4PM1/02 Jun 2019, Question 11: Trigonometry in 3D: Pyramids and Cones

Part (a) · 4 marks

Show that the pyramid's height is 8√2 cm, using the square base of side 16 cm and the 90° angle at V.

1. 1AC = √(16² + 16²) = 16√2, so AP = 8√2 (P the centre of the base)
2. 2Triangle AVC is right-angled and isosceles, so angle VAP = 45°
3. 3VP = AP tan 45° = 8√2 cm

Answerheight = 8√2 cm shown

Part (b) · 3 marks

Calculate the length of the edge VA in cm.

1. 1VA² = (8√2)² + (8√2)² = 256
2. 2VA = 16 cm

AnswerVA = 16 cm

Part (c) · 3 marks

Find the exact length in cm of the perpendicular dropped from D onto VA.

1. 1Triangle ADV is equilateral with side 16
2. 2Let X be the foot of the perpendicular from D: X is the midpoint of VA
3. 3DX² = 16² − 8² = 192, DX = 8√3 cm

Answer8√3 cm

Part (d) · 3 marks

Calculate, to one decimal place in degrees, the angle between plane VAB and the base ABCD.

1. 1Let M be the midpoint of AB: PM = 8, VP = 8√2
2. 2tan θ = 8√2/8 = √2
3. 3θ = 54.7°

Answer54.7°

Part (e) · 3 marks

Calculate, to one decimal place in degrees, the obtuse angle between planes VAB and VAD.

1. 1The perpendiculars from B and from D to VA both meet VA at X, each 8√3
2. 2BD = 16√2
3. 3cos φ = ((8√3)² + (8√3)² − (16√2)²)/(2 × 8√3 × 8√3) = −1/3
4. 4φ = 109.5°

Answer109.5°

More past-paper questions on trigonometry in 3d: pyramids and cones

• 4PM1/01 Jun 2025 Q8 · 13 marks
• 4PM1/02R Jun 2025 Q6 · 6 marks
• 4PM1/02R Jun 2024 Q10 · 13 marks
• 4PM1/01R Jan 2023 Q9 · 13 marks
• 4PM1/02 Jan 2023 Q5 · 7 marks
• 4PM1/02 Jun 2023 Q9 · 6 marks