Higher November 2020 Paper 4 Q14
14 The diagram shows triangle ABC.
Not to scale
AC = 15 cm, BC = 18 cm and angle BAC = 72°.
Calculate length AB, giving your answer correct to 3 significant figures.
Show your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 15.6[1..] with correct working | 6 | M2 for [sin B = ] \(\frac{15 \times \sin 72}{18}\) oe or M1 for \(\frac{\sin B}{15} = \frac{\sin 72}{18}\) oe AND M1 for 180 – 72 – their 52.4 implied by 55.6 or 55.57… and M2 for [AB=] \(\frac{18 \times \sin \textit{their}\ 55.57...}{\sin 72}\) oe or M1 for \(\frac{[...]}{\sin \textit{their}\ 55.57...} = \frac{18}{\sin 72}\) oe If 0 scored award SC2 for 15.6… with insufficient working | Correct working requires evidence of at least M1 AND M1 M2 implied by 0.7925… or 52.4… Alternative cosine rule (AB = \(x\)) M3 for quadratic equation with coefficients evaluated M2 for \(x^2 + (-2 \times 15 \times \cos 72)x + (15^2 - 18^2)\) [=0] oe or M1 for \(18^2 = x^2 + 15^2 - 2 \times x \times 15 \cos 72\) AND M2 for correct use of quadratic formula or M1 for use of quadratic formula with at most two errors |