| 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.
Find the size of angle ADC.
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).
\(\frac{{{{12}^2} + {7^2} - {{12.8557...}^2}}}{{2 \times 12 \times 7}}\) (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).
