18M.2.SL.TZ2.S_2

pestleMathematicsAASLPaper 218M· sl-3-2-2d-and-3d-trig-sine-rule-cosine-rule-areasource ↗

The following diagram shows quadrilateral ABCD.

${\text{AB}} = 11\,{\text{cm,}}\,\,{\text{BC}} = 6\,{\text{cm,}}\,\,{\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{D = 100}}^\circ {\text{, and C}}\mathop {\text{B}}\limits^ \wedge {\text{D = 82}}^\circ$

Find DB.

[3]

Find DC.

[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 $\frac{a}{{{\text{sin }}A}} = \frac{b}{{{\text{sin }}B}} = \frac{c}{{{\text{sin }}C}}$

correct substitution (A1)
eg $\frac{{{\text{DB}}}}{{{\text{sin }}59^\circ }} = \frac{{{\text{11}}}}{{{\text{sin }}100^\circ }}$

9.57429

DB = 9.57 (cm) A1 N2

[3 marks]

evidence of choosing cosine rule (M1)

eg ${a^2} = {b^2} + {c^2} - 2bc\,\,{\text{cos}}\,\,\left( A \right),\,\,\,{\text{D}}{{\text{C}}^2} = \,\,{\text{D}}{{\text{B}}^2}{\text{ + B}}{{\text{C}}^2}{\text{ }} - {\text{ 2DB}} \times \,\,{\text{BC}} \times \,{\text{cos}}\,\left( {{\text{D}}\mathop {\text{B}}\limits^ \wedge {\text{C}}} \right)$

correct substitution into RHS (A1)
eg ${9.57^2} + {6^2} - 2 \times 9.57 \times 6 \times \,\,{\text{cos}}\,\,82^\circ ,\,\,\,111.677$

10.5677

DC = 10.6 (cm) A1 N2

[3 marks]

Examiners’ report

[N/A]