June 2017 Paper 2 Q6
6 A cone has vertex \(V\).
\(C\) is the centre of the base.
The slant height, \(VA\), is 20 cm
The angle between \(VA\) and \(VC\) is \(38^\circ\)
Work out the radius of the base. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sin 38 = \dfrac{r}{20}\) or \(\cos(90 - 38) = \dfrac{r}{20}\) or \(\dfrac{r}{\sin 38} = \dfrac{20}{\sin 90}\) or \(\dfrac{\sin 38}{r} = \dfrac{\sin 90}{20}\) | M1 | oe Any letter |
| \(20 \times \sin 38\) or \(20 \times \cos(90 - 38)\) or \(\dfrac{20}{\sin 90} \times \sin 38\) | M1dep | oe M2 \(\sqrt{20^2 - (20\cos 38)^2}\) M2 \(\dfrac{\sqrt{20^2 + 20^2 - 2 \times 20 \times 20 \times \cos(38 \times 2)}}{2}\) |
| 12.3(…) | A1 | SC2 Angle \(VAC = 38\) seen on diagram and answer 15.76(…) or 15.8 |
Additional guidance
| If trigonometry and Pythagoras are used it must be a fully correct method that would lead to the correct value of \(r\) for M2 | |
| If cosine rule with angle (\(38 \times 2\)) used it must be a fully correct method that would lead to the correct value of \(r\) for M2 | |
| Answer 15.76(…) or 15.8 but angle \(VAC = 38\) not seen | Zero |
| 12.3(…) seen and angle 38 in correct place with further work eg \(20 \sin 38 = 12.3\) \(\sqrt{20^2 - 12.3^2} = 15.8\) | Zero |
| \(\sin 38 \times 20\) (even if subsequently evaluates \(\sin 760\)) | M2 |
| Throughout, accept opp or o for \(r\) eg \(\sin 38 = \dfrac{\text{opp}}{20}\) | M1 |
| \(\sin = \dfrac{r}{20}\) or \(\sin\theta = \dfrac{r}{20}\) (unless recovered) | M0 |
| Answer 12.3(…) coming from scale drawing | M2A1 |
| Answer 12 coming from scale drawing | Zero |
| 12.3(…) seen with no further work followed by answer 12 | M2A1 |