4MA0 Higher June 2017 Paper 4 Q21
21 \(LMNP\) is a quadrilateral.
Diagram NOT accurately drawn
Work out the size of angle \(MLP\).
Give your answer correct to 3 significant figures.
(6)
| Scheme | Marks |
|---|---|
| \(15.6^2 + 4.3^2 - 2 \times 15.6 \times 4.3 \times \cos 72^\circ\) (=220.39...) | M1 |
| \(LN = 14.8(4561\ldots)\) | A1 |
\(\dfrac{\sin 58}{\text{“}14.8..\text{”}} = \dfrac{\sin MLN}{13.7}\) or \(\dfrac{\sin NLP}{4.3} = \dfrac{\sin 72}{\text{“}14.8\text{”}}\) or \(\dfrac{\sin LNP}{15.6} = \dfrac{\sin 72}{\text{“}14.8\text{”}}\) | M1 |
\(MLN = \sin^{-1}\left(\dfrac{\sin 58}{\text{“}14.8..\text{”}} \times 13.7\right)\) (=51.49..) or \(NLP = \sin^{-1}\left(\dfrac{\sin 72}{\text{“}14.8\text{”}} \times 4.3\right)\) (=15.99..) or \(LNP = \sin^{-1}\left(\dfrac{\sin 72}{\text{“}14.8\text{”}} \times 15.6\right)\) (=87.99 or 92.00..) | M1 |
\(MLN = \sin^{-1}\left(\dfrac{\sin 58}{\text{“}14.8..\text{”}} \times 13.7\right)\) (=51.49..) and \(NLP = \sin^{-1}\left(\dfrac{\sin 72}{\text{“}14.8\text{”}} \times 4.3\right)\) (=15.99..) or \(LNP = \sin^{-1}\left(\dfrac{\sin 72}{\text{“}14.8\text{”}} \times 15.6\right)\) (=87.99 or 92.00..) | M1 |
| 67.5 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: substitution into Cosine rule
A1: 14.8(4561…)
M1: ft \(LN\) dep on 1st M1
or correct start to alternative method to find angle \(MLN\) or angle \(NLP\) or angle \(LNP\)
[\(4.3^2 = 14.8..^2 + 15.6^2 - 2 \times 14.8 \times 15.6\cos NLP\)]
M1: ft \(LN\) dep on 1st M1
or complete alternative method to find angle \(MLN\) or angle \(NLP\) or angle \(LNP\)
NB: \(LNP\) = 180 – 87.99 = 92.009...
\(NLP = \cos^{-1}\left(\dfrac{14.8..^2 + 15.6^2 - 4.3^2}{2 \times 14.8.. \times 15.6}\right)\)
M1: ft \(LN\) dep on 1st M1
or complete method to find angle \(MLN\) and angle \(NLP\) (or \(LNP\) acute or obtuse)
A1: for 67.46 – 67.8