| Date | May 2010 | Marks available | 2 | Reference code | 10M.1.sl.TZ2.1 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Calculate | Question number | 1 | Adapted from | N/A |
Question
José stands 1.38 kilometres from a vertical cliff.
Express this distance in metres.
José estimates the angle between the horizontal and the top of the cliff as 28.3° and uses it to find the height of the cliff.
Find the height of the cliff according to José’s calculation. Express your answer in metres, to the nearest whole metre.
José estimates the angle between the horizontal and the top of the cliff as 28.3° and uses it to find the height of the cliff.
The actual height of the cliff is 718 metres. Calculate the percentage error made by José when calculating the height of the cliff.
Markscheme
1380 (m) (A1) (C1)
[1 mark]
\(1380\tan 28.3\) (M1)
\( = 743.05 \ldots \) . (A1)(ft)
\( = 743\) (m) (A1)(ft) (C3)
Notes: Award (M1) for correct substitution in tan formula or equivalent, (A1)(ft) for their 743.05 seen, (A1)(ft) for their answer correct to the nearest m.
[3 marks]
\({\text{percentage error}} = \frac{{743.05 \ldots - 718}}{{718}} \times 100\) (M1)
Note: Award (M1) for correct substitution in formula.
= 3.49 % (% symbol not required) (A1)(ft) (C2)
Notes: Accept 3.48 % for use of 743.
Accept negative answer.
[2 marks]
Examiners report
This question was well answered by the majority of candidates although it was surprising to find some who could not express the given distance in metres. Where working was present, follow through marks could be awarded in the remainder of the question. Most candidates could give their answer correct to the nearest metre and find the percentage error correctly, using the formula. A common error was to use the calculated value in the denominator.
This question was well answered by the majority of candidates although it was surprising to find some who could not express the given distance in metres. Where working was present, follow through marks could be awarded in the remainder of the question. Most candidates could give their answer correct to the nearest metre and find the percentage error correctly, using the formula. A common error was to use the calculated value in the denominator.
This question was well answered by the majority of candidates although it was surprising to find some who could not express the given distance in metres. Where working was present, follow through marks could be awarded in the remainder of the question. Most candidates could give their answer correct to the nearest metre and find the percentage error correctly, using the formula. A common error was to use the calculated value in the denominator.
