Geometry and Trigonometry — IB Math AA SL AA SL Study Guide

For: IB Math AA SL candidates sitting IB Math: Analysis & Approaches SL.

Covers: sine and cosine rules for non-right triangles, trigonometric identities and equation solving, unit circle exact values, 3D shape volume and surface area calculations, and 2D/3D vector basics for Paper 1 and Paper 2.

You should already know: IGCSE / pre-DP math.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Math AA SL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Geometry and Trigonometry?

Geometry and Trigonometry is the study of spatial relationships and the connections between side lengths and angles of shapes, extended to periodic functions and vector-based problem solving for IB AA SL. This topic makes up 20-25% of your final exam mark, with questions appearing on both Paper 1 (no calculator) and Paper 2 (calculator allowed). It connects to core later topics including calculus, kinematics, and statistical modeling, and is frequently tested in real-world context questions about navigation, engineering, and design.

2. Sine and cosine rules

These rules extend right-angled trigonometry to any scalene or obtuse triangle, where no 90° angle exists. We use standard triangle notation: side $a$ is opposite angle $A$, side $b$ opposite angle $B$, side $c$ opposite angle $C$, with the longest side always opposite the largest internal angle.

The sine rule is derived by dropping a perpendicular height from one vertex to the opposite side, giving two equal expressions for the height: $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$ where $R$ is the radius of the circumscribed circle around the triangle. Use the sine rule if you are given: (1) two angles and one side, to find missing sides; or (2) two sides and a non-included angle, to find a missing angle. Note the ambiguous case for the second use case: obtuse angles have the same sine value as their acute supplement ($\sin x=\sin (180^{\circ }-x)$), so always check if two valid angle solutions exist.

The cosine rule is a modified version of the Pythagorean theorem for non-right triangles, also derived from the perpendicular height of the triangle: $$a^2=b^2+c^2-2bc\cos A\text{or rearranged for angles:}\cos A=\frac{b^2+c^2-a^2}{2bc}$$ Use the cosine rule if you are given: (1) two sides and the included angle, to find the missing side; or (2) all three sides, to find any internal angle.

Worked Example

Triangle $ABC$ has angle $A=42^{\circ }$, side $a=12$ cm, side $b=15$ cm. Find all valid missing angle $B$ values.

1. Substitute into the sine rule: $\frac{\sin B}{15}=\frac{\sin 42^{\circ }}{12}$
2. Rearrange: $\sin B=\frac{15\times \sin 42^{\circ }}{12}\approx 0.8365$
3. Calculate principal solution: $B=\arcsin (0.8365)\approx 56.8^{\circ }$
4. Check ambiguous case: $180^{\circ }-56.8^{\circ }=123.2^{\circ }$. Since $42^{\circ }+123.2^{\circ }=165.2^{\circ }<180^{\circ }$, this is also valid. Final answer: $B=56.8^{\circ }$ or $123.2^{\circ }$

Exam tip: Examiners explicitly test the ambiguous case of the sine rule in 1-2 mark questions, so always check for a second valid angle solution when using the sine rule for two sides and a non-included angle.

3. Trig identities and equations

You are required to memorize and apply three core sets of trigonometric identities for IB AA SL, all of which are derived from the unit circle:

1. Pythagorean identity: ${\sin }^2\theta +{\cos }^2\theta =1$, from the equation of the unit circle $x^2+y^2=1$
2. Tangent identity: $\tan \theta =\frac{\sin \theta }{\cos \theta }$, valid for all $\theta$ where $\cos \theta \ne 0$
3. Double angle identities: $\sin 2\theta =2\sin \theta \cos \theta$, $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta =2{\cos }^2\theta -1=1-2{\sin }^2\theta$

To solve trigonometric equations, follow these steps:

1. Rearrange the equation to isolate a single trigonometric function, using identities to rewrite the equation in terms of one function if needed
2. Calculate the principal solution using the inverse trigonometric function
3. Find all solutions in the given domain using periodicity: $\sin \theta$ and $\cos \theta$ have a period of $360^{\circ }$ or $2\pi$, $\tan \theta$ has a period of $180^{\circ }$ or $\pi$

