37 Use theorem. M1 2 M1 for correct Pythagoras statement M Pythagoras’ AC2 = 42 (2)2 A1 A1 for correct value of AC2 – = 16 8 = 8 – BC2 = 8 = (2)2 = 8 M1 M1 for finding BC2 Hence BC = AC, an isosceles triangle. A1 A1 for correct value of BC2 C1 C1 for clear explanation of sides being the same length 5 38 AB2 = 22 12 M1 2 M1 for correct Pythagoras statement M – = 4 1 = 3 A1 A1 for 3 – AB = cm C1 C1 for a clear communication of the method used 3 39 cos 68° = 112° = 248° = 0.3746 cos 338° is the odd one out. B1 2 B1 for cos 338° M –cos –cos cos 338° = 0.9271 All the others have the same numerical C1 C1 for a clear explanation value (ignoring signs). 2 40 a i 2 M sin + 1 = 2 x M1 M1 for sin = 1 x sin = 1 x A1 = 90° x A1 for 90° = sin–11 = 90° x ii M1 2 + 3cos = 1 x M1 for first step of solving equation A1 3cos = 1 2 = x A1 for correct statement of cos – –1 x A1 A1 for correct angle to 1 dp cos = x – A1 A1 for correct angle to 1dp = 109.5° x = 109.5°, 250.5° x b M1 and 360° 109.5° = 250.5° – M1 for method of getting to sin–1 A1 cos 320° = 0.766 044 4 A1 for 50° A1 sin–1 0.766 044 4 = 50° A1 for 130° = 50° and 130° x and 180° 50° = 130° – 9
37 Use Pythagoras’ theorem. AC2 = 42 – (2)2 = 16 – 8 = 8 BC2 = 8 = (2)2 = 8 Hence BC = AC, an isosceles triangle. M1 A1 M1 A1 C1 2 M1 for correct Pythagoras statement A1 for correct value of AC2 M1 for finding BC2 A1 for correct value of BC2 C1 for clear explanation of sides being the same length M 5 38 AB2 = 22 – 12 = 4 – 1 = 3 AB = cm M1 A1 C1 2 M1 for correct Pythagoras statement A1 for 3 C1 for a clear communication of the method used M 3 39 cos 68° = –cos 112° = –cos 248° = 0.3746 cos 338° = 0.9271 cos 338° is the odd one out. All the others have the same numerical value (ignoring signs). B1 C1 2 B1 for cos 338° C1 for a clear explanation M 2 40 a i ii b sin x + 1 = 2 sin x = 1 x = sin–11 = 90° 2 + 3cos x = 1 3cos x = 1 – 2 = –1 cos x = – x = 109.5° and 360° – 109.5° = 250.5° cos 320° = 0.766 044 4 sin–1 0.766 044 4 = 50° and 180° – 50° = 130° x = 90° x = 109.5°, 250.5° x = 50° and 130° M1 A1 M1 A1 A1 A1 M1 A1 A1 2 M1 for sin x = 1 A1 for 90° M1 for first step of solving equation A1 for correct statement of cos x A1 for correct angle to 1 dp A1 for correct angle to 1dp M1 for method of getting to sin–1 A1 for 50° A1 for 130° M 9
Higher_Skills_Geometry_MarkScheme_Edexcel.docx