Topic DSE14 · More about Trigonometry

39 Questions. 11 3D Figures.
Every Trig Pattern HKEAA Has Used.

A forensic decomposition of DSE14 from 2012 to 2024 — the question templates, the 3D geometry traps, and the diagram generation pipeline that makes this topic different from every other.

1. The Raw Numbers

We parsed every Paper 2 from 2012 to 2024. DSE14 (Trigonometry) is one of the heaviest topics — and the only one that requires 3D spatial reasoning.

39
Total Questions
13
Years (2012–2024)
2–5
Questions per Paper
14
Atomic Skills

7 sub-topics, wildly unequal weight. Two sub-topics account for 25 of 39 questions:

14
2D Ratio Chain
11
3D Trig
4
Identity
3
Equations
3
Graphs
2
Bearings
2
2D Cosine
Section A (Foundation) — Q17–Q25
  • Q18–Q23 — 2D trig ratio chains: "find AB/CD" given two connected right triangles
  • Q19–Q23 — Identity simplification or "which must be true?"
Section B (Advanced) — Q37–Q41
  • Q38–Q41 — 3D trigonometry: tetrahedra, cubes, prisms, pyramids
  • Q39 — Trig equation root counting
Key Difference from DSE17
DSE17 (Measures of Dispersion) is pure computation — no spatial reasoning, no diagrams. DSE14 is the opposite: 11 of 39 questions require 3D spatial visualization, and every single one of those 11 is in Section B at the hardest positions (Q38–Q41). A student who can't mentally decompose a tetrahedron into right triangles cannot score on these — no formula will save them.

2. The 4 Question Templates (With Real Exams)

Every DSE14 question falls into one of 4 templates. The first covers Section A (foundation marks); the rest target Section B (the hard half).

Template 1: 2D Trig Ratio Chain

Section A · 14 questions

The foundation template. Appears in every single year without exception. Usually Q18–Q25 in Section A.

The pattern: Given a figure with 2+ connected right triangles and labeled angles (\(\alpha, \beta, \theta\)), express one ratio (e.g., \(\text{AB}/\text{CD}\)) in terms of trig functions of those angles. Chain the ratios through shared sides.

DSE 2024 Paper 2, Q23 — Actual Question
"ABC is a straight line. D is above C with \(\angle BCD = 90°\). \(\angle DAB = \theta\), \(\angle DBC = \varphi\). Find BC/AD."
A. \(\sin\theta / \tan\varphi\) ✓
B. \(\cos\theta / \tan\varphi\)
C. \(\sin\theta \cdot \tan\varphi\)
D. \(\cos\theta \cdot \tan\varphi\)
Step-by-step Solution
1 In right \(\triangle BCD\): \(\text{DC} = \text{BC} \cdot \tan\varphi\)
2 In \(\triangle ABD\): \(\text{BC}/\text{AD} = \sin\theta / \tan\varphi\) (chain through DC)
HKEAA's Trap Design
  • Options C & D — For students who multiply instead of dividing. The chain direction matters: one triangle gives a product, the combined ratio gives a quotient.
  • Option B — For students who swap sin/cos for the angle in the first triangle.
DSE 2019 Paper 2, Q22 — Same Template, Different Dress
"AB = CD (both horizontal). AB/AC = \(\cos\alpha\), CD/CE = \(\cos\beta\). Find CE/AC."
A. \(\sin\alpha / \sin\beta\)
B. \(\cos\alpha / \cos\beta\) ✓
C. \(\sin\beta / \sin\alpha\)
D. \(\cos\beta / \cos\alpha\)
Solution
1 \(\text{AB} = \text{AC}\cos\alpha\) and \(\text{CD} = \text{CE}\cos\beta\)
2 Since AB = CD: \(\text{AC}\cos\alpha = \text{CE}\cos\beta\)
3 \(\text{CE}/\text{AC} = \cos\alpha / \cos\beta\)
The Twist
2024 uses a compound figure with multiple triangles. 2019 uses an equality constraint (AB = CD) to force the chain. Same underlying cognitive move: chain trig ratios through a shared variable and simplify.

