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Date May 2012 Marks available 3 Reference code 12M.1.sl.TZ1.10
Level SL only Paper 1 Time zone TZ1
Command term Calculate Question number 10 Adapted from N/A

Question

The diagram shows quadrilateral ABCD in which AB = 13 m , AD = 6 m and DC = 10 m. Angle ADC =120° and angle ABC = 40°.

Calculate the length of AC.

[3]
a.

Calculate the size of angle ACB.

[3]
b.

Markscheme

AC2 = 62 + 102 – 2×10×6×cos120°     (M1)(A1)

Note: Award (M1) for substitution in cosine formula, (A1) for correct substitutions.

 

AC = 14 (m)     (A1)     (C3)

[3 marks]

a.

\(\frac{{14}}{{\sin 40}} = \frac{{13}}{{\operatorname{sinACB} }}\)     (M1)(A1)(ft)

Note: Award (M1) for substitution in sine formula, (A1) for correct substitutions.

 

Angle ACB = 36.6° (36.6463...)     (A1)(ft)     (C3)

Note: Follow through from their (a).

[3 marks]

b.

Examiners report

In part (a), candidates seemed to be well drilled in the use of the cosine rule and AC = 14 m proved to be a popular, and correct, answer seen. The most popular incorrect answer seemed to be 11.7 m. This seems to have been arrived at by simple Pythagoras on triangle DCA – clearly, a totally incorrect method. Not as many candidates then went on to find the required angle in part (b). Some simply continued with using triangle DCA and found angle DCA. Indeed, some candidates even found angle DCB rather than the required angle ACB. In most cases, correct or otherwise, the sine rule was used and credit was awarded for this process.

a.

In part (a), candidates seemed to be well drilled in the use of the cosine rule and AC = 14 m proved to be a popular, and correct, answer seen. The most popular incorrect answer seemed to be 11.7 m. This seems to have been arrived at by simple Pythagoras on triangle DCA – clearly, a totally incorrect method. Not as many candidates then went on to find the required angle in part (b). Some simply continued with using triangle DCA and found angle DCA. Indeed, some candidates even found angle DCB rather than the required angle ACB. In most cases, correct or otherwise, the sine rule was used and credit was awarded for this process.

b.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.3 » Use of the cosine rule: \({a^2} = {b^2} + {c^2} - 2bc\cos A\) ; \(\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\).
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