1MA0 Higher June 2012 Paper 2 Q16
16
Diagram NOT accurately drawn
\(LMN\) is a right-angled triangle.
\(MN = 9.6\) cm.
\(LM = 6.4\) cm.
Calculate the size of the angle marked \(x^\circ\).
Give your answer correct to 1 decimal place. (3)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(\cos x = \dfrac{6.4}{9.6}\) \(x = \cos^{-1}\dfrac{6.4}{9.6} =\) | 48.2 | 3 | M1 for \(\cos x = \dfrac{6.4}{9.6}\) or \(\cos x = 0.66(6...)\) or \(\cos x = 0.67\) M1 for \(\cos^{-1}\dfrac{6.4}{9.6}\) or \(\cos^{-1} 0.66(6...)\) or \(\cos^{-1} 0.67\) A1 for 48.1 – 48.2 OR Correct use of Pythagoras and then trigonometry, no marks until M1 for \(\sin x = \dfrac{\text{‘}7.155\text{’}}{9.6}\) or \(\tan x = \dfrac{\text{‘}7.155\text{’}}{6.4}\) or \(\sin x = \dfrac{\text{‘}7.155\text{’}}{9.6} \times \sin 90\) or \(\cos x = \dfrac{6.4^2 + 9.6^2 - \text{‘}7.155\text{’}^2}{2 \times 6.4 \times 9.6}\) M1 for \(\sin^{-1}\dfrac{\text{‘}7.155\text{’}}{9.6}\) or \(\tan^{-1}\dfrac{\text{‘}7.155\text{’}}{6.4}\) or \(\sin^{-1}\left(\dfrac{\text{‘}7.155\text{’}}{9.6} \times \sin 90\right)\) or \(\cos^{-1}\left(\dfrac{6.4^2 + 9.6^2 - \text{‘}7.155\text{’}^2}{2 \times 6.4 \times 9.6}\right)\) A1 for 48.1 – 48.2 SC B2 for 0.841... (using rad) or 53.5... (using grad) |