Template 2: Trig Identity Simplification

Section A · 4 questions

Tested in: 2012, 2013, 2014, 2015 (4 of 13 years). Has not appeared since 2016.

The toolkit: Six compound-angle conversions: \(\sin(90°\pm\theta)\), \(\cos(90°\pm\theta)\), \(\sin(180°\pm\theta)\), \(\cos(180°\pm\theta)\), \(\sin(270°\pm\theta)\), \(\cos(270°\pm\theta)\). Convert everything to \(\sin\theta\) and \(\cos\theta\), then simplify.

DSE 2014 Paper 2, Q19 — Actual Question
"Simplify \((\cos(90°+\theta)+1)(\sin(360°-\theta)-1)\)."
A. \(-\cos^2\theta\) ✓
B. \(\cos^2\theta\)
C. \(-\sin^2\theta\)
D. \(\sin^2\theta\)
Step-by-step Solution
1 \(\cos(90°+\theta) = -\sin\theta\) and \(\sin(360°-\theta) = -\sin\theta\)
2 Product = \((-\sin\theta + 1)(-\sin\theta - 1) = \sin^2\theta - 1 = -\cos^2\theta\)
HKEAA's Trap Design
All four options are ±sin² or ±cos². A single sign error in the compound-angle conversion flips the answer. Option D traps students who get both signs wrong (double negative = positive, but wrong function).
Observation
This template disappeared after 2016 and may have migrated to Paper 1. For exam prep, it's still worth drilling (the conversions are mechanical and scorable) but its frequency is declining.

Template 3: 3D Trigonometry

Section B · 11 questions

The hardest template in all of DSE Math Paper 2. Appears in 11 of 13 years (missing only 2015). Always Q38–Q41, the last positions in the exam.

The cognitive demand: Read a 3D figure. Identify the correct right triangle(s) hidden within it. Compute side lengths via 3D Pythagoras. Apply the cosine formula to find the target angle. 4–6 steps, and each wrong triangle choice is unrecoverable.

DSE 2024 Paper 2, Q40 — Actual Question
"Tetrahedron PQRS. Base PQR on horizontal ground. Q vertically below S. \(\angle PQR = 90°\), \(\angle QPS = 30°\), \(\angle QRS = 45°\). Find \(\cos\angle PRS\)."
A. \(\frac{1}{4}\)
B. \(\frac{\sqrt{3}}{4}\)
C. \(\frac{\sqrt{2}}{4}\) ✓
D. \(\frac{\sqrt{6}}{4}\)
Step-by-step Solution
1 Let \(QS = h\). Since \(\angle QPS = 30°\), \(PQ = h/\tan 30° = h\sqrt{3}\). Since \(\angle QRS = 45°\), \(QR = h\).
2 \(\angle PQR = 90°\) → \(PR^2 = PQ^2 + QR^2 = 3h^2 + h^2 = 4h^2\), so \(PR = 2h\)
3 \(PS^2 = PQ^2 + QS^2 = 3h^2 + h^2 = 4h^2\), so \(PS = 2h\)
4 \(RS^2 = QR^2 + QS^2 = 2h^2\), so \(RS = h\sqrt{2}\)
5 Cosine formula: \(\cos\angle PRS = \frac{RS^2 + PR^2 - PS^2}{2 \cdot RS \cdot PR} = \frac{2h^2 + 4h^2 - 4h^2}{2 \cdot h\sqrt{2} \cdot 2h} = \frac{2h^2}{4h^2\sqrt{2}} = \frac{\sqrt{2}}{4}\)
HKEAA's Trap Design
  • Option D (\(\sqrt{6}/4\)) — Tempting because it combines \(\sqrt{2}\) and \(\sqrt{3}\), which both appear in the setup. Students who make an error in the Pythagoras step get here.
  • The real difficulty: Recognizing that \(h\) cancels out entirely. The answer is a pure number despite the problem giving no actual lengths — only angle constraints.
DSE 2014 Paper 2, Q40 — Same Template, Different Figure
"\(\triangle BCD\) with \(BC = 8\), \(BD = 15\), \(CD = 17\). \(AB = 8\text{m}\) perpendicular to plane \(BCD\). \(E\) is foot of perpendicular from \(A\) to \(CD\). Find \(\tan\angle AEB\)."
A. \(\frac{8}{15}\)
B. \(\frac{15}{8}\)
C. \(\frac{8}{17}\)
D. \(\frac{17}{15}\) ✓
Solution
1 \(8^2 + 15^2 = 64 + 225 = 289 = 17^2\). So \(\angle BCD = 90°\) (Pythagorean triple).
2 Area of \(\triangle BCD = \frac{1}{2} \cdot 8 \cdot 15 = 60\). Also \(= \frac{1}{2} \cdot BE \cdot 17\). So \(BE = 120/17\).
3 \(\tan\angle AEB = AB/BE = 8/(120/17) = 136/120 = 17/15\).
The Twist vs 2024
2024 gives angle constraints and no lengths — you set up \(h\) and it cancels. 2014 gives concrete lengths and a Pythagorean triple — you need to recognize it and use the area method to find the foot of perpendicular. Both require 3D decomposition, but the cognitive entry points are completely different.

