IB DP Mathematics: Applications and Interpretation – 3D geometry, triangle trigonometry, bearings and sectors 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 3D geometry, triangle trigonometry, bearings and sectors in IB Diploma Programme Mathematics: Applications and Interpretation. It is aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections 3.1 to 3.4, which are common content for 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.

The guide lists technology as required on every AI paper, so every question is labelled “(calculator allowed)”. Work in degrees. Give answers to 3 significant figures, and angles to 1 decimal place, unless the question says otherwise. Following the guide, 3D questions use right-angled trigonometry only, and the ambiguous case of the sine rule is not used.

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

Questions

1. (calculator allowed) Points P(−3, 4, 2) and Q(5, −2, 8) are given. Find the distance PQ and the midpoint of PQ. [3]

2. (calculator allowed) A right cone has base radius 5 cm and vertical height 12 cm.

(a) Find the volume of the cone. [2] (b) Find the curved surface area of the cone. [3]

3. (calculator allowed) In triangle ABC, angle A = 48°, angle C = 71° and b = 15 cm. Find a. [3]

4. (calculator allowed) A triangular plot of land has sides 6 m, 10 m and 13 m.

(a) Find the size of the largest angle. [3] (b) Find the area of the plot. [2]

5. (calculator allowed) A fan, when open, is a sector of a circle of radius 12 cm. The arc along its outer edge is 30 cm long.

(a) Find the angle of the sector. [2] (b) Find the area of the sector. [2] (c) Find the perimeter of the sector. [1]

6. (calculator allowed) A cliff is 85 m high. From the top, the angles of depression of two boats are 31° and 18°. The boats and the foot of the cliff lie in a straight line, with both boats on the same side of the cliff. Find the distance between the boats, to the nearest metre. [4]

7. (calculator allowed) A drone flies from A on a bearing of 065° for 18 km to B. It then flies on a bearing of 170° for 11 km to C.

(a) Show that angle ABC = 75°. [2] (b) Find the distance AC. [3] (c) Find the bearing of C from A, to the nearest degree. [3]

8. (calculator allowed) ABCD is a horizontal rectangular field with AB = 40 m and BC = 30 m. A vertical flagpole DT, 18 m tall, stands at corner D.

(a) Find BD. [1] (b) Find the angle of elevation of T from B. [2] (c) Find the length TA. [2] (d) Angle TAB = 90°. Find angle TBA. [2]

9. (calculator allowed) A monument is a cuboid 6 m by 6 m by 2 m high, with a right pyramid on top. The pyramid has the same 6 m by 6 m square base as the top of the cuboid, and its apex is 4 m above that base.

(a) Find the total volume of the monument. [3] (b) Find the slant height of one triangular face of the pyramid. [2] (c) The four vertical faces and the four triangular faces are painted. The bottom is not. Find the area painted. [3] (d) Paint costs 14.50 per m². Find the cost of painting the monument. [1] (e) Find the angle between a sloping edge of the pyramid and its base. [3]

10. (calculator allowed) A solid metal sphere has radius 6 cm.

(a) Find its volume. [2] (b) The sphere is melted and recast into solid right cones, each of radius 3 cm and height 4 cm, with no metal wasted. Find the number of cones made. [2] (c) Find the surface area of the original sphere. [1]

11. (calculator allowed) A flower bed OAB is a sector of a circle with centre O, radius 8 m and angle AOB = 115°. A straight path runs along the chord AB. The region between the chord and the arc is planted with roses.

(a) Find the length of the path AB. [3] (b) Find the area planted with roses. [4] (c) Find the perimeter of the rose region. [3]

Answers

1. PQ = √(8² + (−6)² + 6²) [1] = √136 = 11.7 [1]. Midpoint = (1, 1, 5) [1] [3] Examiner insight: A sign slip inside the square root is hidden because squaring removes it, but a slip in the midpoint is not; check each coordinate separately.

2. (a) V = (1/3)π(5²)(12) [1] = 100π = 314 cm³ [1] (b) l = √(5² + 12²) = 13 [1]; curved surface = π(5)(13) [1] = 65π = 204 cm² [1] Examiner insight: Using h = 12 in πrl instead of l = 13 loses the method mark and the answer; the slant height has to be shown.

3. B = 180° − 48° − 71° = 61° [1]; a/sin 48° = 15/sin 61° [1]; a = 15 sin 48°/sin 61° = 12.7 cm [1] [3] Examiner insight: b is opposite B, so you must find B before the sine rule can pair b with an angle; pairing b with C earns no method mark.

4. (a) The largest angle is opposite 13 m: cos θ = (6² + 10² − 13²)/(2 × 6 × 10) [1] = −33/120 = −0.275 [1]; θ = 106.0° [1] (b) Area = (1/2)(6)(10) sin 105.96…° [1] = 28.8 m² [1] Examiner insight: In (b) a wrong angle from (a) still earns follow-through marks if the method is right, so always show the area formula with your numbers in it.

