4MA0 Higher June 2014 Paper 4 Q23
23
Diagram NOT accurately drawn
ABCDEF is a triangular prism.
\(AB = 9\) cm, \(BC = 15\) cm and \(AE = 12\) cm.
Angle \(ABC = 90^\circ\)
M is the midpoint of CD.
Calculate the size of the angle between AM and the plane BCDF.
Give your answer correct to 1 decimal place.
(5)
| Scheme | Marks |
|---|---|
| Angle AMB identified | M1 |
| \((BM^2) = 15^2 + 6^2\) | M1 |
| \((BM =)\ \sqrt{15^2 + 6^2}\) or \(\sqrt{261}\) or \(3\sqrt{29}\) (=16.1…) | M1 |
| \(\tan AMB = \dfrac{9}{\text{“}\sqrt{261}\text{”}}\) | M1 |
| 29.1 | A1 |
| (5 marks) |
Notes
M1: (dep on previous M1)
A1: for 29.1 – 29.25
NB. If angle BAM (60.9) found then maximum of M0M1M1M0A0 unless this is used to go onto find angle AMB
Alternatively: M1 Angle AMB identified; M1 \((AM^2 =)\ 9^2 + 15^2 + 6^2\); M1 \((AM =)\sqrt{9^2 + 15^2 + 6^2}\) or \(\sqrt{342}\) or \(3\sqrt{38}\) (=18.49…); M1 \(\sin AMB = \dfrac{9}{\text{“}18.49\text{”}}(\times \sin 90)\) (= 0.4867) etc or \(\cos AMB = \dfrac{\text{“}16.16\text{”}}{\text{“}18.49\text{”}}\) (= 0.8735) etc or correct method to find AM and BM with correct substitution into Cosine rule and correct rearrangement to make cosAMB the subject
M2 for BM = 16.1 – 16.2 or AM = 18.4 – 18.5