Template 4: Trig Equation Root Counting

Section B · Q39

Tested in: 2014, 2022, 2024 (3 of 13 years). Always Q39 in Section B. Low frequency but completely predictable format.

The algorithm: (1) Convert to a quadratic in \(\sin x\) or \(\cos x\). (2) Factor. (3) Solve for each root. (4) Count solutions in \([0°, 360°)\), checking all quadrants.

DSE 2022 Paper 2, Q39 — Actual Question
"For \(0° \le x < 360°\), how many roots does \(\sin^2 x = 6\cos^2 x\) have?"
A. 2
B. 3
C. 4 ✓
D. 5
Step-by-step Solution
1 Divide both sides by \(\cos^2 x\): \(\tan^2 x = 6\)
2 \(\tan x = +\sqrt{6}\) gives 2 solutions (Q1 and Q3)
3 \(\tan x = -\sqrt{6}\) gives 2 solutions (Q2 and Q4)
4 Total = 4. (Division by \(\cos^2 x\) is safe: \(\cos x = 0\) gives \(\sin^2 x = 0\), contradiction.)
HKEAA's Trap Design
  • Option A (2) — For students who only take the positive root.
  • Option B (3) — For students who count \(\cos x = 0\) as an extra root.
  • Option D (5) — For students who include \(x = 0°\) or \(x = 360°\) incorrectly.

3. 3D Trig Mutation Timeline: 11 Years of 3D Questions

The 3D trigonometry questions are the most valuable to track — hardest, highest-weighted, and the ones Renee's students struggle with most. Here's every single one:

Year Figure Type Core Task Key Technique Difficulty
2012 Regular tetrahedron Find dihedral angle ∠AED Cosine formula on isosceles triangle ADV
2013 Regular tetrahedron Find volume given height 3D Pythagoras + base area ADV
2014 Perpendicular to plane Find tan∠AEB (foot of perp) Pythagorean triple + area method ADV
2015 No 3D question this year
2016 Rectangular box Find sin∠PFQ (midpoint + edge point) 3D Pythagoras + cosine formula, 5 steps HARD
2017 Perpendicular planes Find angle between line and plane Two Pythagorean triples + projection ADV
2018 Cuboid with projection Find cos∠YBX (projection of X on plane) 3D Pythagoras chain: XY→YA→YB→XB ADV
2019 Point above ground Find ∠RPS from elevation angles Cosine formula twice (ground + 3D) HARD
2020 Right triangular prism Find area of △BDP (parametric) Heron's formula + 3D coordinates HARD
2022 Cube (cross-sections) Compare dihedral angles α vs β Coordinate geometry + dot product HARD
2023 Right pyramid (square base) Find cosθ (dihedral along edge BV) Foot of perp + cosine formula HARD
2024 Tetrahedron on ground Find cos∠PRS from angle constraints All lengths cancel — pure angle reasoning HARD
The Clear Trend

2012–2014: regular solids (symmetric, fewer steps). 2016–2024: irregular solids, parametric constraints, side ratios instead of lengths, dihedral angles. The difficulty is escalating. The cosine formula is THE dominant tool — required in 8 of 11 problems.