5. (a) 30 = (θ/360) × 2π(12) [1]; θ = 143.2° [1] (b) Area = (143.239…°/360) × π(12²) [1] = 180 cm² [1] (c) Perimeter = 30 + 12 + 12 = 54 cm [1] Examiner insight: Carry the unrounded angle into (b); rounding θ to 143° first gives 179.7, which only happens to round to 180. Check (b) with area = (1/2) × r × arc = (1/2)(12)(30) = 180.

6. Nearer boat: 85/tan 31° [1] = 141.46… m [1]. Further boat: 85/tan 18° = 261.60… m [1]. Distance = 261.60… − 141.46… = 120 m [1] [4] Examiner insight: The angle of depression equals the angle of elevation from the boat (alternate angles), so each horizontal distance is 85 ÷ tan of the angle; writing 85 × tan instead loses both the method and accuracy marks.

7. (a) Bearing of A from B = 065° + 180° = 245° [1]; angle ABC = 245° − 170° = 75° [1] (b) AC² = 18² + 11² − 2(18)(11) cos 75° [1] = 445 − 102.49… = 342.51… [1]; AC = 18.5 km [1] (c) sin(BAC)/11 = sin 75°/18.506… [1]; angle BAC = 35.0° [1]; bearing = 065° + 35.0° = 100° [1] Examiner insight: (a) is a “show that”, so both the back bearing and the subtraction must be written; stating 75° from a diagram earns nothing.

8. (a) BD = √(40² + 30²) = 50 m [1] (b) tan θ = 18/50 [1]; θ = 19.8° [1] (c) AD = 30 m, so TA = √(30² + 18²) [1] = √1224 = 35.0 m [1] (d) tan(TBA) = 34.985…/40 [1]; angle TBA = 41.2° [1] Examiner insight: In (b) the projection of TB on the field is BD, not BA or BC; naming the right-angled triangle TDB earns the method mark even if the arithmetic slips.

9. (a) Cuboid = 6 × 6 × 2 = 72 m³ [1]; pyramid = (1/3)(36)(4) = 48 m³ [1]; total = 120 m³ [1] (b) Slant height = √(4² + 3²) [1] = 5 m [1] (c) Vertical faces = 4 × 6 × 2 = 48 m² [1]; triangles = 4 × (1/2)(6)(5) = 60 m² [1]; total = 108 m² [1] (d) 108 × 14.50 = 1566 [1] (e) Half the base diagonal = (1/2)√(6² + 6²) = 3√2 = 4.243 m [1]; tan θ = 4/4.243… [1]; θ = 43.3° [1] Examiner insight: In (c), adding the top of the cuboid or the base of the pyramid counts a hidden face and loses the final accuracy mark; list the faces before adding.

10. (a) V = (4/3)π(6³) [1] = 288π = 905 cm³ [1] (b) One cone = (1/3)π(3²)(4) = 12π cm³ [1]; number = 288π/12π = 24 [1] (c) Surface area = 4π(6²) = 144π = 452 cm² [1] Examiner insight: Dividing rounded volumes (905 ÷ 37.7 = 24.005…) still gives 24, but keeping π exact shows the answer is a whole number and avoids a rounding argument.

11. (a) AB² = 8² + 8² − 2(8)(8) cos 115° [1] = 128 + 54.09… = 182.09… [1]; AB = 13.5 m [1] (b) Sector = (115/360)π(8²) = 64.23 m² [1]; triangle = (1/2)(8)(8) sin 115° = 29.00 m² [1]; roses = 64.23 − 29.00 [1] = 35.2 m² [1] (c) Arc = (115/360) × 2π(8) [1] = 16.06 m [1]; perimeter = 16.06 + 13.49 = 29.6 m [1] Examiner insight: The rose region’s perimeter is arc plus chord, not arc plus two radii; adding 16 m of radii loses the final mark even with correct working elsewhere.

Where marks are usually lost

• Writing an answer with no formula or substitution; a wrong GDC answer on its own earns no method marks.
• Mixing up vertical height and slant height in cone and pyramid questions.
• Counting a joined or hidden face in the surface area of a combined solid.
• Using a full diagonal instead of half a diagonal for the angle between a pyramid edge and its base.
• Missing the back bearing at the turning point, so the interior angle is wrong.
• Choosing the sine rule when two sides and the included angle are known.
• Rounding a length or angle mid-question, then losing the accuracy mark at the end.
• Leaving out the radii in a sector perimeter, or adding them to a segment perimeter.
• Calculator in radian mode.

Next steps

• Revision notes for a fast recap of formulas and methods
• Study guide for full explanations and worked examples
• IB DP Maths AI course hub
• Printable syllabus checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: applications and interpretation guide, first assessment 2021 – syllabus sections SL 3.1, SL 3.2, SL 3.3 and SL 3.4.