17M.1.SL.TZ1.S_3

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

The following diagram shows triangle PQR.

M17/5/MATME/SP1/ENG/TZ1/03

Find PR.

Markscheme / solution

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

METHOD 1

evidence of choosing the sine rule (M1)

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

correct substitution A1

eg $\,\,\,\,\,$ $\frac{x}{{\sin 30}} = \frac{{13}}{{\sin 45}},{\text{ }}\frac{{13\sin 30}}{{\sin 45}}$

$\sin 30 = \frac{1}{2},{\text{ }}\sin 45 = \frac{1}{{\sqrt 2 }}$ (A1)(A1)

correct working A1

eg $\,\,\,\,\,$ $\frac{1}{2} \times \frac{{13}}{{\frac{1}{{\sqrt 2 }}}},{\text{ }}\frac{1}{2} \times 13 \times \frac{2}{{\sqrt 2 }},{\text{ }}13 \times \frac{1}{2} \times \sqrt 2$

correct answer A1 N3

eg $\,\,\,\,\,$ ${\text{PR}} = \frac{{13\sqrt 2 }}{2},{\text{ }}\frac{{13}}{{\sqrt 2 }}{\text{ (cm)}}$

METHOD 2 (using height of ΔPQR)

valid approach to find height of ΔPQR (M1)

eg $\,\,\,\,\,$ $\sin 30 = \frac{x}{{13}},{\text{ }}\cos 60 = \frac{x}{{13}}$

$\sin 30 = \frac{1}{2}{\text{ or }}\cos 60 = \frac{1}{2}$ (A1)

${\text{height}} = 6.5$ A1

correct working A1

eg $\,\,\,\,\,$ $\sin 45 = \frac{{6.5}}{{{\text{PR}}}},{\text{ }}\sqrt {{{6.5}^2} + {{6.5}^2}}$

correct working (A1)

eg $\,\,\,\,\,$ $\sin 45 = \frac{1}{{\sqrt 2 }},{\text{ }}\cos 45 = \frac{1}{{\sqrt 2 }},{\text{ }}\sqrt {\frac{{169 \times 2}}{4}}$

correct answer A1 N3

eg $\,\,\,\,\,$ ${\text{PR}} = \frac{{13\sqrt 2 }}{2},{\text{ }}\frac{{13}}{{\sqrt 2 }}{\text{ (cm)}}$

[6 marks]

Examiners’ report

[N/A]