72 a Draw a triangle ABC. 2 H B1 B1 for drawing diagram correctly labelled Using trigonometric functions: sin = C Therefore: sin = h b C Then using the basic formula for area of a triangle: a h Area = × B1 B1 for trigonometric expression linking C, and Substituting for gives: h b h a b Area = × sin C B1 B1 for = b sin C h ab Area = sin as required. C Using the given triangle: B1 B1 formula for area of triangle a x b x = 45°, = + 2 = 6, = 2 = 2 C – M1 M1 substitution of h ab Area = sin C =× 6 × 2 × sin 45° A1 A1 cao = 6 × M1 M1 for complete correct proof with correct mathematical = notation throughout Multiply numerator and denominator by . b This gives = = 3as required. M1 M1 for expressing and a, b C M1 M1 for substituting for and a, b C A1 A1 cao M1 for dealing with the in denominator M1 A1 for full explanation showing given result A1 12
72 a b Draw a triangle ABC. Using trigonometric functions: sin C = Therefore: h = b sin C Then using the basic formula for area of a triangle: Area = a × h Substituting for h gives: Area = a × b sin C Area = ab sin C as required. Using the given triangle: C = 45°, a = x + 2 = 6, b = x – 2 = 2 Area = ab sin C =× 6 × 2 × sin 45° = 6 × = Multiply numerator and denominator by . This gives = = 3as required. B1 B1 B1 B1 M1 A1 M1 M1 M1 A1 M1 A1 2 B1 for drawing diagram correctly labelled B1 for trigonometric expression linking C, h and b B1 for h = b sin C B1 formula for area of triangle M1 substitution of h A1 cao M1 for complete correct proof with correct mathematical notation throughout M1 for expressing a, b and C M1 for substituting for a, b and C A1 cao M1 for dealing with the in denominator A1 for full explanation showing given result H 12
Higher_Skills_Algebra_MarkScheme_AQA.docx