagoras' Theorem: \(a^2 + b^2 = c^2\) (when you know two sides)

⚠ Common Mistakes:

Wrong side identification: Opposite/adjacent depends on which angle you're using!
Calculator mode: Ensure calculator is in degrees (or radians if specified)
Inverse function confusion: \(\sin^{-1}\) finds angle from ratio, not reciprocal
Hypotenuse: Always the longest side, opposite the right angle

4. Sine and Cosine Rules (Non-Right Triangles)

The Sine Rule

Sine Rule

\(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)

Each side divided by sine of opposite angle

When to use:

Given two angles and one side (find another side)
Given two sides and one non-included angle (find another angle)

The Cosine Rule

Cosine Rule

To find a side:

\(c^2 = a^2 + b^2 - 2ab\cos C\)

To find an angle:

\(\cos C = \frac{a^2 + b^2 - c^2}{2ab}\)

When to use:

Given two sides and the included angle (find the third side) - SAS
Given all three sides (find any angle) - SSS

Choosing the Right Rule

Decision Guide:

Right triangle? Use SOH-CAH-TOA or Pythagoras
Two angles known? Use Sine Rule
Two sides and included angle? Use Cosine Rule
Three sides known? Use Cosine Rule
Two sides and opposite angle? Use Sine Rule (watch for ambiguous case!)

💡 Strategy Tips:

Sine rule involves pairs: one side with its opposite angle
Cosine rule involves three sides and one angle
Always label triangle clearly: sides as \(a, b, c\) opposite angles \(A, B, C\)
Check angle sum: \(A + B + C = 180°\) for all triangles

Example 2: Sine and Cosine Rules (IB-Style)

Problem: In triangle ABC, \(AB = 7\) cm, \(BC = 9\) cm, and angle \(ABC = 65°\)

(a) Find the length of AC

(b) Find angle BAC

Solution:

(a) Finding AC:

We have two sides and the included angle (SAS) → use Cosine Rule

Let \(AB = c = 7\), \(BC = a = 9\), \(ABC = B = 65°\)

We want to find \(AC = b\)

Using: \(b^2 = a^2 + c^2 - 2ac\cos B\)

\(b^2 = 9^2 + 7^2 - 2(9)(7)\cos(65°)\)

\(b^2 = 81 + 49 - 126\cos(65°)\)

\(b^2 = 130 - 126(0.4226)\)

\(b^2 = 130 - 53.25 = 76.75\)

\(b = \sqrt{76.75} = 8.76\) cm

AC = 8.76 cm (to 3 s.f.)

(b) Finding angle BAC:

Now we know all three sides: \(a = 9\), \(b = 8.76\), \(c = 7\)

We want angle \(A\) (angle BAC), opposite side \(a\)

Method 1: Sine Rule

\(\frac{a}{\sin A} = \frac{c}{\sin B}\)

\(\frac{9}{\sin A} = \frac{7}{\sin 65°}\)

\(\sin A = \frac{9 \times \sin 65°}{7} = \frac{9 \times 0.9063}{7} = 1.165\)

Wait! \(\sin A > 1\) is impossible. Let me recalculate AC...

Actually, using proper calculation: \(b = 8.76\) cm is correct.

Using Sine Rule correctly:

\(\frac{9}{\sin A} = \frac{8.76}{\sin 65°}\)

\(\sin A = \frac{9 \times \sin 65°}{8.76} = \frac{9 \times 0.9063}{8.76} = 0.931\)

\(A = \sin^{-1}(0.931) = 68.6°\)

Angle BAC = 68.6° (to 3 s.f.)

Check: \(65° + 68.6° + C = 180°\) → \(C = 46.4°\) ✓

5. Triangle Areas and Bearings

Area of a Triangle - Multiple Methods

Three Area Formulas:

Formula 1: Base × Height

\(A = \frac{1}{2}bh\)

When perpendicular height is known

Formula 2: Two Sides and Included Angle

\(A = \frac{1}{2}ab\sin C\)

When you know two sides and angle between them

This is the most useful for IB exams!

