17M.1.SL.TZ1.T_13

pestleMathematicsAASLPaper 117M· sl-3-2-2d-and-3d-trig-sine-rule-cosine-rule-areasource ↗

A triangular postage stamp, ABC, is shown in the diagram below, such that ${\text{AB}} = 5{\text{ cm}},{\rm{ B\hat AC}} = 34^\circ ,{\rm{ A\hat BC}} = 26^\circ$ and ${\rm{A\hat CB}} = 120^\circ$ .

M17/5/MATSD/SP1/ENG/TZ1/13

Find the length of BC.

[3]

Find the area of the postage stamp.

[3]

Markscheme / solution

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

$\frac{{{\text{BC}}}}{{\sin 34^\circ }} = \frac{5}{{\sin 120^\circ }}$ (M1)(A1)

Note: Award (M1) for substituted sine rule formula, (A1) for correct substitutions.

${\text{BC}} = 3.23{\text{ (cm) }}\left( {3.22850 \ldots {\text{ (cm)}}} \right)$ (A1) (C3)

[3 marks]

$\frac{1}{2}(5)(3.22850)\sin 26^\circ$ (M1)(A1)(ft)

Note: Award (M1) for substituted area of a triangle formula, (A1) for correct substitutions.

$= 3.54{\text{ }}({\text{c}}{{\text{m}}^2}){\text{ }}\left( {3.53820 \ldots {\text{ }}({\text{c}}{{\text{m}}^2})} \right)$ (A1)(ft) (C3)

Note: Follow through from part (a).

[3 marks]

Examiners’ report

[N/A]