4MA0 Higher January 2015 Paper 3R Q23
23 Here is triangle LMN, where angle LMN is an obtuse angle.
Diagram NOT accurately drawn
\(ML = 13.8\) cm
\(MN = 8.5\) cm
Angle \(MNL = 47^\circ\)
Work out the area of triangle LMN.
Give your answer correct to 3 significant figures.
(6)
| Scheme | Marks |
|---|---|
| \(\dfrac{\sin 47}{13.8} = \dfrac{\sin MLN}{8.5}\) | M1 |
| \(MLN = \sin^{-1}\left(\dfrac{\sin 47 \times 8.5}{13.8}\right)\) | M1 |
| \(MLN = 26.7(73\ldots)\) | A1 |
| \(LMN = 180 - 47 - \text{“}26.7\ldots\text{”}\) or 106(.2260622…) | M1 |
| \(\dfrac{1}{2} \times 8.5 \times 13.8 \times \sin(\text{“}106\text{”})\) | M1 |
| 56.3 | A1 |
| (6 marks) |
Notes
M1: Or method using a right angled triangle to find length \(MX\) (\(MX\) is perpendicular to \(LN\))
\(\sin 47 = \dfrac{MX}{8.5}\)
M1: Or \(\cos^{-1}\left(\dfrac{8.5\sin 47}{13.8}\right)\)
A1: \(LMX = 63.232..\)
M1: \(LMN = 63.232 + (180 - (90 + 47))\)… or
106(.2260622…)
A1: Accept an answer that rounds to 56.3 or 56.4 unless clearly obtained from incorrect working.