4PM1/02R Jun 2024, Question 10: Trigonometry in 3D: Pyramids and Cones

Part (a) · 3 marks

Show that the sloping edge VA equals (√10/2)x cm using the 45° angle with the base.

1. 1AC = √((2x)² + x²) = √5 x
2. 2Triangle VAC has base angles 45°, so it is right-angled at V
3. 3VA = AC sin 45° = √5 x × (√2/2) = (√10/2)x

AnswerVA = (√10/2)x cm shown

Part (b) · 2 marks

Find the exact perpendicular height of the pyramid in cm in terms of x.

1. 1Height² = VA² − (AC/2)²
2. 2= (10/4)x² − (5/4)x² = (5/4)x²
3. 3Height = (√5/2)x

Answer(√5/2)x cm

Part (c) · 2 marks

Calculate the size of angle VBA in degrees to one decimal place.

1. 1Triangle VAB is isosceles with VA = VB, so the foot of the perpendicular from V is the midpoint of AB
2. 2cos VBA = x ÷ (√10/2)x = 2/√10
3. 3∠VBA = 50.8°

Answer50.8°

Part (d) · 4 marks

Find the obtuse angle between planes AVC and BVD, in degrees to one decimal place.

1. 1Let M be where the diagonals meet; the planes meet along VM, which is perpendicular to the base
2. 2So the angle between the planes is angle AMB
3. 3AM = MB = AC/2 = (√5/2)x and AB = 2x
4. 4cos AMB = (5/4 + 5/4 − 4)/(2 × 5/4) = −3/5
5. 5∠AMB = 126.9°

Answer126.9°

Part (e) · 2 marks

Given volume 9√5 cm³, find the value of x.

1. 1V = (1/3) × 2x × x × (√5/2)x = (√5/3)x³
2. 2(√5/3)x³ = 9√5, so x³ = 27
3. 3x = 3

Answerx = 3

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