Higher June 2017 Paper 3 Q19
19 A, B, C, D and E are points on a circle.
BFD and AFC are straight lines.
DC = DF
Not drawn accurately
Work out the size of angle \(x\).
You must show your working which may be on the diagram. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| BDC = 24 | B1 | May be on the diagram |
| DFC \(= \dfrac{180 - 24}{2}\) or DCF \(= \dfrac{180 - 24}{2}\) or \(\dfrac{156}{2}\) or 78 | B1dep | May be on the diagram Finding a base angle in triangle CDF |
| (\(3x =\)) 180 – their 78 or (\(3x =\)) 24 + their 78 or (\(3x =\)) 102 | M1 | oe May be on the diagram |
| 34 | A1 | May be on the diagram |
| Alternative method 2 | ||
| BDC = 24 | B1 | May be on the diagram |
| DFC \(= 180 - 3x\) | M1 | May be on the diagram |
| \(2(180 - 3x) + 24 = 180\) or \(360 - 6x + 24 = 180\) or \(3x + 78 = 180\) or (\(3x =\)) 102 | M1dep | oe |
| 34 | A1 | May be on the diagram |
Additional guidance
| If angles in the same segment are not used ie all the working is using triangle ABF then award maximum of 2 marks | |
| If triangle ABF is assumed to be isosceles and there is no evidence of angle BDC = 24 being used then award maximum of 2 marks | |
| If triangle ABF is used as isosceles and correctly justified then all marks are available eg ‘triangle ABF is similar to triangle CDF’ | |
| Answer of 34 does not imply full marks | |
| Answer of 34 with no working | B0B0M1A1 |
| ‘their 78’ must come from an attempt to calculate \(\dfrac{180 - 24}{2}\) | |
| Angles must be clearly identified eg D = 24 24 (unless shown on diagram) | B1 B0 |