M1 for method of finding β A1 C1 A1 for to 4 or more sf β A1 for final three-figure bearing 20 C1 for effective use of diagrams and good use of mathematical language 80 Use the sine rule to work out angle B. M1 3 M1 for use of sine rule H sin = B = sin–1 = 62° B But we know ABC is obtuse, so: = 180° 62° = 118° B – A1 A1 for 62°, accept any rounding to 2sf or more Then = 180° (118° + 32°) = 30° C – Use the sine area rule. Area = A1 A1 for 118° and any rounding to 3sf or more = × 10 × 6 × sin 30° B1 B1 for 30° with any rounding to 2sf or more = 15 cm2 M1 M1 for use of area sine rule M1 M1 for correct substitution 15 cm2 A1 A1 for 15 rounded to 2 or 3 sf C1 C1 for a complete solution clearly set out with correct mathematical language and symbols 8 81 a 3 H Use theorem to work out C1 C1 for use of clear diagram H. Pythagoras’ = H = 26 Use the cosine rule to work out angle D. cos = = 0.275 D = 74.037 986° D Use the area sine rule to work out the area M1 M1 for use of theorem Pythagoras’ of triangle ADC. Area = × 20 × 23 × sin 74.037 986° A1 A1 for 26 cao = 221.13217 m2 Area of triangle ABC = × 24 × 10 = 120 m2 M1 M1 for use of cosine rule Total area = 341.132 17 m2
A1 C1 M1 for method of finding β A1 for β to 4 or more sf A1 for final three-figure bearing C1 for effective use of diagrams and good use of mathematical language 20 80 Use the sine rule to work out angle B. sin B = B = sin–1 = 62° But we know ABC is obtuse, so: B = 180° – 62° = 118° Then C = 180° – (118° + 32°) = 30° Use the sine area rule. Area = = × 10 × 6 × sin 30° = 15 cm2 15 cm2 M1 A1 A1 B1 M1 M1 A1 C1 3 M1 for use of sine rule A1 for 62°, accept any rounding to 2sf or more A1 for 118° and any rounding to 3sf or more B1 for 30° with any rounding to 2sf or more M1 for use of area sine rule M1 for correct substitution A1 for 15 rounded to 2 or 3 sf C1 for a complete solution clearly set out with correct mathematical language and symbols H 8 81 a Use Pythagoras’ theorem to work out H. H = = 26 Use the cosine rule to work out angle D. cos D = = 0.275 D = 74.037 986° Use the area sine rule to work out the area of triangle ADC. Area = × 20 × 23 × sin 74.037 986° = 221.13217 m2 Area of triangle ABC = × 24 × 10 = 120 m2 Total area = 341.132 17 m2 C1 M1 A1 M1 3 C1 for use of clear diagram M1 for use of Pythagoras’ theorem A1 for 26 cao M1 for use of cosine rule H
Higher_Skills_Geometry_MarkScheme_Edexcel.docx