| Date | May 2010 | Marks available | 1 | Reference code | 10M.1.sl.TZ1.7 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Write down | Question number | 7 | Adapted from | N/A |
Question
The diagram shows a triangle ABC in which AC = 17 cm. M is the midpoint of AC.
Triangle ABM is equilateral.
Write down the size of angle MCB.
Write down the length of BM in cm.
Write down the size of angle BMC.
Calculate the length of BC in cm.
Markscheme
30° (A1) (C3)
[1 mark]
8.5 (cm) (A1)
[1 mark]
120° (A1)
[1 mark]
\(\frac{{{\text{BC}}}}{{\sin 120}} = \frac{{8.5}}{{\sin 30}}\) (M1)(A1)(ft)
Note: Award (M1) for correct substituted formula, (A1) for correct substitutions.
\({\text{BC}} = {\text{14}}{\text{.7}}\left( {\frac{{17\sqrt 3 }}{2}} \right)\) (A1)(ft)
[3 marks]
Examiners report
Part (a) was generally well answered with many candidates gaining full marks. Some candidates went on to make incorrect assumptions about triangle BMC being right angled and used Pythagorus theorem incorrectly. Those who used either the Sine rule or the Cosine rule correctly were generally able to substitute correctly and gain at least two marks.
Part (a) was generally well answered with many candidates gaining full marks. Some candidates went on to make incorrect assumptions about triangle BMC being right angled and used Pythagorus theorem incorrectly. Those who used either the Sine rule or the Cosine rule correctly were generally able to substitute correctly and gain at least two marks.
Part (a) was generally well answered with many candidates gaining full marks. Some candidates went on to make incorrect assumptions about triangle BMC being right angled and used Pythagorus theorem incorrectly. Those who used either the Sine rule or the Cosine rule correctly were generally able to substitute correctly and gain at least two marks.
Part (a) was generally well answered with many candidates gaining full marks. Some candidates went on to make incorrect assumptions about triangle BMC being right angled and used Pythagorus theorem incorrectly. Those who used either the Sine rule or the Cosine rule correctly were generally able to substitute correctly and gain at least two marks.
