| Date | May 2011 | Marks available | 3 | Reference code | 11M.1.sl.TZ2.10 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Calculate | Question number | 10 | Adapted from | N/A |
Question
In the diagram, \({\text{AD}} = 4{\text{ m}}\), \({\text{AB}} = 9{\text{ m}}\), \({\text{BC}} = 10{\text{ m}}\), \({\text{B}}\hat {\text{D}}{\text{A}} = {90^ \circ }\) and \({\text{D}}\hat {\text{B}}{\text{C}} = {100^ \circ }\) .
Calculate the size of \({\text{A}}\hat {\text{B}}{\text{C}}\).
Calculate the length of AC.
Markscheme
\(\sin {\text{A}}\hat {\text{B}}{\text{D}} = \frac{4}{9}\) (M1)
\(100 + {\text{their }}({\text{A}}\hat {\text{B}}{\text{D}})\) (M1)
\(126\% \) (A1) (C3)
Notes: Accept an equivalent trigonometrical equation involving angle ABD for the first (M1).
Radians used gives \(100\% \) . Award at most (M1)(M1)(A0) if working shown.
\({\text{BD}} = 8{\text{ m}}\) leading to \(127\% \) . Award at most (M1)(M1)(A0) (premature rounding).
[3 marks]
\({\text{A}}{{\text{C}}^2} = {10^2} + {9^2} - 2 \times 10 \times 9 \times \cos (126.38 \ldots )\) (M1)(A1)
Notes: Award (M1) for substituted cosine formula. Award (A1) for correct substitution using their answer to part (a).
\({\text{AC}} = 17.0{\text{ m}}\) (A1)(ft) (C3)
Notes: Accept \(16.9{\text{ m}}\) for using \(126\). Follow through from their answer to part (a). Radians used gives \(5.08\). Award at most (M1)(A1)(A0)(ft) if working shown.
[3 marks]
Examiners report
Although many candidates were able to calculate the size of angle ABD correctly, a significant number then simply stopped, failing to add on \(100\% \) and consequently losing the last two marks in part (a). Recovery was seen on many scripts in part (b) as candidates seemed to be well drilled in the use of the cosine rule and much correct working was seen. Indeed, despite many incorrect final answers of \(26.4\% \) seen in part (a), many used the correct angle of \(126\% \) in part (b).
Although many candidates were able to calculate the size of angle ABD correctly, a significant number then simply stopped, failing to add on \(100\% \) and consequently losing the last two marks in part (a). Recovery was seen on many scripts in part (b) as candidates seemed to be well drilled in the use of the cosine rule and much correct working was seen. Indeed, despite many incorrect final answers of \(26.4\% \) seen in part (a), many used the correct angle of \(126\% \) in part (b).