Formula 3: Heron's Formula (all three sides)

\(A = \sqrt{s(s-a)(s-b)(s-c)}\)

where \(s = \frac{a+b+c}{2}\) (semi-perimeter)

When all three sides are known

Bearings

Understanding Bearings:

Definition: Bearing is direction measured clockwise from North
Format: Three-digit number (e.g., 045°, 120°, 270°)
Range: 000° to 360°
North = 000° or 360°
East = 090°
South = 180°
West = 270°

Key Bearing Concepts:

Back bearing: Bearing from B to A is bearing from A to B ± 180°
In triangles: Use bearings to find angles, then apply sine/cosine rules
Angle from bearing: Often need to find angle in triangle from given bearings

⚠ Bearing Pitfalls:

Always from North: Bearings are measured from North, not from previous direction
Three digits: Write 45° as 045°, 5° as 005°
Clockwise only: Always measure clockwise from North
Drawing diagrams: Always sketch a diagram with North marked

Example 3: Area and Bearings

Problem: A ship sails from port P on a bearing of 040° for 8 km to point Q. It then sails on a bearing of 130° for 6 km to point R.

(a) Find the angle PQR

(b) Find the distance PR

Solution:

(a) Finding angle PQR:

At Q, the ship changes direction

Original bearing to Q: 040°

New bearing from Q: 130°

The angle between the paths (at Q) is the angle PQR

Bearing from Q back to P would be: 040° + 180° = 220°

Angle from back bearing to new bearing: 220° - 130° = 90°

Angle PQR = 90°

(b) Finding distance PR:

We have: \(PQ = 8\) km, \(QR = 6\) km, angle \(PQR = 90°\)

Since angle is 90°, we can use Pythagoras' Theorem:

\(PR^2 = PQ^2 + QR^2\)

\(PR^2 = 8^2 + 6^2 = 64 + 36 = 100\)

\(PR = 10\) km

Distance PR = 10 km

(c) Area of triangle PQR:

Using \(A = \frac{1}{2}ab\sin C\) with angle 90°:

\(A = \frac{1}{2}(8)(6)\sin(90°)\)

\(A = \frac{1}{2}(8)(6)(1)\)

\(A = 24\) km²

Area = 24 km²

Note: For a right triangle, \(\frac{1}{2}ab\sin(90°) = \frac{1}{2}ab\)

📋 Geometry & Trigonometry Quick Reference

Concept	Formula/Rule	When to Use
Circle Area	\(A = \pi r^2\)	Given radius
Sector Area	\(A = \frac{\theta}{360°} \times \pi r^2\)	Angle in degrees
Arc Length	\(l = \frac{\theta}{360°} \times 2\pi r\)	Angle in degrees
Sine Rule	\(\frac{a}{\sin A} = \frac{b}{\sin B}\)	2 angles + 1 side (ASA/AAS)
Cosine Rule	\(c^2 = a^2 + b^2 - 2ab\cos C\)	SAS or SSS
Triangle Area	\(A = \frac{1}{2}ab\sin C\)	2 sides + included angle

🎯 IB Exam Strategy

Common Question Types:

"Find the area of the sector": Use sector formula with correct angle unit
"Find side/angle in triangle": Choose sine rule (ASA/AAS) or cosine rule (SAS/SSS)
"Find area of triangle": Use \(\frac{1}{2}ab\sin C\) when given two sides and angle
"A ship sails on bearing...": Draw diagram, find angles in triangle, apply rules
"Find volume/surface area": Identify shape, apply correct formula

Key Reminders:

Always draw and label a diagram
Check calculator mode (degrees vs radians)
Use exact answers (\(\pi\), \(\sqrt{}\)) when possible, then approximate
For bearings, always measure clockwise from North
Show all working—method marks are crucial

🎉 Master Geometry & Trigonometry!

Geometry and trigonometry connect abstract mathematics to the physical world. From measuring circles and calculating volumes to navigating by bearings and solving triangle problems, these tools are essential for engineering, physics, architecture, and countless other fields. Master these concepts for IB success!

Key Success Factors:

✓ Know your formulas: areas, volumes, surface areas
✓ Sector area and arc length use angle as fraction of circle
✓ Sine rule: 2 angles + 1 side or 2 sides + opposite angle
✓ Cosine rule: SAS (2 sides + included angle) or SSS (all 3 sides)
✓ Triangle area: \(\frac{1}{2}ab\sin C\) (most useful!)
✓ Bearings: always measured clockwise from North (000°-360°)

Draw Diagrams • Label Clearly • Choose the Right Rule

Master geometry and excel in IB Mathematics! 🚀