DSE14 · More about Trigonometry

3D Trigonometry:
Generated Questions

Seven Section B questions with diagrams, covering the figure types that appear at Q38–Q41 of HKDSE Paper 2.

How These Were Made

Each question was authored by an LLM, which also wrote geometry-accurate TikZ source code for the accompanying diagram. node-tikzjax rendered the TikZ to SVG server-side. Solutions were verified independently. The diagrams went through two improvement passes based on human feedback (right-angle mark alignment, degree symbol rendering, label placement).

Figure types: perpendicular-to-plane, tetrahedron, rectangular box, cube, equilateral-base pyramid, tower with ground points, cuboid.

Generated Questions

Q1 — Perpendicular to Plane perpendicular-to-plane
In the figure, triangle BCD lies on a horizontal plane with BC = 6, BD = 8 and CD = 10. AB is perpendicular to the plane BCD and AB = 9. E is the foot of the perpendicular from B to CD. Find \(\tan(\angle AEB)\).
A. \(\dfrac{3}{4}\)
B. \(\dfrac{5}{4}\)
C. \(\dfrac{15}{8}\) ✓
D. \(\dfrac{9}{10}\)
Trap Analysis
  • A (3/4): BC/BD ratio trap
  • B (5/4): Confused ratio
  • D (9/10): AB/CD — skipping BE step
generated ABCDE68109
Step-by-step Solution
1 \(6^2 + 8^2 = 100 = 10^2\). Triangle BCD is right-angled at B.
2 \(\text{Area}(BCD) = \tfrac{1}{2}(6)(8) = 24 = \tfrac{1}{2}(BE)(10)\). So \(BE = \tfrac{24}{5}\).
3 AB is perpendicular to the ground, so AB ⊥ BE. \(\tan(\angle AEB) = \dfrac{AB}{BE} = \dfrac{9}{24/5} = \dfrac{45}{24} = \dfrac{15}{8}\).
Verification
\(AE = 51/5\). \(\sin = 15/17\), \(\cos = 8/17\), \(\tan = 15/8\). Consistent.
Pass 2 Fix
Right angle mark at B now follows edges BA and BC in projected coords (was screen-axis-aligned in Pass 1).
Q2 — Tetrahedron on Ground tetrahedron-on-ground
PQRS is a tetrahedron. The triangular base PQR lies on a horizontal plane. S is vertically above Q (SQ perpendicular to ground). \(\angle PQR = 90°\), \(\angle QPS = 60°\) and \(\angle QRS = 45°\). Find \(\cos(\angle PRS)\).
A. \(\dfrac{\sqrt{2}}{4}\)
B. \(\dfrac{\sqrt{6}}{4}\) ✓
C. \(\dfrac{\sqrt{3}}{4}\)
D. \(\dfrac{1}{4}\)
Trap Analysis
  • A (\(\sqrt{2}/4\)): DSE 2024 Q40 similar setup, confusable
  • C (\(\sqrt{3}/4\)): Partial simplification error
  • D (1/4): Missing factor in denominator
generated 60°45°QPRS
Step-by-step Solution
1 Let \(QS = h\). \(\tan 60° = h/PQ\), so \(PQ = h/\sqrt{3}\). \(\tan 45° = h/QR\), so \(QR = h\).
2 \(\angle PQR = 90°\). \(PR^2 = h^2/3 + h^2 = 4h^2/3\). \(PR = 2h/\sqrt{3}\).
3 \(PS^2 = h^2/3 + h^2 = 4h^2/3\). \(PS = 2h/\sqrt{3}\). \(RS^2 = 2h^2\). \(RS = h\sqrt{2}\).
4 \(\cos(\angle PRS) = \dfrac{RS^2 + PR^2 - PS^2}{2 \cdot RS \cdot PR} = \dfrac{2h^2 + 4h^2/3 - 4h^2/3}{2 \cdot h\sqrt{2} \cdot 2h/\sqrt{3}} = \dfrac{2h^2}{4h^2\sqrt{2}/\sqrt{3}} = \dfrac{\sqrt{6}}{4}\).
Verification
\(h\) cancels. \(\cos(\angle PRS) = \sqrt{6}/4 \approx 0.612\). Angle \(\approx 52.2°\). Reasonable.
Pass 2 Fixes
Both right angle marks now follow edge directions in projected coords. PQR = 90° uses L-mark (not arc). Degree symbol post-processed.
Q3 — Rectangular Box rectangular-box
ABCDEFGH is a rectangular box where ABCD is the bottom face and EFGH is the top face. AB = 3, BC = 4, and BF = 12. Find \(\tan\) of the angle between the space diagonal AG and the base plane ABCD.
A. \(\dfrac{12}{5}\) ✓
B. \(\dfrac{5}{13}\)
C. \(\dfrac{12}{13}\)
D. \(\dfrac{5}{12}\)
Trap Analysis
  • B (5/13): cos — wrong trig function
  • C (12/13): sin — wrong trig function
  • D (5/12): Inverted ratio
generated 3412ABCDEFGH
Step-by-step Solution
1 Coordinates: \(A = (0,0,0)\), \(G = (3,4,12)\). \(\vec{AG} = (3,4,12)\), \(|AG| = \sqrt{169} = 13\).
2 Projection on base = \((3,4,0)\), \(|\text{proj}| = 5\).
3 \(\tan\theta = 12/5\).
Verification
3-4-5 base diagonal, height 12, space diagonal 13 (5-12-13 triple). \(\sin = 12/13\), \(\cos = 5/13\), \(\tan = 12/5\).
