June 2023 Paper 1 Q20
20 \(P\), \(Q\) and \(R\) are points on a circle, centre \(O\).
Angle \(POR = 120^\circ \qquad PQ = 4\) cm \(\qquad QR = 5\) cm
Not drawn accurately
Work out the radius of the circle.
Give your answer in the form \(\sqrt{k}\) where \(k\) is an integer. [6 marks]
| Answer | Mark | Comments |
|---|---|---|
| Angle \(PQR = 60^\circ\) | B1 | no reason required (angle at circumference is half angle at centre) could be on diagram |
| (\(PR^2 =\)) \(4^2 + 5^2 - 2(4)(5)\cos(\text{their } 60^\circ)\) | M1 | cosine rule applied to work out \(PR\) |
| \(\cos 60^\circ = 0.5\) or \(\dfrac{1}{2}\) or \(\dfrac{\sqrt{1}}{2}\) | B1 | must be used as part of an attempted cosine rule that requires \(60^\circ\) |
| \(PR^2 = 21\) or \(PR = \sqrt{21}\) | M1dep | would imply the B marks |
| (\(PO =\)) \(\left(\dfrac{\frac{1}{2}(\text{their } PR)}{\sin 60^\circ}\right)\) | M1 | \(PR\) must have a value oe eg \(PO = \dfrac{\sqrt{21}}{2} \div \dfrac{\sqrt{3}}{2}\) could use cosine rule again or sine rule at this stage but it would need to lead directly in one calculation to \(\sqrt{7}\) eg \(\left(\dfrac{\text{their}\sqrt{21}\sin 30^\circ}{\sin 120^\circ}\right)\) must be explicit: \(\left(\dfrac{PO}{\sin 30^\circ} = \dfrac{\text{their}\sqrt{21}}{\sin 120^\circ}\right)\) is not M1 but this then followed by \(\left(\dfrac{PO}{\frac{1}{2}} = \dfrac{\text{their}\sqrt{21}}{\frac{1}{3}}\right)\) (error in denominator) and \(\left(PO = \dfrac{\frac{1}{2}\text{their}\sqrt{21}}{\frac{1}{3}}\right)\) (rearranged correctly) would be M1 |
| (\(PO =\)) \(\sqrt{7}\) | A1 |
Additional guidance
| \(POM = 60^\circ\) where \(M\) is the midpoint of \(PR\) | B0 |
| Final answer coming from double errors in surd manipulation | B2 M3 A0 or B2 M2 A0 |