19M.2.SL.TZ2.S_3

pestleMathematicsAASLPaper 219M· sl-3-2-2d-and-3d-trig-sine-rule-cosine-rule-areasource ↗

The following diagram shows the quadrilateral ABCD.

AB = 6.73 cm, BC = 4.83 cm, BĈD = 78.2° and CD = 3.80 cm.

Find BD.

[3]

The area of triangle ABD is 18.5 cm². Find the possible values of θ.

[4]

Markscheme / solution

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

choosing cosine rule (M1)

eg ${c^2} = {a^2} + {b^2} - 2ab\,{\text{cos}}\,C$

correct substitution into RHS (A1)

eg ${4.83^2} + {3.80^2} - 2 \times 4.83 \times 3.80 \times {\text{cos}}\,78.2$ , $30.2622$ ,

${4.83^2} + {3.80^2} - 2\left( {4.83} \right)\left( {3.80} \right) \times {\text{cos}}\,1.36$

5.50111

5.50 (cm) A1 N2

[3 marks]

correct substitution for area of triangle ABD (A1)

eg $\frac{1}{2} \times 6.73 \times 5.50111\,{\text{sin}}\,\theta$

correct equation A1

eg $\frac{1}{2} \times 6.73 \times 5.50111\,{\text{sin}}\,\theta = 18.5$ , ${\text{sin}}\,\theta = {\text{0}}{\text{.999393}}$

88.0023, 91.9976, 1.53593, 1.60566

θ = 88.0 (degrees) or 1.54 (radians)

θ = 92.0 (degrees) or 1.61 (radians) A1A1 N2

[4 marks]

Examiners’ report

[N/A]