📐 Trigonometry in 3D
IGCSE Year 8 — Intermediate Level · SOH CAH TOA in Space
Instructions: Answer all questions. Show your working clearly. Use a calculator where needed — give answers to 3 significant figures unless stated otherwise. Leave units in every final answer.
📖 Key Concepts & Formulas
- Sine: sin θ = opposite ÷ hypotenuse
- Cosine: cos θ = adjacent ÷ hypotenuse
- Tangent: tan θ = opposite ÷ adjacent (SOH CAH TOA)
- Angle of elevation: angle measured from the horizontal, looking up.
- Angle of depression: angle measured from the horizontal, looking down (equal to the alternate angle at the base).
- Cuboid space diagonal: D = √(l² + w² + h²) — use it when a length is missing before finding angles.
- Hidden triangles in cones & pyramids: the axis (perpendicular height), radius or half-base, and slant height form right-angled triangles through the centre.
SECTION A
Cuboids — Angles of Space Diagonals
[14 marks]
💡 Tip: In a cuboid, the space diagonal and two edges form right-angled triangles hidden inside the shape. To find an angle, first identify which triangle contains it — then use tan θ = opposite ÷ adjacent.
Q1 [2 marks]
A cuboid has dimensions 6 cm × 8 cm × 10 cm. Find the angle that the space diagonal makes with the base.
Q2 [2 marks]
A room is 4 m long, 3 m wide, and 2.5 m high. A spider is in one corner on the floor and a fly is in the opposite corner on the ceiling. Find the angle of elevation from the spider to the fly.
Q3 [2 marks]
A rectangular box has a base of 9 cm × 12 cm. The space diagonal is 17 cm. Find the angle that the space diagonal makes with the base.
Q4 [2 marks]
A cuboid has dimensions 7 cm × 24 cm × h cm. The space diagonal is 26 cm. Find the angle that the space diagonal makes with the base.
Q5 [2 marks]
A cuboid has dimensions 15 cm × 20 cm × 8 cm. Find the angle that the space diagonal makes with the base, giving your answer to 3 significant figures.
Q6 [2 marks]
A storage container is 3 m long, 2 m wide, and 4 m high. The longest straight line that fits inside the container is its space diagonal. Find the angle this line makes with the base.
Q7 [2 marks]
The base of a cuboid measures 10 cm by 24 cm. If the space diagonal is 29 cm, find the angle that the space diagonal makes with a vertical edge.
SECTION B
Cones & Pyramids — Finding Angles
[26 marks]
💡 Key idea: Every cone and pyramid contains hidden right-angled triangles running through the centre (axis, radius or half-base, slant height). Identify the triangle that contains your unknown angle, then use tan θ = opposite ÷ adjacent.
Q8 [3 marks]
A cone has radius 6 cm and slant height 10 cm. Find the angle that the slant height makes with the perpendicular height.
Q9 [3 marks]
A cone has radius 9 cm and perpendicular height 12 cm. Find the angle of elevation from a point on the edge of the base to the apex.
Q10 [3 marks]
A square-based pyramid has base sides of 8 cm and perpendicular height of 6 cm. Find the angle that a slant edge makes with the base.
Q11 [3 marks]
A square-based pyramid has base sides of 10 cm and slant height (of each triangular face) of 13 cm. Find the angle that a triangular face makes with the base.
Q12 [2 marks]
A cone has volume 377 cm3 and radius 5 cm. Find the angle that the slant height makes with the base.
Q13 [3 marks]
A cone has diameter 16 cm and slant height 10 cm. Find the angle that the slant height makes with the perpendicular height.
Q14 [3 marks]
A square-based pyramid has base sides of 12 cm and perpendicular height of 8 cm. Find the angle that a slant edge makes with the base.
Q15 [3 marks]
A cone has radius 7 cm and slant height 25 cm. Find the angle that the slant height makes with the base.
Q16 [3 marks]
A square-based pyramid has base sides of 24 cm and slant height of 13 cm. Find the angle that a triangular face makes with the base.
SECTION C
Mixed Challenge Problems
[25 marks]Q17 [3 marks]
A tent is in the shape of a triangular prism. The triangular face has base 4 m and height 1.5 m. The length of the tent is 6 m.
(a) Find the angle that each sloping side of the roof makes with the horizontal ground.
(b) Hence find the length of each sloping side.
(a) Find the angle that each sloping side of the roof makes with the horizontal ground.
(b) Hence find the length of each sloping side.
Q18 [3 marks]
A cone has radius 15 cm and perpendicular height 20 cm.
(a) Find the angle that the slant height makes with the base.
(b) Hence find the slant height of the cone.
(a) Find the angle that the slant height makes with the base.
(b) Hence find the slant height of the cone.
Q19 [4 marks]
A square-based pyramid has base sides of 16 cm and perpendicular height of 15 cm.
(a) Find the angle that a slant edge makes with the base.
(b) Hence find the length of each slant edge.
(a) Find the angle that a slant edge makes with the base.
(b) Hence find the length of each slant edge.
Q20 [3 marks]
A cone has radius 8 cm and perpendicular height 15 cm.
(a) Find the angle that the slant height makes with the perpendicular height.
(b) Hence find its total surface area, giving your answer to 3 significant figures.
(a) Find the angle that the slant height makes with the perpendicular height.
(b) Hence find its total surface area, giving your answer to 3 significant figures.
Q21 [4 marks]
A square-based pyramid has base sides of 10 cm. The perpendicular height is 12 cm.
(a) Find the angle that a triangular face makes with the base.
(b) Hence find the slant height of each triangular face.
(c) Find the total surface area.
(a) Find the angle that a triangular face makes with the base.
