In this question all lengths are in centimetres. The diagram shows triangle ABC which has area 2√5/3 cm². Angle A is acute. [Figure 9.1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

1. [M1] 1/2 (√5-1)(√5+1)sin A = 2√5/3
2. [A1] Simplifies to 1/2 × (5-1) × sin A = 2√5/3 oe
3. [A1] sin A = √5/3 isw
4. [B2] cos A = √(1 - (√5/3)²) = 2/3 or exact equivalent (B1 for cos A = √(1 - (their sin A)²))
5. [M1] (√5-1)² + (√5+1)² - 2(√5-1)(√5+1)cos A soi
6. [A1] (√5-1)² + (√5+1)² - 2(√5-1)(√5+1) × (2/3) soi
7. [A1] x = √(20/3) or 2√(5/3) or 2√15/3 oe, isw
8. [M1] (their x)/(their sin A) = (√5+1)/sin B oe, soi (FT their x and their sinA)
9. [A1] Correct expression (√(20/3))/(√5/3) = (√5+1)/sin B oe
10. [A1] sin B = (√15+√3)/6 or (√5+1)/√(20/3) or (√5+1)/(2√5/√3) oe, isw