| Date | November 2007 | Marks available | 3 | Reference code | 07N.1.sl.TZ0.9 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 9 | Adapted from | N/A |
Question
On a map three schools A, B and C are situated as shown in the diagram.
Schools A and B are 625 metres apart.
Angle ABC = 102° and BC = 986 metres.
Find the distance between A and C.
Find the size of angle BAC.
Markscheme
Unit penalty (UP) is applicable in question part (a) only.
\({\text{AC}}^2 = 625^2 + 986^2 - 2 \times 625 \times 986 \times \cos102^\circ\) (M1)(A1)
( = 1619072.159)
AC = 1272.43
(UP) =1270 m (A1) (C3)
[3 marks]
\(\frac{{986}}{{\operatorname{sinA} }} = \frac{{1270}}{{\sin 102^\circ }}\) (M1)(A1)(ft)
\({\text{A}} = 49.4^\circ\) (A1)(ft)
OR
\(\frac{{986}}{{\operatorname{sinA} }} = \frac{{1272.43}}{{\sin 102^\circ }}\) (M1)(A1)(ft)
\({\text{A}} = 49.3^\circ\) (A1)(ft)
OR
\(\cos {\text{A}} = \left( {\frac{{{{625}^2} + {{1270}^2} - {{986}^2}}}{{2 \times 625 \times 1270}}} \right)\) (M1)(A1)(ft)
\({\text{A}} = 49.5^\circ\) (A1)(ft) (C3)
[3 marks]
Examiners report
The candidates who used the cosine and sine rules for this question were successful on the whole. Some had their calculators in radian mode (and hence the second answer for the angle was unrealistic) but this was less frequent than in previous sessions. Those candidates who used right-angled trigonometry scored no marks. Many candidates lost an accuracy penalty in this question.
The candidates who used the cosine and sine rules for this question were successful on the whole. Some had their calculators in radian mode (and hence the second answer for the angle was unrealistic) but this was less frequent than in previous sessions. Those candidates who used right-angled trigonometry scored no marks. Many candidates lost an accuracy penalty in this question.
