Mathematics · May / June 2026 · Paper 43 · Question 13

[sin BAC =] 15sin120 oe 15 3 angle BAC = 30 180 – 120 – 30 = 30 with one of: BCA = 30 or BAC = BCA or a comment that two angles are the same. Alternative method A correct cosine rule statement with involving AB that would lead to a quadratic in AB Correct rearrangement to a 3-term quadratic in AB AB = 15 AB = BC or [Triangle has] two equal sides [30]

M2 Accept use of A or C for the angles. For M2, must or M1 15 3 15 sin120 = sin BAC oe be angle BAC being calculated. 3 Condone 2 for sin120. Following an M1 start, M2 can be awarded for seeing a correct rearrangement before reaching sinA = ½, i.e. at least one intermediate step is required. Note: After an M1 start, sinA = ½ as the next step scores M1 and not M2. A1 Dep on M1 nfww Accept use of x for BAC A1 Dep on previous A1 Note: M1 A1 A1 is possible. M1 e.g. 2 = 2 + 2 −× × (15 3 ) 15 AB 2 15 AB cos1 2 0 Accept use of a variable for AB. 2 For both M1 allow recovery from 15 3 . M1 e.g. AB2 + 15AB – 450 = 0 Must see = 0. For M1 accept 450 written as a product, e.g. 2(225). A1 dep on second M1 awarded Second solution not required. A1 dep on A1 Note: M1 A1 A1 is not possible.

7.5 [3]

Mark at most accurate. Accept 15 2 . 2 15 3 For M2 condone 12.99… or 13[.0] for 15 3 15 2 − .

M2 for oe 2 2

M2 for e.g. 15sin 30 oe or for 15cos 60 oe or 1 1 × × × = × × 15 15 3 sin( their 30) 15 3 h 2 2

M1 for distance required is the perpendicular M1 may be on diagram, must see the right angle from B to AC soi symbol. M1 may be implied by correct implicit trig d = statement, e.g. cos60 or correct implicit 15 2 15 3 Pythagoras statement, d 2 + = 15 . 2 2