IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~8 min read

Pythagoras & Right-Angled Trigonometry

Right-angled triangles unlock almost everything in trigonometry. Pythagoras’ theorem handles the side lengths; SOH CAH TOA handles the angles. Together they form the toolkit you’ll lean on whenever a problem can be reduced to a right-angled triangle — which is most of geometry, plenty of physics, and even 3D problems where you slice the solid into 2D right-triangle cross-sections.

📘 What you need to know

Pythagoras’ theorem: a² + b² = c², where c is the hypotenuse (the side opposite the right angle).
Pythagoras is NOT in the formula booklet — memorise it.
SOH CAH TOA: sin θ = O/H, cos θ = A/H, tan θ = O/A. Also not in the formula booklet.
O = side opposite the angle θ; A = side adjacent to θ; H = hypotenuse.
Inverse functions for finding angles: θ = sin⁻¹(O/H), cos⁻¹(A/H), or tan⁻¹(O/A).
Converse of Pythagoras: if a² + b² = c², the triangle is right-angled — handy for proofs.
3D Pythagoras: the diagonal of a cuboid with sides a, b, c has length √(a² + b² + c²).
3D problems: split into 2D right-triangles. Find lengths first, then use trig for angles between lines and planes.

Pythagoras’ theorem

The square of the hypotenuse equals the sum of the squares of the two shorter sides. This holds only for right-angled triangles — and the converse is also true: if the side lengths satisfy this equation, the triangle has a right angle.

Pythagoras’ theorem a² + b² = c² (c is the hypotenuse)

Find the hypotenuse

c = √(a² + b²)

add the squares, then take the root

Find a shorter side

a = √(c² − b²)

subtract the smaller square from the bigger

Pythagorean triples appear constantly in IB exams: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), and any integer multiple of these (e.g., 6-8-10, 9-12-15, 10-24-26). Spotting one lets you skip the calculator and gives a clean exact answer.

Right-angled trigonometry — SOH CAH TOA

Once a triangle has a right angle, the remaining angles can be linked to side ratios. There are three ratios — sine, cosine, and tangent — each connecting a different pair of sides.

Before applying any ratio, label the three sides relative to the angle θ you’re working with:

Label	Meaning
H — Hypotenuse	longest side, opposite the right angle
O — Opposite	side directly across from θ
A — Adjacent	side next to θ (and not the hypotenuse)

The three trig ratios sin θ = OH cos θ = AH tan θ = OA

The mnemonic SOH CAH TOA packs all three: Sin = O/H; Cos = A/H; Tan = O/A. Some teachers also use “Some Old Horse Caught Another Horse Trotting On Asphalt” — whatever helps it stick.

Picking the right ratio

Look at which two sides are involved in your problem. The two sides you have (or want) tell you which ratio:

You have / want	Use
O and H	sin θ = O/H
A and H	cos θ = A/H
O and A	tan θ = O/A

Finding angles — use the inverse functions

If both sides are known and you want the angle, apply the inverse: sin⁻¹, cos⁻¹, or tan⁻¹. On most calculators these are the “shift” or “second function” of sin, cos, tan.

Extending to 3D problems

3D problems often look intimidating, but the trick is always the same: find a 2D right-angled triangle inside the solid, draw it separately, and apply Pythagoras or SOH CAH TOA on that 2D triangle. Most 3D problems decompose into one or two such triangles.

3D Pythagoras — diagonal of a cuboid d = √(a² + b² + c²)

This is just Pythagoras applied twice. The base diagonal has length √(a² + b²); combine that with the height c to get the full space diagonal.

Angle between a line and a plane

For an angle between a line (e.g., the space diagonal) and a flat plane (e.g., the base), drop a perpendicular from the line down to the plane. The right-angled triangle has: the original line as hypotenuse, the perpendicular drop as one leg, and the projected length on the plane as the other leg. Apply SOH CAH TOA to find the angle.

🧭 Recipe — solving right-angled triangle problems

Sketch the triangle with the right angle marked. Label all known sides and angles.
Choose your tool: Pythagoras (only sides, no angles); SOH CAH TOA (one angle plus two sides involved); inverse trig (two sides, want the angle).
Label H, O, A with respect to the angle θ. Hypotenuse is opposite the right angle; opposite is across from θ; adjacent is next to θ.
Substitute and solve. For an unknown side, multiply or divide; for an unknown angle, use the inverse function.
For 3D problems: decompose into 2D right-angled triangles. Find lengths first, then angles.
Round to 3 s.f. unless asked otherwise. Keep exact values (surds) for intermediate steps to avoid rounding errors.

Worked examples

WE 1

Pythagoras — find the hypotenuse

A right-angled triangle has shorter sides of length 9 cm and 12 cm. Find the length of the hypotenuse.

Step 1: Apply a² + b² = c² c² = 9² + 12² c² = 81 + 144 = 225 Step 2: Take the positive square root c = √225 = 15 c = 15 cm a 9-12-15 triangle is just 3 × (3-4-5) — Pythagorean triple in disguise

WE 2

Pythagoras — find a shorter side

A right-angled triangle has hypotenuse 26 cm and one shorter side of 10 cm. Find the length of the other shorter side.