Pass 2 Fix
12 label moved to BF (right-front vertical edge). Was labeled on AE edge in Pass 1.
Q4 — Cube Diagonals cube
ABCDEFGH is a cube where ABCD is the bottom face and EFGH is the top face. Find \(\cos\) of the angle between the space diagonals AG and BH.
A. \(\dfrac{1}{3}\) ✓
B. \(\dfrac{1}{2}\)
C. \(\dfrac{2}{3}\)
D. \(\dfrac{1}{\sqrt{3}}\)
Trap Analysis
  • B (1/2): Confusion with face diagonal angle
  • C (2/3): Numerator error
  • D (\(1/\sqrt{3}\)): Confusing with direction cosine of a space diagonal
generated µABCDEFGH
Step-by-step Solution
1 Let side \(= a\). \(A = (0,0,0)\), \(G = (a,a,a)\). \(B = (a,0,0)\), \(H = (0,a,a)\).
2 \(\vec{AG} = (a,a,a)\), \(\vec{BH} = (-a,a,a)\).
3 \(\vec{AG} \cdot \vec{BH} = -a^2 + a^2 + a^2 = a^2\). \(|AG| = |BH| = a\sqrt{3}\).
4 \(\cos\theta = \dfrac{a^2}{a^2 \cdot 3} = \dfrac{1}{3}\).
Verification
Answer is independent of side length. \(\cos = 1/3\), angle \(\approx 70.5°\). Well-known result for cube space diagonals.
Q5 — Equilateral Base Pyramid pyramid-equilateral-base
VABC is a pyramid where the base ABC is an equilateral triangle of side 6 lying on a horizontal plane. V is directly above the centroid G of triangle ABC, and VG = 4. Find \(\cos(\angle AVB)\).
A. \(\dfrac{5}{14}\) ✓
B. \(\dfrac{5}{7}\)
C. \(\dfrac{3}{7}\)
D. \(\dfrac{2}{7}\)
Trap Analysis
  • B (5/7): Divides by VA instead of \(2 \cdot VA \cdot VB\) (factor of 2 error)
  • C (3/7): Wrong numerator
  • D (2/7): Wrong numerator
generated 6664ABCVG
Step-by-step Solution
1 Centroid distance from vertex of equilateral triangle side 6: \(GA = 6/\sqrt{3} = 2\sqrt{3}\).
2 \(VA = \sqrt{GA^2 + VG^2} = \sqrt{12 + 16} = \sqrt{28} = 2\sqrt{7}\). Similarly \(VB = 2\sqrt{7}\).
3 \(AB = 6\). Cosine formula: \(\cos(\angle AVB) = \dfrac{VA^2 + VB^2 - AB^2}{2 \cdot VA \cdot VB} = \dfrac{28 + 28 - 36}{2 \cdot 28} = \dfrac{20}{56} = \dfrac{5}{14}\).
Verification
\(5/14 \approx 0.357\). Angle \(\approx 69.1°\). Reasonable for a squat pyramid.
Q6 — Tower & Two Points tower-two-points
OT is a vertical tower on horizontal ground, with O at the base. A and B are points on the ground with OA = OB = OT = 10 and \(\angle AOB = 90°\). Find \(\cos(\angle ATB)\).
A. \(\dfrac{1}{4}\)
B. \(\dfrac{1}{3}\)
C. \(\dfrac{1}{2}\) ✓
D. \(\dfrac{\sqrt{2}}{2}\)
Trap Analysis
  • A (1/4): Students who square incorrectly
  • B (1/3): Confusion with cube diagonal result
  • D (\(\sqrt{2}/2\)): Students who get \(\cos 45°\) instead
generated 101010µOTAB
Step-by-step Solution
1 \(O = (0,0,0)\), \(T = (0,0,10)\). \(A = (10,0,0)\), \(B = (0,10,0)\).
2 \(\vec{TA} = (10,0,-10)\), \(|TA| = 10\sqrt{2}\). \(\vec{TB} = (0,10,-10)\), \(|TB| = 10\sqrt{2}\).
3 \(\vec{TA} \cdot \vec{TB} = 0 + 0 + 100 = 100\).
4 \(\cos(\angle ATB) = \dfrac{100}{10\sqrt{2} \cdot 10\sqrt{2}} = \dfrac{100}{200} = \dfrac{1}{2}\).
Verification
\(\cos = 1/2\), angle \(= 60°\). OA = OB = OT creates symmetric configuration. Clean answer.
Q7 — Cuboid Diagonal cuboid
ABCDEFGH is a cuboid where ABCD is the bottom face and EFGH is the top face. AB = 4, BC = 4, CG = 7. Find \(\sin\) of the angle between the diagonal BH and the base plane ABCD.
A. \(\dfrac{7}{9}\) ✓
B. \(\dfrac{4\sqrt{2}}{9}\)
C. \(\dfrac{7}{4\sqrt{2}}\)
D. \(\dfrac{4}{9}\)
Trap Analysis
  • B (\(4\sqrt{2}/9\)): \(\cos\theta\) — wrong trig function
  • C (\(7/4\sqrt{2}\)): \(\tan\theta\) — wrong trig function
  • D (4/9): Confused base measurement
generated 447ABCDEFGH
Step-by-step Solution
1 \(B = (4,0,0)\), \(H = (0,4,7)\). \(\vec{BH} = (-4,4,7)\). \(|BH| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9\).
2 Projection on base = \((-4,4,0)\). \(|\text{proj}| = \sqrt{32} = 4\sqrt{2}\).
3 \(\sin\theta = \text{vertical}/|BH| = 7/9\).
Verification
\(4^2 + 4^2 + 7^2 = 81 = 9^2\). \(\sin = 7/9\), \(\cos = 4\sqrt{2}/9\), \(\tan = 7/(4\sqrt{2})\). Consistent.