1MA0 Higher November 2012 Paper 2 Q6
*6
Diagram NOT accurately drawn
\(ABCD\) is a parallelogram.
Angle \(ADB = 38^\circ\).
Angle \(BEC = 41^\circ\).
Angle \(DAB = 120^\circ\).
Calculate the size of angle \(x\).
You must give reasons for your answer. (4)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| Angle \(DEC = 180 - 41 = 139\) Angles on a straight line sum to 180° Angle \(EDC = 60 - 38\) or Angle \(ABD = 180 - 120 - 38\) (=22) Co-interior/allied angles of parallel lines sum to 180° or Angles in a triangle sum to 180° and Alternate angles \(x = 180 -\) ‘139’ – ‘22’ (=19) Angles in a triangle sum to 180° OR Angle \(ADC = 180^\circ - 120^\circ = 60^\circ\) Co-interior/allied angles of parallel lines sum to 180° Angle \(EDC = 22^\circ\) Angle \(ECD = 41^\circ - 22^\circ = 19^\circ\) Exterior angle of triangle equals sum of the two opposite interior angles OR Angle \(DBC = 38^\circ\) Alternate angles Angle \(BCE = 101^\circ\) Angle sum of a triangle is 180° Angle \(BCD = 120^\circ\) Opposite angles of a parallelogram are equal Angle \(ECD = 120^\circ - 101^\circ = 19^\circ\) | \(x = 19^\circ\) and reasons | 4 | M1 for \(DBC = 38^\circ\) or \(ADC = 60^\circ\) (can be implied by \(BDC = 22^\circ\)) or \(ABC = 60^\circ\) or \(DCB = 120^\circ\) or (\(ABD =\)) \(180 - 120 - 38\) (=22) M1 for (\(BDC =\)) \(60 - 38\) (=22) or \(BDC =\) ‘22’ or (\(DEC =\)) \(180 - 41\) (=139) or (\(BCE =\)) \(180 - 41 - 38\) (=101) M1 (dep on both previous M1) for complete correct method to find \(x\) or (\(x =\)) 19 C1 for \(x = 19^\circ\) AND Co-interior/allied angles of parallel lines sum to 180° or Opposite angles of a parallelogram are equal or Alternate angles AND Angles on a straight line sum to 180° or Angles in a triangle sum to 180° or Exterior angle of triangle equals sum of the two opposite interior angles or Angles in a quadrilateral sum to 360° |