4. Inside the Knowledge Graph: Real Data

Every question is decomposed into structured data. Here are three entries — a 2D ratio chain, a 3D problem with diagram spec, and an identity question:

// DSE 2024 Q40 — 3D Tetrahedron (with diagramSpec) { "year": 2024, "q": 40, "section": "B", "stem": "Tetrahedron PQRS. ∠PQR=90°, ∠QPS=30°, ∠QRS=45°. Find cos∠PRS.", "skillsTested": ["3d-cosine-formula", "3d-perpendicular-projection", "3d-figure-decomposition"], "toAce": "Set QS=h. PQ=h√3, QR=h. PR=PS=2h, RS=h√2. Cosine formula → √2/4.", "errorTraps": ["Wrong elevation-to-side relationship", "Arithmetic errors in cosine formula"], "diagramSpec": { "figureType": "tetrahedron-on-ground", "vertices": ["P", "Q", "R", "S"], "markedAngles": ["∠PQR=90°", "∠QPS=30°", "∠QRS=45°"], "constraints": ["PQR on ground", "SQ ⊥ ground"] } }
// DSE 2019 Q22 — 2D Ratio Chain { "year": 2019, "q": 22, "section": "A", "stem": "AB = CD. AB/AC = cosα, CD/CE = cosβ. Find CE/AC.", "skillsTested": ["trig-ratio-chain"], "toAce": "AB=CD → AC cosα = CE cosβ → CE/AC = cosα/cosβ.", "errorTraps": ["Inverting the ratio", "Wrong trig function"] }
// DSE 2014 Q19 — Identity Transform { "year": 2014, "q": 19, "section": "A", "stem": "Simplify (cos(90°+θ)+1)(sin(360°-θ)-1).", "skillsTested": ["trig-identity-transform"], "toAce": "cos(90°+θ)=-sinθ, sin(360°-θ)=-sinθ. Product = sin²θ-1 = -cos²θ.", "errorTraps": ["Wrong sign for cos(90°+θ)", "Expanding incorrectly"] }

DSE14 vs DSE17: Schema Comparison

FieldDSE17 (Dispersion)DSE14 (Trig)
stemText only — no figures neededOften references a figure — requires diagram
diagramSpecN/A — never neededPresent on 11 questions — figure type, vertices, angles, constraints
skillsTestedSame schemaSame schema
errorTrapsComputational traps (sign errors, formula confusion)Spatial traps (wrong triangle, wrong projection, wrong angle)
toAceSame schemaSame schema, but often longer (4–6 steps vs 2–3)
The key difference
DSE17's knowledge graph is complete and self-contained — every question can be fully understood from the stem text alone. DSE14's graph has a diagram dependency: 11 questions cannot be properly studied, generated, or presented without the accompanying 3D figure. This is what the diagram pipeline addresses.

5. The Diagram Challenge: 3D Figure Generation

This is the capability DSE17 doesn't need and DSE14 can't do without. Every 3D trig question comes with a figure that students must read. Any question engine for this topic must generate correct, exam-style diagrams.