Worked Example

Solve $2{\cos }^{2}x+\sin x=1$ for $0\le x\le 2\pi$

1. Substitute ${\cos }^{2}x=1-{\sin }^{2}x$ from the Pythagorean identity: $2(1-{\sin }^{2}x)+\sin x=1$
2. Rearrange to a quadratic in $\sin x$: $2{\sin }^{2}x-\sin x-1=0$
3. Factor: $(2\sin x+1)(\sin x-1)=0$
4. Solve for $\sin x$: $\sin x=1$ or $\sin x=-\frac{1}{2}$
5. Find solutions in the domain: $\sin x=1⟹x=\frac{\pi }{2}$; $\sin x=-\frac{1}{2}⟹x=\frac{7\pi }{6},\frac{11\pi }{6}$ Final answer: $x=\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6}$

4. Unit circle, exact values

The unit circle is a circle of radius 1 centered at the origin of a coordinate plane. For any angle $\theta$ measured counterclockwise from the positive x-axis, the coordinates of the point where the terminal side of $\theta$ intersects the circle are $(\cos \theta ,\sin \theta )$.

The sign of trigonometric functions in each quadrant follows the mnemonic All Students Take Calculus:

• Quadrant 1 ($0<\theta <\frac{\pi }{2}$): All functions positive
• Quadrant 2 ($\frac{\pi }{2}<\theta <\pi$): $\sin \theta$ positive, $\cos \theta ,\tan \theta$ negative
• Quadrant 3 ($\pi <\theta <\frac{3\pi }{2}$): $\tan \theta$ positive, $\sin \theta ,\cos \theta$ negative
• Quadrant 4 ($\frac{3\pi }{2}<\theta <2\pi$): $\cos \theta$ positive, $\sin \theta ,\tan \theta$ negative

You must memorize exact values for $\theta =0,\frac{\pi }{6},\frac{\pi }{4},\frac{\pi }{3},\frac{\pi }{2}$ and their multiples for Paper 1 (no calculator):

Worked Example

Find the exact value of $\cos \left(\frac{5\pi }{6}\right)$ and $\sin \left(\frac{7\pi }{4}\right)$

1. $\frac{5\pi }{6}$ is in Quadrant 2, so $\cos$ is negative, reference angle $\frac{\pi }{6}$: $\cos \left(\frac{5\pi }{6}\right)=-\cos \left(\frac{\pi }{6}\right)=-\frac{\sqrt{3}}{2}$
2. $\frac{7\pi }{4}$ is in Quadrant 4, so $\sin$ is negative, reference angle $\frac{\pi }{4}$: $\sin \left(\frac{7\pi }{4}\right)=-\sin \left(\frac{\pi }{4}\right)=-\frac{\sqrt{2}}{2}$

5. Volumes and surface areas of 3D shapes

All volume and surface area formulas are provided in the IB formula booklet, but you must know how to apply them to simple and composite 3D shapes, and distinguish between total surface area (all exposed faces) and curved/lateral surface area (only curved faces, excluding flat bases for cylinders and cones).

Core formulas for IB AA SL:

1. Right prism: $V=\text{cross-sectional area}\times \text{length}$, $\text{Total SA}=2\times \text{cross-sectional area}+\text{perimeter of cross-section}\times \text{length}$
2. Right cylinder: $V=\pi r^{2}h$, $\text{Curved SA}=2\pi rh$, $\text{Total SA}=2\pi r(r+h)$
3. Right pyramid/cone: $V=\frac{1}{3}\times \text{base area}\times \text{perpendicular height}$, cone curved SA = $\pi rl$ where $l$ is slant height
4. Sphere: $V=\frac{4}{3}\pi r^3$, $\text{SA}=4\pi r^2$

Worked Example

A solid composite shape is made by attaching a hemisphere of radius 5 cm to the base of a right cone of radius 5 cm and height 12 cm. Find the total surface area of the shape.

