Date | November 2008 | Marks available | 7 | Reference code | 08N.2.hl.TZ0.1 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 1 | Adapted from | N/A |
Question
In a triangle ABC, \(\hat A = 35^\circ \), BC = 4 cm and AC = 6.5 cm. Find the possible values of \(\hat B\) and the corresponding values of AB.
Markscheme
\(\frac{{\sin B}}{{6.5}} = \frac{{\sin 35^\circ }}{4}\) M1
\(\hat B = 68.8^\circ {\text{ or }}111^\circ \) A1A1
\(\hat C = 76.2^\circ \) or \(33.8^\circ \) (accept \(34^\circ \)) A1
\(\frac{{{\text{AB}}}}{{\sin C}} = \frac{{{\text{BC}}}}{{\sin A}}\)
\(\frac{{{\text{AB}}}}{{\sin 76.2^\circ }} = \frac{4}{{\sin 35^\circ }}\) (M1)
AB = 6.77 cm A1
\(\frac{{{\text{AB}}}}{{\sin 33.8^\circ }} = \frac{4}{{\sin 35^\circ }}\)
AB = 3.88 cm\(\,\,\,\,\,\)(accept 3.90) A1
[7 marks]
Examiners report
Most candidates realised that the sine rule was the best option although some used the more difficult cosine rule which was an alternative method. Many candidates failed to realise that there were two solutions even though the question suggested this. Many candidates were given an arithmetic penalty for giving one of the possible of values \({\hat B}\) as 112.2° instead of 111°.