Current Pipeline: v3 Eval Results

We evaluated diagram generation on 9 past-paper 3D questions (ground truth from official exam papers):

3
Visual PASS
5
Visual PARTIAL
1
Visual FAIL
3
5
1
YearFigureProgrammaticVisualIssue
2014 Q40Perp to plane11/11PASS—
2019 Q40Point above ground10/10PASS—
2024 Q40Tetrahedron on ground10/10PASS—
2012 Q40Regular tetrahedron9/9PARTIALProjection angle slightly off
2016 Q39Rectangular box9/9PARTIALEdge proportions inaccurate
2017 Q39Perp planes11/11PARTIALPlane intersection unclear
2018 Q41Cuboid projection8/8PARTIALX/Y positioning off
2020 Q38Triangular prism7/7PARTIALBF line rendered at wrong angle
2022 Q40Cube cross-sections7/7FAILCube shape structurally deformed

Question Generation Pipeline: Working

We flipped the approach: generate questions first, then diagrams. The LLM writes exam-style questions with full solutions, then writes geometry-accurate TikZ code. node-tikzjax renders TikZ to SVG. Triple validation: programmatic checks + LLM review + human review.

7
Questions Generated
7
Figure Types
75/75
Auto Checks Pass
2
Improvement Passes

Here are 3 examples from the 7 generated questions, each with its AI-rendered diagram. See all 7 →

Generated — Perpendicular to Plane AI-GENERATED
"△BCD lies on a horizontal plane with BC = 6, BD = 8, CD = 10. AB ⊥ plane BCD, AB = 9. E is foot of ⊥ from B to CD. Find \(\tan\angle AEB\)."
A. \(\frac{3}{4}\)
B. \(\frac{5}{4}\)
C. \(\frac{15}{8}\) ✓
D. \(\frac{9}{10}\)
Uses 6-8-10 Pythagorean triple + area method for foot of perpendicular. Skill: 3d-perpendicular-projection.
AI
ABCDE68109
Generated — Tetrahedron on Ground AI-GENERATED
"PQRS tetrahedron. Base PQR on horizontal plane. S vertically above Q. \(\angle PQR=90°\), \(\angle QPS=60°\), \(\angle QRS=45°\). Find \(\cos\angle PRS\)."
A. \(\frac{\sqrt{2}}{4}\)
B. \(\frac{\sqrt{6}}{4}\) ✓
C. \(\frac{\sqrt{3}}{4}\)
D. \(\frac{1}{4}\)
Pure angle reasoning — h cancels. Skills: 3d-cosine-formula, 3d-figure-decomposition.
AI
60°45°QPRS
Generated — Equilateral Base Pyramid AI-GENERATED
"VABC pyramid. Base ABC equilateral, side 6, horizontal. V above centroid G, VG = 4. Find \(\cos\angle AVB\)."
A. \(\frac{5}{14}\) ✓
B. \(\frac{5}{7}\)
C. \(\frac{3}{7}\)
D. \(\frac{2}{7}\)
Centroid distance + 3D Pythagoras + cosine formula. Skill: 3d-cosine-formula.
AI
6664ABCVG
Pipeline: LLM → TikZ → node-tikzjax → SVG

The generation pipeline runs entirely without manual code: LLM generates the question and writes geometry-accurate TikZ, node-tikzjax renders to SVG, programmatic validators check structure. 7 figure types covered across 2 improvement passes. Full generation showcase →

6. The Complete Skill Map: All 14 Atomic Skills

These are the 14 irreducible skills we extracted. Every DSE14 question from 2012–2024 can be solved using some combination of these.

