19M.1.SL.TZ1.S_3

pestleMathematicsAASLPaper 119M· sl-3-2-2d-and-3d-trig-sine-rule-cosine-rule-areasource ↗

The following diagram shows a right triangle ABC. Point D lies on AB such that CD bisects AĈB.

AĈD = θ and AC = 14 cm

Given that ${\text{sin}}\,\theta = \frac{3}{5}$ , find the value of ${\text{cos}}\,\theta$ .

[3]

Find the value of ${\text{cos}}\,2\theta$ .

[3]

Hence or otherwise, find ${\text{BC}}$ .

[2]

Markscheme / solution

valid approach (M1)

eg labelled sides on separate triangle, ${\text{si}}{{\text{n}}^2}\,x + {\text{co}}{{\text{s}}^2}\,x = 1$

correct working (A1)

eg missing side is 4, $\sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}}$

${\text{cos}}\,\theta = \frac{4}{5}$ A1 N3

[3 marks]

correct substitution into ${\text{cos}}\,2\theta$ (A1)

eg $2\left( {\frac{{16}}{{25}}} \right) - 1$ , $1 - 2{\left( {\frac{3}{5}} \right)^2}$ , $\frac{{16}}{{25}} - \frac{9}{{25}}$

${\text{cos}}\,2\theta = \frac{7}{{25}}$ A1 N2

[2 marks]

correct working (A1)

eg $\frac{7}{{25}} = \frac{{14}}{{{\text{BC}}}}$ , ${\text{BC}} = \frac{{14 \times 25}}{7}$

${\text{BC}} = 50$ (cm) A1 N2

[2 marks]

Examiners’ report

[N/A]