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Date May 2017 Marks available 6 Reference code 17M.1.sl.TZ1.3
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 3 Adapted from N/A

Question

The following diagram shows triangle PQR.

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

Find PR.

Markscheme

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]

Syllabus sections

Topic 3 - Circular functions and trigonometry » 3.6 » The sine rule, including the ambiguous case.
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