4MA0 Higher January 2018 Paper 3R Q2
2 \(ABCD\) is a parallelogram.
Diagram NOT accurately drawn
Angle \(DCB = 124^\circ\)
\(E\) is the point on \(DC\) such that \(AE = DE\).
Angle \(AEC = x^\circ\)
Work out the value of \(x\).
(4)
| Scheme | Marks |
|---|---|
| \(\angle ADE = 180 - 124\;(= 56)\) or \(\angle ADE = \dfrac{360 - 2 \times 124}{2}\;(= 56)\) | M1 |
| \(\angle DAE = \angle ADE = \text{‘}56\text{’}\) | M1 |
| \(\angle AEC = 2 \times \text{‘}56\text{’}\) | M1 |
| 112 | A1 |
| (4) |
Notes
M1: for 2 × ‘56’ or for \(\angle AED\) = 180 – 2 × ‘56’ (=68) and \(\angle AEC\) = 180 – ‘68’