4PM1/01 Jun 2025, Question 8: Trigonometry in 3D: Pyramids and Cones

Part (a) · 2 marks

Show that the pyramid's height is 2√3 cm using the 30° angle between plane VAB and the base.

1. 1Let P be the centre of the base; the distance from P to AB is 6
2. 2tan 30° = VP/6
3. 3VP = 6 tan 30° = 6 × (√3/3) = 2√3

AnswerHeight = 2√3 cm shown

Part (b) · 3 marks

Find the exact length of VA in cm.

1. 1AC = √(12² + 12²) = 12√2, so AP = 6√2
2. 2VA² = (2√3)² + (6√2)² = 12 + 72 = 84
3. 3VA = √84 = 2√21

AnswerVA = 2√21 cm

Part (c) · 3 marks

Find the exact area of triangle VAD in cm².

1. 1Let X be the midpoint of AD
2. 2VX = √((2√3)² + 6²) = √48 = 4√3
3. 3Area of VAD = ½ × 12 × 4√3 = 24√3

Answer24√3 cm²

Part (d) · 2 marks

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

1. 1Area of VAD = ½ × VA × DY
2. 224√3 = ½ × 2√21 × DY
3. 3DY = 24√3/√21 = 24/√7 = 24√7/7

Answer24√7/7 cm

Part (e) · 3 marks

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

1. 1The perpendiculars from D and from B to VA meet VA at the same point Y, each of length 24√7/7
2. 2BD = 12√2
3. 3cos θ = (DY² + BY² − BD²)/(2 × DY × BY) = −3/4
4. 4θ = 138.6°

Answer138.6°

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