64 C1 2 C1 for clear use of diagram H Draw triangle BCP, where P is on the circumference in the opposite segment . Then angle BPC is 180° as BAC x – and BPC are opposite angles in a cyclic quadrilateral. Angle BOC is double BPC, which is 360° 2x. – Angle BCO will also be as BCO is an y, isosceles triangle. Angles in triangle OBC sum to 180°, M1 M1 for using point P so 2y + 360° 2x = 180° – A1 and 2y = 2x 180° A1 for 180 – – x Divide both sides by 2 to give: = 90°. y x – A1 A1 for identifying 360° 2x – B1 B1 for identifying BCO M1 M1 for adding known angles together A1 A1 for simplifying C1 C1 for a clear, well presented solution 8 As ABCD is a cyclic quadrilateral, the C1 2 C1 for explanation of cyclic quadrilateral H 65 opposite angles will sum to 180°. M1 M1 for adding opposite angles So 2x 5° + 5y 20° = 180° – – A1 A1 for first correct equation 2x + 5y = 205° ………(1) And 3y + 5 + 2x + 20° = 180° A1 A1 for second correct equation M1 M1 for method of eliminating one variable 2x + 3y = 155° ……..(2) Subtract (2) from (1): A1 A1 for 25° = 25° y 2y = 50° M1 M1 for method of substitution = 25° y Substitute into (2): A1 A1 for 40° y 2x + 75° = 155° 8 = 40° x 2x = 80° = 40° x
64 Draw triangle BCP, where P is on the circumference in the opposite segment . Then angle BPC is 180° – x as BAC and BPC are opposite angles in a cyclic quadrilateral. Angle BOC is double BPC, which is 360° – 2x. Angle BCO will also be y, as BCO is an isosceles triangle. Angles in triangle OBC sum to 180°, so 2y + 360° – 2x = 180° and 2y = 2x – 180° Divide both sides by 2 to give: y = x – 90°. C1 M1 A1 A1 B1 M1 A1 C1 2 C1 for clear use of diagram M1 for using point P A1 for 180 – x A1 for identifying 360° – 2x B1 for identifying BCO M1 for adding known angles together A1 for simplifying C1 for a clear, well presented solution H 8 65 As ABCD is a cyclic quadrilateral, the opposite angles will sum to 180°. So 2x – 5° + 5y – 20° = 180° 2x + 5y = 205° ………(1) And 3y + 5 + 2x + 20° = 180° 2x + 3y = 155° ……..(2) Subtract (2) from (1): 2y = 50° y = 25° Substitute y into (2): 2x + 75° = 155° 2x = 80° x = 40° y = 25° x = 40° C1 M1 A1 A1 M1 A1 M1 A1 2 C1 for explanation of cyclic quadrilateral M1 for adding opposite angles A1 for first correct equation A1 for second correct equation M1 for method of eliminating one variable A1 for 25° M1 for method of substitution A1 for 40° H 8
Higher_Skills_Geometry_MarkScheme_Edexcel.docx