Higher June 2018 Paper 2 Q18
18 \(ABCDEFGH\) is a cuboid.
\(AB = 7.3\) cm
\(CH = 8.1\) cm
Angle \(BCA = 48^\circ\)
Find the size of the angle between \(AH\) and the plane \(ABCD\).
Give your answer correct to 1 decimal place. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 39.5 | P1 | for a start to a process eg, for a correct trigonometric statement, eg \(\sin 48 = \dfrac{7.3}{AC}\) or \(\cos 42 = \dfrac{7.3}{AC}\) or \(\dfrac{AC}{\sin 90} = \dfrac{7.3}{\sin 48}\) OR angle \(CAH\) unambiguously identified on a diagram |
| P1 | for a complete correct process to find \(AC\), eg (\(AC =\)) \(\dfrac{7.3}{\sin(48)}\) (=9.8..) or (\(AC =\)) \(\dfrac{7.3}{\cos(42)}\) (=9.8..) or (\(AC =\)) \(\sin 90 \times \dfrac{7.3}{\sin 48}\) (=9.8..) | |
| P1 | for a correct statement using angle \(CAH\), eg \(\tan(CAH) = \dfrac{8.1}{\text{``}9.8...\text{''}}\) OR \(\sqrt{8.1^2 + \text{``}9.8\text{''}^2}\) (=12.7…) and \(\dfrac{\sin CAH}{8.1} = \dfrac{\sin 90}{\text{``}12.7\text{''}}\) | |
| A1 | for answer in the range 39.5 – 39.51 |
Additional guidance
P1 (start): Must include correct values
If an answer is given in the range but then incorrectly rounded award full marks.