O-LevelAdditional MathematicsTrigonometry4037/11 · May/June 2023 · Q58 Marks

5 DO NOT USE A CALCULATOR IN THIS QUESTION. In this question, all lengths are in centimetres.

(a)You are given that cos 120° = -½, sin 120° = √3/2 and tan 120° = -√3. In the triangle ABC, AB = 5√3 - 6, BC = 5√3+6 and angle ABC = 120°. Find AC, giving your answer in the form a√b where a and b are integers greater than 1.[4]
(b)You are given that cos30° = √3/2, sin 30° = ½ and tan 30° = 1/√3. In the triangle PQR, PQ = 3+2√5 and angle PQR = 30°. Given that the area of this triangle is (2+5√5)/4, find QR, giving your answer in the form c+d√5, where c and d are integers.[4]

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

[4] M1 For the correct use of the cosine rule
Condone missing brackets if intention is clear
75 + 36 - 60√3 + 75 + 36 + 60√3 + 75 - 36
M1 dep must see sufficient detail to be sure that a calculator is not being used. This is the minimum acceptable 75 + 36 - 60√3 + 75 + 36 + 60√3 + 39
A1 Maybe implied by √261
A1 3√29
[4] M1 For the correct use of the area of the triangle. Condone missing brackets if intention is clear
1/2 (3 + 2√5) × QR × sin 30° = (2 + 5√5)/4
M1 dep for a correct attempt to rationalise their QR. must be the same two terms in the numerator and denominator to rationalise
M1 dep, must see sufficient detail to be sure that a calculator is not being used. This is the minimum acceptable 6+15√5-4√5-50 / -11
A1 4-√5

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