Higher January 2019 Paper 2 Q22
22 The diagram shows a cone.
Diagram NOT accurately drawn
\(AB\) is a diameter of the cone.
\(V\) is the vertex of the cone.
Given that
the area of the base of the cone : the total surface area of the cone = 3 : 8
work out the size of angle \(AVB\).
Give your answer correct to 1 decimal place.
(6)
| Scheme | Marks |
|---|---|
| \(\dfrac{\pi r^2}{\pi r^2 + \pi rl} = \dfrac{3}{8}\) or \(\pi r^2 : \pi r^2 + \pi rl = 3 : 8\) or | M1 |
\(\pi r^2 : \pi rl = 3 : 5\) or \(\pi r^2 = 3\) and \(\pi rl = 5\) \(8\pi r^2 = 3(\pi r^2 + \pi rl)\) or \(5\pi r^2 = 3\pi rl\) or \(\left[r = \sqrt{\dfrac{3}{\pi}}\ (= 0.9772...)\right.\) and \(\left.l = \dfrac{5}{\pi r}\right]\) | M1 |
| \(\dfrac{r}{l} = \dfrac{3}{5}\) oe or \(l = \dfrac{5}{\pi\sqrt{\dfrac{3}{\pi}}}\) (=1.62...) | M1 |
| e.g. \(\sin\left(\dfrac{AVB}{2}\right) = \dfrac{3}{5}\) oe eg \(\sin\left(\dfrac{AVB}{2}\right) = \dfrac{\sqrt{3/\pi}}{\dfrac{5}{\pi\sqrt{3/\pi}}}\) | M1 |
| \(2 \times \sin^{-1}\left(\dfrac{3}{5}\right)\) oe | M1 |
| 73.7 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: \(\sin^{-1}\left(\dfrac{3}{5}\right) = 36.86....\)
A1: awrt