1. Calculate slant height of the cone: $l=\sqrt{r^2+h^2}=\sqrt{25+144}=13$ cm
2. The flat base of the cone and flat face of the hemisphere are glued together, so they are not exposed. Only count curved surfaces:
3. Curved SA of cone: $\pi rl=\pi \times 5\times 13=65\pi$ cm²
4. Curved SA of hemisphere: $\frac{1}{2}\times 4\pi r^2=2\pi \times 25=50\pi$ cm²
5. Total SA: $65\pi +50\pi =115\pi \approx 361$ cm²

6. 2D and 3D vectors basics

A vector is a quantity with both magnitude (size) and direction, while a scalar only has magnitude. Vectors can be represented as column vectors $\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$, unit vector form $x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$, or directed line segments $\overrightarrow{AB}$ from point $A$ to point $B$.

Core vector operations for IB AA SL:

1. Magnitude: For vector $\mathbf{v}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$, magnitude $|\mathbf{v}|=\sqrt{x^2+y^2+z^2}$ (derived from Pythagoras)
2. Addition/subtraction: Add or subtract corresponding components of the vectors; geometrically, add vectors tip-to-tail, reverse the direction of the second vector for subtraction
3. Scalar multiplication: Multiply each component by a scalar $k$, which scales the magnitude by $|k|$ and reverses direction if $k<0$
4. Scalar (dot) product: For vectors $\mathbf{v}$ and $\mathbf{w}$, $\mathbf{v}\cdot \mathbf{w}=v_x{w}_x+v_y{w}_y+v_z{w}_z=|\mathbf{v}||\mathbf{w}|\cos \theta$, where $\theta$ is the angle between the vectors when placed tail-to-tail. Use the dot product to find the angle between two vectors, or check if vectors are perpendicular (dot product = 0).

Worked Example

Find the angle between vectors $\mathbf{a}=2\mathbf{i}+3\mathbf{j}-\mathbf{k}$ and $\mathbf{b}=\mathbf{i}-4\mathbf{j}+2\mathbf{k}$, to the nearest degree.

1. Calculate dot product: $\mathbf{a}\cdot \mathbf{b}=(2\times 1)+(3\times -4)+(-1\times 2)=2-12-2=-12$
2. Calculate magnitudes: $|\mathbf{a}|=\sqrt{4+9+1}=\sqrt{14}\approx 3.74$, $|\mathbf{b}|=\sqrt{1+16+4}=\sqrt{21}\approx 4.58$
3. Rearrange dot product formula: $\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\approx \frac{-12}{17.15}\approx -0.70$
4. Calculate angle: $\theta =\arccos (-0.70)\approx 134^{\circ }$

7. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting the ambiguous case of the sine rule, only giving one solution when two exist. Why: Students assume all triangle solutions are unique, not realizing obtuse angles share sine values with their acute supplements. Correct move: Every time you solve for an angle using the sine rule, check if $180^{\circ }$ minus your calculated angle plus the other given angles is less than $180^{\circ }$; if yes, include both solutions.
• Wrong move: Mixing degrees and radians when solving trig equations, giving answers in the wrong unit. Why: Students forget to check the domain specified in the question, or leave their calculator in the wrong mode. Correct move: Circle the domain unit (deg/rad) at the start of the question, and double-check your calculator mode before starting calculations.
• Wrong move: Including hidden overlapping faces when calculating total surface area of composite 3D shapes. Why: Students just add the total surface area of each individual shape without accounting for glued faces. Correct move: Sketch the composite shape, cross out any faces that are attached to another shape, and only count exposed faces.
• Wrong move: Using the side adjacent to an angle in the sine/cosine rule instead of the side opposite. Why: Students mix up standard triangle notation. Correct move: Always label your triangle first: side $a$ = opposite angle $A$, side $b$ = opposite angle $B$, side $c$ = opposite angle $C$ before plugging values into formulas.
• Wrong move: Calculating the angle between two vectors tip-to-tail instead of tail-to-tail, getting the supplement of the correct angle. Why: Students use the vectors as given without adjusting their orientation. Correct move: If vectors are not tail-to-tail, reverse one vector (multiply by -1) before calculating the dot product to find the correct angle between them.

