17N.2.SL.TZ0.S_1

pestleMathematicsAASLPaper 217N· sl-3-2-2d-and-3d-trig-sine-rule-cosine-rule-areasource ↗

The following diagram shows a triangle ABC.

N17/5/MATME/SP2/ENG/TZ0/01

${\text{AB}} = 5{\rm{ cm, C\hat AB}} =$ 50° and ${\rm{A\hat CB}} =$ 112°

Find BC.

[3]

Find the area of triangle ABC.

[3]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

evidence of choosing sine rule (M1)

eg $\,\,\,\,\,$ $\frac{{\sin A}}{a} = \frac{{\sin B}}{b}$

correct substitution (A1)

eg $\,\,\,\,\,$ $\frac{{{\text{BC}}}}{{\sin 50}} = \frac{5}{{\sin 112}}$

4.13102

${\text{BC}} = 4.13{\text{ (cm)}}$ A1 N2

[3 marks]

correct working (A1)

eg $\,\,\,\,\,$ ${\rm{\hat B}} = 180 - 50 - 112$ , 18°, ${\text{AC}} = 1.66642$

correct substitution into area formula (A1)

eg $\,\,\,\,\,$ $\frac{1}{2} \times 5 \times 4.13 \times \sin 18,{\text{ }}0.5(5)(1.66642)\sin 50,{\text{ }}\frac{1}{2}(4.13)(1.66642)\sin 112$

3.19139

${\text{area}} = 3.19{\text{ (c}}{{\text{m}}^2}{\text{)}}$ A1 N2

[3 marks]

Examiners’ report

[N/A]