June 2016 Paper 1 Q5
5 \(A\), \(B\), \(C\) and \(D\) are points on a circle, centre \(O\).
Not drawn accurately
Work out the value of \(x\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| (Angle at circumference =) \(x + 24\) | M1 | oe |
| \(x + 24 + 3x = 180\) | M1 dep | oe |
| 39 | A1 | |
| Alternative method 2 | ||
| (Reflex angle at centre =) \(6x\) | M1 | oe |
| \(2x + 48 + 6x = 360\) | M1dep | oe |
| 39 | A1 | |
| Alternative method 3 | ||
| (Angle at circumference =) \(180 - 3x\) | M1 | oe |
| \(2x + 48 = 2(180 - 3x)\) | M1 dep | oe |
| 39 | A1 | |
| Alternative method 4 | ||
| (Reflex angle at centre =) \(360 - (2x + 48)\) | M1 | oe |
| \(360 - (2x + 48) = 2(3x)\) | M1 dep | oe |
| 39 | A1 | |
Additional guidance
Look on the diagram for evidence of the 1st M1
\(39^\circ\) seen on the answer line ... check to see that it has come from correct working
Reasons are not necessary
Here are two examples of equations coming from incorrect geometrical reasoning.
\(2(3x) = 2x + 48\) scores M0 M0
\(2x + 48 + 6x = 180\) scores M1 M0 ... because the \(6x\) in this equation implies the reflex angle at the centre, so it scores M1 in this particular case.