IB DP Mathematics: Analysis and Approaches – 3D geometry, triangle trigonometry and radian measure Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the 3D geometry, triangle trigonometry and radian measure unit of IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 3.1–3.4. This is SL content, so every question suits both SL and HL. It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 sessions.

Paper 1 allows no technology; Paper 2 requires it (and HL Paper 3). Questions are labelled “(calculator-free)” or “(calculator allowed)” to match. Give calculator-allowed answers to 3 significant figures unless stated. Angles are in radians unless a question says degrees.

Learn the methods first in the study guide and the revision notes. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator-free) The points A(2, −1, 5) and B(6, 3, −3) are given. Find the midpoint of [AB] and the exact length AB. [3]

2. (calculator-free)

(a) Write 150° in radians as an exact multiple of π. [1] (b) Write 7π/12 radians in degrees. [1]

3. (calculator-free) A sector of a circle has radius 6 cm and angle 5π/9 at the centre.

(a) Find the exact arc length. [2] (b) Find the exact area of the sector. [2]

4. (calculator-free) In triangle PQR, PQ = 5 cm, PR = 8 cm and QR = 7 cm.

(a) Show that cos QPR = ½. [3] (b) Given that sin QPR = √3/2, find the exact area of the triangle. [2]

5. (calculator allowed) In triangle ABC, angle BAC = 42°, angle ABC = 67° and BC = 12.5 cm.

(a) Find AC. [2] (b) Find the area of the triangle. [3]

6. (calculator allowed) From a point A on level ground, the angle of elevation of the top T of a vertical tower is 28°. A surveyor walks 40 m directly towards the tower to a point B, where the angle of elevation of T is 47°. Find the height of the tower. [6]

7. (calculator allowed) A ship sails 18 km from port P on a bearing of 065° to a point Q. It then sails 11 km on a bearing of 150° to a point R.

(a) Show that angle PQR = 95°. [2] (b) Find the distance PR. [2] (c) Find the bearing of R from P. [3]

8. (calculator-free) A solid consists of a right cone of radius 3 cm and height 4 cm fixed to a hemisphere of radius 3 cm, so that the base of the cone matches the flat face of the hemisphere.

(a) Find the exact volume of the solid. [3] (b) Find the exact surface area of the solid. [3]

9. (calculator allowed) VABCD is a right pyramid with a square base ABCD of side 10 cm. The apex V is 12 cm vertically above M, the centre of the base. N is the midpoint of [BC]. Give angles in degrees.

(a) Find the volume of the pyramid. [2] (b) Find the angle between the edge [VA] and the base. [3] (c) Find the angle between [VN] and the base. [2] (d) Find the total surface area of the pyramid. [3]

10. (calculator allowed) A circle has centre O and radius 9 cm. Points A and B lie on the circle with angle AOB = 1.8 radians.

(a) Find the length of the minor arc AB. [1] (b) Find the area of triangle OAB. [2] (c) Find the area of the minor segment cut off by the chord [AB]. [2] (d) Find the perimeter of this minor segment. [3]

11. (calculator-free) A sector of radius 15 cm and angle θ is rolled up so that its two straight edges meet exactly, forming the curved surface of a right cone with base radius 6 cm.

(a) Show that θ = 4π/5. [3] (b) Find the exact height of the cone. [2] (c) Find the exact volume of the cone. [2]

12. (calculator allowed) A and B are points on level ground, with B 150 m due east of A. T is the foot of a vertical mast. The bearing of T from A is 040° and the bearing of T from B is 330°. Give angles in degrees.

(a) Show that angle ATB = 70°. [2] (b) Find the distance AT. [2] (c) The angle of elevation of the top of the mast from A is 12°. Find the height of the mast. [2] (d) Find the angle of elevation of the top of the mast from B. [2]

Answers

1. Midpoint = ((2 + 6)/2, (−1 + 3)/2, (5 − 3)/2) = (4, 1, 1) [1]. AB = √(4² + 4² + (−8)²) [1] = √96 = 4√6 [1] Examiner insight: “Exact” means √96 or 4√6 earns the accuracy mark; 9.80 does not on a calculator-free question.

2. (a) 150 × π/180 = 5π/6 [1] (b) 7π/12 × 180/π = 105° [1] Examiner insight: A one-mark conversion has no method mark to fall back on, so an unsimplified 150π/180 may not be accepted as the final answer.

3. (a) l = rθ = 6 × 5π/9 [1] = 10π/3 cm [1] (b) A = ½r²θ = ½ × 36 × 5π/9 [1] = 10π cm² [1] Examiner insight: Converting 5π/9 to 100° and putting it into l = rθ gives 600, which scores no marks: the formula needs radians.

4. (a) cos QPR = (PQ² + PR² − QR²)/(2 × PQ × PR) [1] = (25 + 64 − 49)/(2 × 5 × 8) [1] = 40/80 = ½ [1] (b) Area = ½ × 5 × 8 × √3/2 [1] = 10√3 cm² [1] Examiner insight: In a “show that”, starting from cos P = ½ and working backwards earns nothing; the substitution and the 40/80 line must both appear.

