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Date May 2013 Marks available 3 Reference code 13M.1.sl.TZ2.5
Level SL only Paper 1 Time zone TZ2
Command term Find Question number 5 Adapted from N/A

Question

The quadrilateral ABCD has AB = 10 cm, AD = 12 cm and CD = 7 cm.

The size of angle ABC is 100° and the size of angle ACB is 50°.

Find the length of AC in centimetres.

[3]
a.

Find the size of angle ADC.

[3]
b.

Markscheme

\(\frac{{{\text{AC}}}}{{\sin 100^\circ }} = \frac{{10}}{{\sin 50^\circ }}\)     (M1)(A1)


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


\(=12.9 (12.8557...)\)     (A1)     (C3)


Note: Radian answer is 19.3, award (M1)(A1)(A0).

a.

\(\frac{{{{12}^2} + {7^2} - {{12.8557...}^2}}}{{2 \times 12 \times 7}}\)     (M1)(A1)(ft)


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


= 80.5° (80.4994...°)     (A1)(ft)     (C3)


Notes: Follow through from their answer to part (a). Accept 80.9° for using 12.9. Using the radian answer from part (a) leads to an impossible triangle, award (M1)(A1)(ft)(A0).

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.3 » Use of the sine rule: \(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}\).
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