Higher November 2017 Paper 2 Q21
21. The diagram shows a cone with a base diameter of 8·4 cm.
Diagram not drawn to scale
Calculate the perpendicular height of the cone. [5]
| Scheme | Marks | Comments |
|---|---|---|
| \(\cos A = \dfrac{6.3^2 + 8.4^2 - 12.4^2}{2 \times 6.3 \times 8.4}\) \((= -0.41109\ldots)\) | M2 | M1 for sight of \(12.4^2 = 6.3^2 + 8.4^2 - 2 \times 6.3 \times 8.4 \times \cos A\) |
| \(A = 114.27(\ldots^\circ)\) or \(114.3(^\circ)\) | A1 | |
| Use of acute \(A\) in trig ratio \(\sin 65.7(^\circ) = \dfrac{\text{perpendicular}}{6.3}\) or perpendicular \(= 6.3 \times \sin 65.7(^\circ)\) | m1 | FT from M2 for ‘their acute A’ |
| Perpendicular height 5.7(43…cm) | A1 | |
| (5) |
Alternatives
| \(\cos B = \dfrac{12.4^2 + 8.4^2 - 6.3^2}{2 \times 12.4 \times 8.4}\) \((= 0.886\ldots)\) (OR M1 for sight of \(6.3^2 = 12.4^2 + 8.4^2 - 2 \times 12.4 \times 8.4 \times \cos B\)) | M2 |
| \(B = 27.59(\ldots^\circ)\) or \(27.6(^\circ)\) | A1 |
| \(\sin 27.59(\ldots^\circ) = \dfrac{\text{perpendicular}}{12.4}\) or perpendicular \(= 12.4 \times \sin 27.59(^\circ)\) (FT provided M2 previously awarded) | m1 |
| Perpendicular height 5.7(43…cm) | A1 |
OR
| \(12.4^2 = h^2 + (x + 8.4)^2\) and \(6.3^2 = h^2 + x^2\) (OR M1 for sight of either equation) | M2 |
| \(12.4^2 = 6.3^2 - x^2 + (x + 8.4)^2\) or \(x = \dfrac{12.4^2 - 6.3^2 - 8.4^2}{2 \times 8.4}\) | m1 |
| \(x = 2.58988\ldots\)(cm) or 2.6(cm) and \(h^2 = 6.3^2 - 2.58988\ldots^2\) \((= 32.98\ldots)\) (FT ‘their \(x\)’ provided M2 previously awarded) | m1 |
| Perpendicular height 5.7(43…cm) | A1 |
(A = obtuse base angle, B = acute base angle)