SkillLevelWhat It MeansTested
trig-ratio-basicFOUND.Apply sin/cos/tan in a single right triangle. Direct application.8×
trig-ratio-chainCOREChain trig ratios across 2+ connected triangles. Express composite ratio (AB/CD) through shared sides.8×
3d-pythagorasCOREApply Pythagoras across perpendicular planes to find 3D distances.8×
3d-figure-decompositionADV.Mentally extract right triangles from tetrahedra, cubes, prisms, pyramids.7×
3d-cosine-formulaADV.Compute 3D side lengths via Pythagoras, then apply \(a^2 = b^2 + c^2 - 2bc\cos A\).5×
3d-perpendicular-projectionADV.Find foot of perpendicular from point to plane. Angle between line and plane = angle to projection.5×
trig-identity-transformCORESimplify using complementary, supplementary, and Pythagorean identities.4×
3d-dihedral-angleADV.Find angle between two planes sharing a common edge via foot-of-perpendicular construction.4×
trig-equation-solvingADV.Convert to quadratic in sinx or cosx, factor, count roots in all quadrants.3×
trig-graph-readingFOUND.Determine amplitude, frequency, phase, and vertical shift from graph of \(y = h + k\cos(nx°)\).3×
cosine-formula-2dCOREApply \(a^2 = b^2 + c^2 - 2bc\cos A\) in 2D non-right triangles.3×
bearing-directionCOREConvert between compass bearings and angles. Use alternate angles with parallel North lines.2×
trig-identity-propertiesCOREGiven constraints, determine which trig statements must hold. Test edge cases.1×
sine-formula-2dCOREApply \(a/\sin A = b/\sin B\) in circle/sector problems.1×
The 80/20
If a student masters trig-ratio-basic + 3d-pythagoras, they can attempt 16 of 39 questions. Add 3d-cosine-formula + 3d-figure-decomposition and they cover 30 of 39. The remaining 9 are identity, equation, and graph — formulaic templates that drill well.

7. Exam Intelligence: What Our Data Reveals

Patterns We Extracted From 13 Years

  • 3D trig appears in 11 of 13 years — only missing 2015. Always Q38–Q41 (hardest positions). This is effectively guaranteed on the exam.
  • Cosine formula is THE dominant 3D tool — needed in 8 of 11 3D problems. A student who can't apply cosine formula in 3D is locked out of Section B's hardest marks.
  • Figure types are diversifying. 2012–2014 tested regular solids (symmetric tetrahedra). 2020–2024 test irregular solids with parametric constraints, side ratios, and dihedral angles. Practice with only regular shapes is insufficient.
  • Trig identity questions disappeared after 2016. Four appearances in 2012–2015, zero since. Likely migrated to Paper 1.
  • Trig equation root counting is sparse but predictable: 3 appearances in 13 years, always Q39, always "how many roots" — never "find the roots." The algorithm (convert to quadratic, factor, count per quadrant) is identical every time.
  • Side ratios replacing concrete lengths: 2023 and 2024 give "AB:AV = 5:4" instead of "AB = 20cm." This makes the answer a pure number (like \(\sqrt{2}/4\)) and prevents students from relying on calculators.

The Teaching Implication

What Most Tutors Do
  • Teach the cosine formula
  • Give 2D practice problems
  • Show one 3D example and hope
  • Skip dihedral angles ("too hard")
What Our System Does
  • Diagnose: can they decompose a 3D figure?
  • Generate 3D figures with correct diagrams
  • Drill the specific figure type that's failing
  • Escalate: tetrahedron → cuboid → prism → pyramid

8. What This Enables

DSE14 is the proof case for diagram-dependent topic mastery. Everything we build here applies to any topic that requires spatial reasoning.

The DSE14 Deliverable Pipeline

✓
Competency Map
39 questions, 14 skills, 13 years
✓
Knowledge Graph
Full decomposition with diagramSpec
✓
Diagram Pipeline
7/7 generated, 75/75 checks
✓
Question Engine
7 questions, 7 figure types

Compared to DSE17

CapabilityDSE17DSE14
Competency mapDone (47 Qs)Done (39 Qs)
Knowledge graphDoneDone + diagramSpec
Skill map20 skills14 skills
Exam intelligenceDoneDone
Mutation timelineDoneDone (3D trig focus)
Deep dive reportDoneThis page
Generated questions5 demonstrated7 with diagrams
Diagram generationN/A7/7 generated, 75/75 checks

The Moat for DSE14

DSE17 (Dispersion) proved the competency-map methodology works for text-only topics. DSE14 (Trigonometry) proves it for diagram-dependent topics — and that's where the moat deepens, because:

  • No competitor can generate correct 3D exam figures programmatically today
  • The diagramSpec schema makes every 3D question machine-readable
  • Once the pipeline works for DSE14, it extends to coordinate geometry, loci, and circle theorems
  • The question engine + diagram pipeline together enable unlimited 3D practice — the one thing tutors can't provide