Higher January 2020 Paper 1 Q19
19 The diagram shows a cuboid \(ABCDEFGH\).
Diagram NOT accurately drawn
\(EH = 9\) cm, \(HG = 5\) cm and \(GB = 6\) cm.
Work out the size of the angle between \(AH\) and the plane \(EFGH\).
Give your answer correct to 3 significant figures.
(4)
| Scheme | Marks |
|---|---|
\((AH =)\; \sqrt{6^2 + 5^2 + 9^2}\;\; (= \sqrt{142})\) or \((FH = GE =)\; \sqrt{5^2 + 9^2}\;\; (= \sqrt{106})\) | M1 |
E.g. \(\sin AHF = \dfrac{6}{\text{‘}\sqrt{142}\text{’}}\) or \(\tan AHF = \dfrac{6}{\text{‘}\sqrt{106}\text{’}}\) or \(\cos AHF = \dfrac{\text{‘}\sqrt{106}\text{’}}{\text{‘}\sqrt{142}\text{’}}\) or \(\sin FAH = \dfrac{\text{‘}\sqrt{106}\text{’}}{\text{‘}\sqrt{142}\text{’}}\) or \(\cos FAH = \dfrac{6}{\text{‘}\sqrt{142}\text{’}}\) or \(\tan FAH = \dfrac{\text{‘}\sqrt{106}\text{’}}{6}\) | M1 |
E.g. \(\sin^{-1}\left(\dfrac{6}{\text{‘}\sqrt{142}\text{’}}\right)\) or \(\tan^{-1}\left(\dfrac{6}{\text{‘}\sqrt{106}\text{’}}\right)\) or \(\cos^{-1}\left(\dfrac{\text{‘}\sqrt{106}\text{’}}{\text{‘}\sqrt{142}\text{’}}\right)\) or \(90 - \sin^{-1}\left(\dfrac{\text{‘}\sqrt{106}\text{’}}{\text{‘}\sqrt{142}\text{’}}\right)\) or \(90 - \cos^{-1}\left(\dfrac{6}{\text{‘}\sqrt{142}\text{’}}\right)\) or \(90 - \tan^{-1}\left(\dfrac{\text{‘}\sqrt{106}\text{’}}{6}\right)\) | M1 |
| 30.2 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for working out \(AH\) or \(FH\) or \(GE\)
M1: for a correct method for finding angle \(AHF\) or finding angle \(FAH\)
Allow
\(\cos AHF = \left(\dfrac{\text{‘}\sqrt{142}\text{’}^2 + \text{‘}\sqrt{106}\text{’}^2 - 6^2}{2 \times \text{‘}\sqrt{142}\text{’} \times \text{‘}\sqrt{106}\text{’}}\right)\) oe or
\(\sin AHF = \dfrac{\sin 90}{\text{‘}\sqrt{142}\text{’}} \times 6\) oe
M1: for a complete method
Allow
\(\cos^{-1}\left(\dfrac{\text{‘}\sqrt{142}\text{’}^2 + \text{‘}\sqrt{106}\text{’}^2 - 6^2}{2 \times \text{‘}\sqrt{142}\text{’} \times \text{‘}\sqrt{106}\text{’}}\right)\) oe or
\(\sin^{-1}\left(\dfrac{\sin 90}{\text{‘}\sqrt{142}\text{’}} \times 6\right)\) oe
A1: for 30.2 – 30.3