Step 1: Rearrange Pythagoras for the missing side a² = c² − b² a² = 26² − 10² Step 2: Compute a² = 676 − 100 = 576 a = √576 = 24 a = 24 cm 10-24-26 = 2 × (5-12-13) — another Pythagorean triple

WE 3

Right-angled trigonometry — find a missing side

A right-angled triangle has a hypotenuse of 16 cm and one of its acute angles is 35°. Find the length of the side opposite the 35° angle, correct to 3 s.f.

Step 1: Label the sides relative to the 35° angle H = 16; O = ?; angle = 35° Step 2: We have H, want O — use sin (SOH) sin(35°) = O/16 Step 3: Solve for O O = 16 × sin(35°) O = 16 × 0.5736… O = 9.177… O ≈ 9.18 cm (3 s.f.) make sure the calculator is in degree mode — radians here would give a wildly wrong answer

WE 4

Right-angled trigonometry — find a missing angle

In a right-angled triangle, the side opposite an acute angle θ is 7 cm and the side adjacent to it is 11 cm. Find the value of θ, correct to 3 s.f.

Step 1: Identify which sides are given O = 7; A = 11 (no hypotenuse involved) Step 2: Use tan (TOA) tan(θ) = O/A = 7/11 Step 3: Apply the inverse tangent θ = tan⁻¹(7/11) θ = tan⁻¹(0.6364…) θ = 32.471…° θ ≈ 32.5° (3 s.f.) since both legs are positive and θ is acute, only one solution exists

WE 5

3D Pythagoras — diagonal of a cuboid

A cuboid has dimensions 6 cm × 6 cm × 7 cm. Find the length of the longest internal diagonal.

Step 1: Apply 3D Pythagoras: d = √(a² + b² + c²) d² = 6² + 6² + 7² d² = 36 + 36 + 49 = 121 Step 2: Take the positive square root d = √121 = 11 d = 11 cm 36 + 36 + 49 = 121 — clean Pythagorean structure giving an integer answer; the formula is just Pythagoras applied twice (base diagonal, then up to the top corner)

WE 6

3D — diagonal length and angle with the base

A rectangular box has dimensions 8 cm × 6 cm × 5 cm (length × width × height). Find (a) the length of the space diagonal in exact form, and (b) the angle that the space diagonal makes with the base, correct to 3 s.f.

Step 1: Find the base diagonal first (2D Pythagoras) d_base² = 8² + 6² = 64 + 36 = 100 d_base = 10 cm (a) Step 2: Find the space diagonal (Pythagoras with height) d² = 10² + 5² = 100 + 25 = 125 d = √125 = 5√5 (a) d = 5√5 cm ≈ 11.2 cm (b) Step 3: The angle θ with the base — right triangle with O = 5 (height), A = 10 (base diag) tan(θ) = 5/10 = 0.5 θ = tan⁻¹(0.5) = 26.565…° (b) θ ≈ 26.6° (3 s.f.) classic 3D pattern: 2D Pythagoras gives the projection on the base, then trig finds the angle of elevation of the diagonal

💡 Top tips

Always sketch the triangle with the right angle marked. Almost every error in this topic comes from mis-identifying which side is which.
Memorise common Pythagorean triples: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25). Spotting one means a clean exact answer with no calculator needed.
Check your calculator mode — degree mode for degree problems. A misplaced RAD/DEG flips every trig answer.
Label H first, then O, then A. The hypotenuse is always opposite the right angle; “opposite” and “adjacent” depend on which acute angle you’re working with.
For 3D problems, draw the 2D triangle separately. Once it’s isolated on its own diagram, the problem reduces to standard right-triangle work.
Keep exact values during intermediate steps. Rounding too early — especially squaring decimals — accumulates error fast.
Check answers are physically sensible. The hypotenuse should be the longest side; angles should be between 0° and 90° in a right triangle.

⚠ Common mistakes

Adding when you should subtract (or vice versa). Hypotenuse: square + square. Shorter side: square − square.
Forgetting to take the square root. c² = 169 means c = 13, not 169.
Mis-labelling opposite and adjacent. They depend on which acute angle you’ve chosen — reread the question and pick the angle deliberately.
Using sin when you should use cos (or vice versa). Each ratio uses a specific pair of sides — look at which two sides are involved.
Calculator in radian mode. sin(30°) = 0.5; sin(30 rad) ≈ −0.988. Wildly different answers.
Forgetting the inverse function when finding an angle. tan(θ) = 7/11 means θ = tan⁻¹(7/11), not θ = 7/11.
Mixing units. If one length is in cm and another in mm, convert first. The trig ratios work only if the units agree.
For 3D, treating the space diagonal as the angle’s hypotenuse without justification. The right triangle for the angle uses the projection on the base — sketch it carefully.

Pythagoras and SOH CAH TOA only handle right-angled triangles. The next note — Sine Rule, Cosine Rule & Area of a Triangle — extends the toolkit to any triangle. The sine rule handles “side-angle” pairs; the cosine rule handles three sides or two-sides-and-an-included-angle; the area formula gives ½ab sin C for any triangle, not just right-angled ones. With both notes combined, you can solve any triangle problem the IB throws at you.

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