Higher November 2017 Paper 1 Q3
3
\(ABCD\) is a parallelogram.
\(EDC\) is a straight line.
\(F\) is the point on \(AD\) so that \(BFE\) is a straight line.
Angle \(EFD = 35^\circ\)
Angle \(DCB = 75^\circ\)
Show that angle \(ABF = 70^\circ\)
Give a reason for each stage of your working. (4)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(CB\) extended to form \(CG\) | Reasoning | B1 | for 35 or 75 or 145 or 105 or \(DEF = 70\), marked on the diagram or 3 letter description |
| M1 | for \(180-70-35\) or \(180-75-35\) or a correct pair of angles that would lead to 75 or 70, eg \(AFB = 35\) and \(FAB = 75\) or \(AFB = 35\) and \(ABG = 75\) or \(FBC = 35\) and \(ABG = 75\) or \(EDF = 75\) and \(DEF = 70\) or \(FDC = 105\) and \(FBC = 35\) or \(ABC = 105\) and \(FBC = 35\) | ||
| C2 | (dep on B1M1) All figures correct with all appropriate reasons stated. Angles must be clearly labelled or on the diagram. Full solution must be seen | ||
| (C1 | (dep on B1 or M1) for one reason clearly used and stated.) Corresponding angles are equal, alternate angles are equal, opposite angles in a parallelogram are equal, angles in a triangle sum to 180, angles on a straight line sum to 180, vertically opposite angles are equal, vertically opposite angles are equal, angles in a quadrilateral sum to 360, co-interior angles sum to 180, allied angles sum to 180, angles around a point sum to 360 |