Higher June 2022 Paper 2 Q28
28 \(AB\), \(BC\) and \(CD\) are sides of a regular 12-sided polygon.
\(CDMN\) is a square.
Not drawn accurately
Prove that points \(A\), \(B\) and \(N\) lie on a straight line. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| interior angle = 150 or exterior angle = 30 or angle \(BCN\) = 120 | B1 | method not required may be seen on diagram |
| interior angle = 150 with a valid method shown or exterior angle = 30 with a valid method shown or angle \(BCN\) = 120 with a valid method shown | B1dep | angles may be seen on diagram but methods will be in working lines eg \(180 - \dfrac{360}{12} = 150\) or \(\dfrac{1800}{12} = 150\) or \(360 - 120 - 90 = 150\) or \(\dfrac{360}{12} = 30\) or \(\dfrac{180 - 120}{2} = 30\) or \(180 - 150 = 30\) or \(360 - 150 - 90 = 120\) or \(360 - 240 = 120\) or \(180 - 2 \times 30 = 120\) |
| interior angle = 150 with a valid method shown and exterior angle = 30 with a valid method shown and angle \(BCN\) = 120 with a valid method shown | B1dep | angles may be seen on diagram but methods will be in working lines eg \(\dfrac{1800}{12} = 150\) and \(\dfrac{180 - 120}{2} = 30\) and \(360 - 240 = 120\) angles worked out in any order |
| Fully correct working that must show correct progression and show all valid methods Valid methods shown must be appropriate for the approach used A reason must be included in the final step | B1dep | examples of the final step are (i) angle \(ABC\) + angle \(CBN\) = 180 (ii) interior angle = 150 in two different ways (iii) exterior angle = 30 in two different ways (iv) angle \(BCN\) = 120 in two different ways (v) sum of three angles at \(C\) = 360 (vi) sum of angles of triangle \(BCN\) = 180 |
Additional guidance
| Condone incorrect use of equals signs throughout | |
| eg interior angle \(= 12 - 2 = 10 \times 180 = 1800 \div 12 = 150\) | B1B1 |
| interior angle may be seen as angle \(ABC\) or angle \(BCD\) exterior angle may be seen as angle \(CBN\) | |
| It must be clear which angle they are working out eg1 Do not accept 150 if it is not correctly identified or not in the correct position on diagram eg2 Do accept 150 if it is identified as an interior angle or angle \(ABC\) or is in the correct position on the diagram | |
| Do not accept incorrect statements eg1 exterior angle = 150 (even if 150 in correct position on the diagram) eg2 angle \(ACB\) = 150 (even if 150 in correct position on the diagram) | |
| Ignore reasons for the first three marks | |
| Angles on the diagram with no valid methods can score a maximum of B1B0B0B0 | |
| For the 2nd and 3rd marks the methods shown do not have to show progression | |
| Example of fully correct working for (i) | |
| interior angle \(= \dfrac{1800}{12} = 150\) | B1B1 |
| angle \(BCN = 360 - 150 - 90 = 120\) | |
| angle \(CBN = \dfrac{180 - 120}{2} = 30\) | B1 |
| \(150 + 30 = 180 \quad\) angles on a (straight) line | B1 |
| Example of fully correct working for (ii) | |
| exterior angle \(= \dfrac{360}{12} = 30\) | B1B1 |
| angle \(BCN = 180 - 2 \times 30 = 120\) | |
| interior angle \(= 360 - 120 - 90 = 150\) | B1 |
| interior angle \(= \dfrac{1800}{12} = 150 \quad\) (interior) angle of polygon | B1 |