Higher June 2019 Paper 5 Q15
15 OAB is a sector of a circle, centre O.
OA = 6 cm and AX is perpendicular to OB.
Not to scale
The area of sector OAB is \(6\pi\) cm\(^2\).
Show that \(\text{AX} = 3\sqrt{3}\) cm. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{x}{360} \times \pi \times 6^2\) or \(\dfrac{[\pi \times]6^2}{[\pi \times]6}\) or shows \(\pi \times 6^2\) and \(\dfrac{1}{6}\) oe | M1 | Accept 36 for \(6^2\) | \(x\) = angle AOX , condone any variable used For M1 may be seen in stages e.g. M1 for \(36\pi \div 6\) |
| \(\dfrac{x}{360}[\times \pi] \times 6^2 = 6[\pi]\) or 360 ÷ 6 | M1dep | Dep on previous M1 | |
| [\(x\) = ] 60 | A1 | Must earn M1M1 before awarding A1 | |
| \(\dfrac{\text{AX}}{6} = \sin\) their 60 oe | M1 | Dep on 0 < their 60 < 90 Accept use of cos 30 or cos 60 and Pythagoras’ or sine rule with 90 | Do not accept assumption that OX = 3 without any evidence |
| \(\text{AX} = 6 \times \dfrac{\sqrt{3}}{2} = 3\sqrt{3}\) or \(\dfrac{3\sqrt{3}}{6} = \dfrac{\text{AX}}{6}\), \(\text{AX} = 3\sqrt{3}\) | M2 | or M1 for \(\sin 60 = \dfrac{\sqrt{3}}{2}\) or \(\cos 30 = \dfrac{\sqrt{3}}{2}\) To award 6 marks, there must be no errors seen | Beware circular methods using \(3\sqrt{3}\) leading to 60, this can only score M1 maximum for \(\sin 60 = \dfrac{\sqrt{3}}{2}\) but ignore circular methods if alongside a correct method |