8. Practice Questions (IB Math AA SL Style)

Question 1 (Paper 2, calculator allowed)

A surveyor measures two sides of a triangular plot of land: $AB=220$ m, $BC=180$ m, and angle at $A$ is $48^{\circ }$. a) Find the two possible values of angle $C$, to 1 decimal place. b) Find the smaller possible area of the plot, giving your answer to the nearest square meter.

Solution

a) Use sine rule: $\frac{BC}{\sin A}=\frac{AB}{\sin C}⟹\frac{180}{\sin 48^{\circ }}=\frac{220}{\sin C}$ $\sin C=\frac{220\times \sin 48^{\circ }}{180}\approx 0.9082$ Possible values: $C=\arcsin (0.9082)\approx 65.2^{\circ }$, or $180^{\circ }-65.2^{\circ }=114.8^{\circ }$ b) Smaller area corresponds to the smaller total internal angle sum for the third angle, so use $C=114.8^{\circ }$, angle $B=180-48-114.8=17.2^{\circ }$ Area = $\frac{1}{2}\times AB\times BC\times \sin B=0.5\times 220\times 180\times \sin 17.2^{\circ }\approx 5855$ m²

Question 2 (Paper 1, no calculator)

Solve $2\sin 2\theta =\tan \theta$ for $0\le \theta \le 2\pi$, giving exact solutions.

Solution

Rewrite using identities: $2\times 2\sin \theta \cos \theta =\frac{\sin \theta }{\cos \theta }$ Multiply both sides by $\cos \theta$ (note $\cos \theta \ne 0$, so $\theta \ne \frac{\pi }{2},\frac{3\pi }{2}$): $4\sin \theta {\cos }^2\theta =\sin \theta$ Rearrange: $\sin \theta (4{\cos }^2\theta -1)=0$ Solutions:

1. $\sin \theta =0⟹\theta =0,\pi ,2\pi$
2. $4{\cos }^2\theta =1⟹\cos \theta =\pm \frac{1}{2}$: $\cos \theta =\frac{1}{2}⟹\theta =\frac{\pi }{3},\frac{5\pi }{3}$; $\cos \theta =-\frac{1}{2}⟹\theta =\frac{2\pi }{3},\frac{4\pi }{3}$ All valid solutions: $\theta =0,\frac{\pi }{3},\frac{2\pi }{3},\pi ,\frac{4\pi }{3},\frac{5\pi }{3},2\pi$

Question 3 (Paper 2, calculator allowed)

A closed cylindrical can of height 10 cm has total surface area of $200\pi$ cm². a) Find the radius $r$ of the can, to 2 decimal places. b) A spherical ball of radius 4 cm is placed inside the can, find the volume of empty space left in the can, to the nearest cm³.

Solution

a) Total SA of cylinder: $2\pi r^2+2\pi rh=200\pi$ Divide by $2\pi$, substitute $h=10$: $r^2+10r=100⟹r^2+10r-100=0$ Solve quadratic: $r=\frac{-10+\sqrt{100+400}}{2}=-5+5\sqrt{5}\approx 6.18$ cm (take positive root) b) Volume of can: $\pi r^{2}h=\pi \times (6.18{)}^2\times 10\approx 1200$ cm³ Volume of sphere: $\frac{4}{3}\pi (4{)}^3\approx 268$ cm³ Empty space: $1200-268=932$ cm³

9. Quick Reference Cheatsheet

10. What's Next

This Geometry and Trigonometry topic is a foundational building block for the rest of the IB Math AA SL syllabus. Trigonometric functions are used extensively in the calculus topic, where you will learn to differentiate and integrate $\sin$, $\cos$ and $\tan$ functions to solve problems involving rates of change and area under periodic curves. Vectors are extended to cover vector equations of lines, intersections of lines, and applications to kinematics (motion of objects) in later units, which appear frequently on both Paper 1 and Paper 2 of your final exam.

To reinforce your understanding of these concepts, practice with as many exam-style questions as possible, and make sure you review the common pitfalls listed above to avoid losing easy marks. If you have any questions about specific problems, concepts, or exam strategy, you can ask Ollie anytime on the homepage, where you will also find more topic-specific study guides, full mock exams, and personalized feedback to help you hit your target grade.