(b) Hence find the slant height of each triangular face.
(c) Find the total surface area.
Q22 [4 marks]
A cone has radius 10 cm and slant height 26 cm.
(a) Find the perpendicular height of the cone.
(b) Hence find the angle that the slant height makes with the base.
(c) Find its volume, giving your answer in terms of π.
(a) Find the perpendicular height of the cone.
(b) Hence find the angle that the slant height makes with the base.
(c) Find its volume, giving your answer in terms of π.
Q23 [4 marks]
A square-based pyramid has base sides of 18 cm and slant height of 15 cm.
(a) Find the perpendicular height of the pyramid.
(b) Hence find the angle that a triangular face makes with the base.
(c) Find its volume.
(a) Find the perpendicular height of the pyramid.
(b) Hence find the angle that a triangular face makes with the base.
(c) Find its volume.
ANSWERS
Answer Key
Q1: Base diagonal d = √(6² + 8²) = 10 cm. tan θ = h ÷ d = 10 ÷ 10 → θ = 45°
Q2: Base diagonal d = √(4² + 3²) = 5 m. tan θ = 2.5 ÷ 5 = 0.5 → θ ≈ 26.6°
Q3: Base diagonal d = √(9² + 12²) = 15 cm. tan θ = h ÷ d, where h = √(17² − 15²) = 8 → tan θ = 8/15 → θ ≈ 28.1°
Q4: Base diagonal d = √(7² + 24²) = 25 cm. h = √(26² − 25²) = √51 ≈ 7.14 cm. tan θ = 7.14 ÷ 25 → θ ≈ 16.0°
Q5: Base diagonal d = √(15² + 20²) = 25 cm. tan θ = 8 ÷ 25 → θ ≈ 17.7°
Q6: Base diagonal d = √(3² + 2²) = √13 m. tan θ = 4 ÷ √13 → θ ≈ 54.4°
Q7: h = √(29² − 26²) = √165 ≈ 12.8 cm. Angle with vertical: cos θ = h ÷ D = 12.8 ÷ 29 → θ ≈ 63.4°
Q8: sin α = r ÷ l = 6 ÷ 10 → α ≈ 36.9°
Q9: tan θ = h ÷ r = 12 ÷ 9 → θ ≈ 53.1°
Q10: Half diagonal = √(4² + 4²) = 4√2 ≈ 5.66 cm. tan θ = h ÷ half-diagonal = 6 ÷ 5.66 → θ ≈ 46.7°
Q11: cos θ = (s/2) ÷ l = 5 ÷ 13 → θ ≈ 67.4°
Q12: V = (1/3)πr²h → h = 377 ÷ ((1/3)π × 5²) ≈ 14.4 cm. tan θ = h ÷ r = 14.4 ÷ 5 → θ ≈ 70.9°
Q13: r = 8 cm. sin α = r ÷ l = 8 ÷ 10 → α ≈ 53.1°
Q14: Half diagonal = √(6² + 6²) = 6√2 ≈ 8.49 cm. tan θ = 8 ÷ 8.49 → θ ≈ 43.4°
Q15: cos θ = r ÷ l = 7 ÷ 25 → θ ≈ 73.7°
Q16: cos θ = (s/2) ÷ l = 12 ÷ 13 → θ ≈ 22.6°
Q17(a): tan θ = h ÷ (base/2) = 1.5 ÷ 2 → θ ≈ 36.9°
(b) s = √(2² + 1.5²) = 2.5 m
(b) s = √(2² + 1.5²) = 2.5 m
Q18(a): tan θ = h ÷ r = 20 ÷ 15 → θ ≈ 53.1°
(b) l = √(15² + 20²) = 25 cm
(b) l = √(15² + 20²) = 25 cm
Q19(a): Half diagonal = √(8² + 8²) = 8√2 ≈ 11.31 cm. tan θ = 15 ÷ 11.31 → θ ≈ 53.1°
(b) Slant edge = √(11.31² + 15²) = √360 ≈ 18.97 cm
(b) Slant edge = √(11.31² + 15²) = √360 ≈ 18.97 cm
Q20(a): sin α = r ÷ l, where l = √(8² + 15²) = 17 → sin α = 8/17 → α ≈ 28.1°
(b) SA = π × 8² + π × 8 × 17 = 64π + 136π = 200π ≈ 628 cm²
(b) SA = π × 8² + π × 8 × 17 = 64π + 136π = 200π ≈ 628 cm²
Q21(a): cos θ = (s/2) ÷ l, where l = √(5² + 12²) = 13 → cos θ = 5/13 → θ ≈ 67.4°
(b) Slant height = 13 cm
(c) SA = 10² + 4 × (½ × 10 × 13) = 100 + 260 = 360 cm²
(b) Slant height = 13 cm
(c) SA = 10² + 4 × (½ × 10 × 13) = 100 + 260 = 360 cm²
Q22(a): h = √(26² − 10²) = √576 = 24 cm
(b) cos θ = r ÷ l = 10 ÷ 26 → θ ≈ 67.4°
(c) V = (1/3)π × 10² × 24 = 800π cm³
(b) cos θ = r ÷ l = 10 ÷ 26 → θ ≈ 67.4°
(c) V = (1/3)π × 10² × 24 = 800π cm³
Q23(a): h = √(15² − 9²) = √144 = 12 cm
(b) cos θ = (s/2) ÷ l = 9 ÷ 15 → θ ≈ 53.1°
(c) V = (1/3) × 18² × 12 = 1296 cm³
(b) cos θ = (s/2) ÷ l = 9 ÷ 15 → θ ≈ 53.1°
(c) V = (1/3) × 18² × 12 = 1296 cm³