5. (a) AC/sin 67° = 12.5/sin 42° [1], so AC = 17.195… = 17.2 cm [1] (b) Angle ACB = 180° − 42° − 67° = 71° [1]. Area = ½ × 12.5 × 17.195… × sin 71° [1] = 102 cm² [1] Examiner insight: The area must use the angle between the two sides you multiply; ½ × 12.5 × 17.2 × sin 42° uses the wrong angle and loses both remaining marks.

6. In triangle ABT, angle ABT = 180° − 47° = 133° [1]. So angle ATB = 180° − 28° − 133° = 19° [1]. Sine rule: BT/sin 28° = 40/sin 19° [1], so BT = 57.68… m [1]. Height = BT sin 47° [1] = 42.2 m [1] Examiner insight: The angle at T (19°) is the step most often missed; a labelled sketch showing the exterior angle at B makes this method mark visible to the examiner.

7. (a) The bearing of P from Q is 065° + 180° = 245° [1]. Angle PQR = 245° − 150° = 95° [1] (b) PR² = 18² + 11² − 2(18)(11)cos 95° [1], so PR = 21.9 km [1] (c) sin QPR = 11 sin 95°/21.897… [1], so angle QPR = 30.0° [1]. Bearing of R from P = 065° + 30.0° = 095.0° [1] Examiner insight: The final mark is for the bearing, not the angle QPR; stopping at 30.0° or writing 95° without three figures can cost it.

8. (a) Cone: (1/3)π(3²)(4) = 12π [1]. Hemisphere: (2/3)π(3³) = 18π [1]. Total = 30π cm³ [1] (b) Slant height l = √(3² + 4²) = 5 [1]. Surface = π(3)(5) + 2π(3²) = 15π + 18π [1] = 33π cm² [1] Examiner insight: Adding the flat circle πr² (giving 42π) counts a face that is inside the solid, and the accuracy mark is lost.

9. (a) V = (1/3)(10²)(12) [1] = 400 cm³ [1] (b) MA = half the diagonal = 5√2 = 7.07… cm [1]. tan θ = 12/(5√2) [1], so θ = 59.5° [1] (c) MN = 5 cm, so tan φ = 12/5 [1] and φ = 67.4° [1] (d) VN = √(12² + 5²) = 13 cm [1]. Each triangular face = ½ × 10 × 13 = 65 cm² [1]. Total = 100 + 4 × 65 = 360 cm² [1] Examiner insight: The height of each triangular face is VN (13), not the vertical height VM (12); using 12 gives 340 and loses the last two marks.

10. (a) l = 9 × 1.8 = 16.2 cm [1] (b) Area = ½ × 9 × 9 × sin 1.8 [1] = 39.4 cm² [1] (c) Sector = ½ × 81 × 1.8 = 72.9 cm² [1]. Segment = 72.9 − 39.44… = 33.5 cm² [1] (d) AB² = 9² + 9² − 2(9)(9)cos 1.8 [1], so AB = 14.09… cm [1]. Perimeter = 16.2 + 14.09… = 30.3 cm [1] Examiner insight: sin 1.8 in degree mode gives 0.0314 and a triangle area of 1.27; a follow-through mark may survive, but every later accuracy mark goes.

11. (a) The arc of the sector becomes the base circumference, 2π × 6 = 12π [1]. The arc length is 15θ [1]. So 15θ = 12π and θ = 12π/15 = 4π/5 [1] (b) The slant height is 15, so h² = 15² − 6² = 189 [1] and h = 3√21 cm [1] (c) V = (1/3)π(6²)(3√21) [1] = 36√21π cm³ [1] Examiner insight: The radius of the sector becomes the slant height, not the vertical height; the method mark in (b) depends on this link being stated.

12. (a) Angle TAB = 90° − 40° = 50° [1]. Angle TBA = 330° − 270° = 60°, so angle ATB = 180° − 50° − 60° = 70° [1] (b) AT/sin 60° = 150/sin 70° [1], so AT = 138.24… = 138 m [1] (c) Height = AT tan 12° [1] = 29.4 m [1] (d) BT = 150 sin 50°/sin 70° = 122.28… m [1]. Angle = tan⁻¹(29.38…/122.28…) = 13.5° [1] Examiner insight: Using the rounded 29.4 and 122 still gives 13.5° here, but carrying stored values is the only safe habit; early rounding often shifts the third figure.

Where marks are usually lost

• Using degrees in l = rθ or ½r²θ, as in questions 3 and 10.
• GDC in degree mode for sin 1.8 (question 10), or in radian mode for the degree questions.
• Counting a hidden face in the surface area of a combined solid (question 8).
• Using the vertical height instead of the slant height for a pyramid face or a cone’s curved surface.
• Missing the interior angle at T in elevation problems with two observation points (question 6).
• Bearings given as an angle inside the triangle, or without three figures (questions 7 and 12).
• “Show that” answers that start from the given result instead of reaching it.
• Decimal answers on calculator-free questions that ask for exact values.
• Rounding intermediate lengths before using them again.

Next steps

• Recap the formulas and methods in the revision notes.
• Re-learn any weak section in the study guide.
• See the rest of the course on the IB DP Maths AA course hub and tick off the printable syllabus checklist.
• Plan your final weeks with the AA exam preparation guide.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020).