Higher June 2017 Paper 2 Q18
18.
Diagram not drawn to scale
The diagram shows a cuboid.
\(A\), \(B\) and \(C\) are all vertices of the cuboid.
Calculate the size of \(C\hat{A}B\).
Give your answer correct to 3 significant figures. [5]
| Scheme | Marks | Comments |
|---|---|---|
| (\(AC^2 =\)) \(2.6^2 + 13.4^2\) OR (\(AC =\)) \(\sqrt{186.32}\) or \(AC^2 = 186.32\) | M1 | |
| \(AC = 13.6499\ldots\) (cm) | A1 | Accept rounded or truncated May be implied in further calculation |
| \(\tan A = 8.7/13.6499\ldots\ (=0.637\ldots)\) | M1 | FT ‘their derived \(AC\)’ provided M1 previously awarded |
| \(A = 32.5\)(°) | A2 | CAO A1 for \(\tan^{-1}0.637\ldots\) or FT angle correct but not given to 3 sig. figs. |
| (5) |
Alternative
| Scheme | Marks |
|---|---|
| \(AB^2 = 2.6^2 + 13.4^2 + 8.7^2\ (=262.01)\) | M1 |
| \(AB = 16.186\ldots\) (cm) | A1 |
| \(\sin A = 8.7/16.186\ldots\ (=0.537\ldots)\) FT ‘their derived \(AB\)’ provided M1 previously awarded | M1 |
| \(A = 32.5\)(°) CAO with FT and A1 as above | A2 |
Alternative
| Scheme | Marks |
|---|---|
| \(AB^2 = 2.6^2 + 13.4^2 + 8.7^2\ (=262.01)\) | M1 |
| \(AB = 16.186\ldots\) (cm) | A1 |
| \(\cos A = 13.6499\ldots/16.186\ldots\) \((=0.843\ldots)\) FT ‘their derived \(AB\)’ provided M1 previously awarded and attempt (\(AC^2 =\)) \(2.6^2 + 13.4^2\) | M1 |
| \(A = 32.5\)(°) CAO with FT and A1 as above | A2 |
Corrected by Westie's Workshop
- The board's mark scheme printed “(=0.537…)”. Changed to “(=0.843…)” because 13.6499… ÷ 16.186… = 0.843…; the value 0.537… belongs to the sine method above it.
Original: see the board's mark scheme PDF (link above).