IB Maths AI HL Trigonometry Paper 1 & 2 Pythagoras & SOHCAHTOA ~8 min read

Pythagoras & Right-Angled Trigonometry

Two tools cover everything right-angle: Pythagoras (a²+b²=c²) links the three sides, and SOHCAHTOA links one acute angle to a pair of sides. For 3D problems, break the figure into 2D right triangles and apply the same two tools.

📘 What you need to know

Pythagoras’ theorem: a²+b²=c², where c is the hypotenuse (longest side, opposite the right angle).
Trig ratios (SOHCAHTOA): sin θ = O/H, cos θ = A/H, tan θ = O/A.
O, A, H labels depend on θ: opposite = across from θ, adjacent = next to θ (not the hypotenuse), hypotenuse = opposite the right angle.
To find a missing angle, use inverse: θ = sin⁻¹(O/H), etc. Calculator must be in degree mode for AI HL.
3D Pythagoras: the diagonal of a cuboid a×b×c is d = √(a²+b²+c²).
Pythagoras & SOHCAHTOA are NOT in the formula booklet — memorise both.

Pythagoras and SOHCAHTOA at a glance

The hypotenuse is the longest side — always opposite the right angle. The other two sides are labelled opposite and adjacent relative to the angle θ you’ve chosen. Once labelled, pick the ratio that matches what you have and what you want: identify two letters from O, A, H and let SOHCAHTOA tell you which function to use.

The right-angled triangle, with sides labelled relative to the chosen angle θ. The hypotenuse is fixed (opposite the right angle); opposite and adjacent flip if you switch to the other acute angle.

Right-angled toolkit Pythagoras: a²+b²=c² · sin θ = OH · cos θ = AH · tan θ = OA 3D cuboid diagonal: d = √(a²+b²+c²)

3D problems: break into 2D triangles

For any 3D problem — cuboid diagonals, pyramids, prisms — the trick is to slice out a 2D right triangle that lies inside the figure. Often you need two Pythagoras applications: first to find the floor (or face) diagonal, then to find the body diagonal using that floor diagonal as one leg and the height as the other. To find the angle between a body diagonal and a face, the right triangle has the face diagonal as the adjacent side and the height (perpendicular to the face) as the opposite side.

Two-step trick: for a cuboid a×b×c, the floor diagonal is √(a²+b²), and the body diagonal is √(a²+b²+c²). The 3D formula is just two 2D Pythagoras steps combined.

🧭 Recipe — right-angled trig

Sketch and label: mark the right angle, the known angle θ, and label sides O, A, H relative to θ.
Pick the tool: if you have/want only sides, use Pythagoras. If an angle is involved, use SOHCAHTOA.
Match the ratio: pick the two of O, A, H you have/need — sin (OH), cos (AH), or tan (OA).
Substitute & rearrange: solve for the unknown. Use sin⁻¹, cos⁻¹, tan⁻¹ for angles.
For 3D: break into 2D right triangles; usually two Pythagoras steps, or one step + a trig ratio for an angle.

Worked examples

WE 1

Pythagoras — find the hypotenuse

A rectangular flag is 12 cm wide and 9 cm tall. Find the length of its diagonal.

apply a² + b² = c² c² = 12² + 9² = 144 + 81 = 225 take the positive root c = √225 diagonal = 15 cm 3-4-5 triple scaled by 3 — clean integer answer.

WE 2

Pythagoras — find a shorter side

A loading ramp is 13 m long. Its upper end rests on a platform 5 m above ground level. Find the horizontal distance from the base of the ramp to the platform.

ramp = hypotenuse; rise = shorter side x² + 5² = 13² x² = 169 − 25 = 144 x = 12 m 5-12-13 triple — another clean integer.

WE 3

SOHCAHTOA — find a missing length

A kite is flown on a string 25 m long. The string makes an angle of 40° with the horizontal ground. Find the horizontal distance from the kite-flyer to the point directly below the kite. Give your answer to 1 d.p.

label the triangle H = 25, want A, angle θ = 40° pick cos (A and H) cos(40°) = A/25 A = 25 × cos(40°) = 25 × 0.766… A ≈ 19.2 m (1 d.p.)

WE 4

SOHCAHTOA — find a missing angle

A wheelchair ramp rises 2.4 m vertically over a horizontal distance of 3.2 m. Find the angle the ramp makes with the ground, giving your answer to 1 d.p.

label O = 2.4 (rise), A = 3.2 (run) use tan (O and A) tan(θ) = 2.4/3.2 = 0.75 inverse tangent θ = tan⁻¹(0.75) = 36.869… θ ≈ 36.9° calculator in DEG mode — keep full value in memory for any follow-up parts.

WE 5

Isosceles split — height of a tent

A symmetric ridge tent has two equal sloping sides of 13 m meeting at the top, and the base across the ground is 10 m wide. Find the vertical height of the tent at its centre.

drop a perpendicular from the apex it splits the base into two halves of 5 m right triangle: leg 5, hyp 13, height h 5² + h² = 13² h² = 169 − 25 = 144 h = 12 m isosceles symmetry → split into two right triangles → use Pythagoras.

WE 6

3D — diagonal and angle in a fish tank

A rectangular fish tank has internal dimensions 6 cm by 8 cm (base) and 24 cm tall. Find: (a) the length of the longest straight rod that fits inside the tank; (b) the angle this rod makes with the base of the tank. Give answers exactly where possible and to 1 d.p. otherwise.

(a) first the floor diagonal f² = 6² + 8² = 36 + 64 = 100 f = 10 cm then the body diagonal d² = f² + 24² = 100 + 576 = 676 d = 26 cm (exact) (b) right triangle: floor diag (A) and height (O) tan(θ) = 24/10 = 2.4 θ = tan⁻¹(2.4) = 67.380… θ ≈ 67.4° 6-8-10 on the floor, then 10-24-26 vertically — two Pythagorean triples back-to-back.

💡 Top tips

Memorise both formulas — Pythagoras and SOHCAHTOA are NOT in the formula booklet.
Always sketch and label O, A, H relative to the chosen angle θ before picking a ratio.
Look for Pythagorean triples: (3,4,5), (5,12,13), (8,15,17), (7,24,25). Multiples count: (6,8,10), (9,12,15), (10,24,26).
For 3D, draw the 2D triangle separately with the sides clearly labelled — it removes ambiguity.
Store full calculator values between parts of a question; only round at the very last step.

⚠ Common mistakes

Subtracting the wrong way around in Pythagoras — for a shorter side, it’s hypotenuse² − other side², never the reverse.
Mislabelling O and A — they depend on the chosen angle. Switching the angle swaps them.
Calculator in radian mode when the answer should be in degrees — check the mode indicator.
Forgetting to use inverse trig when finding an angle — sin(θ) = 0.5 means θ = sin⁻¹(0.5), not θ = 0.5.
Rounding mid-calculation in multi-step 3D problems — carry full accuracy until the end.

Next up — Sine Rule, Cosine Rule & Area of a Triangle. When the triangle is not right-angled, SOHCAHTOA is out and these three rules take over.

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