20N.1.SL.TZ0.S_2

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

METHOD 1 – (sine rule)

evidence of choosing sine rule       (M1)

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

correct substitution       (A1)

eg   $\frac{{\displaystyle \raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{10}=\frac{\sin {\displaystyle }{\displaystyle \theta }}{15} , \frac{\sqrt{3}}{30}=\frac{{\displaystyle \sin \theta }}{15} , \frac{{\displaystyle \sqrt{3}}}{{\displaystyle 30}}=\frac{{\displaystyle \sin \text{B}}}{15}$

$\sin \theta =\frac{\sqrt{3}}{2}$        A1  N2

 

METHOD 2 – (perpendicular from vertex $\mathbf{\text{C}}$)

valid approach to find perpendicular length (may be seen on diagram)       (M1)

eg    , $\frac{h}{15}=\frac{\sqrt{3}}{3}$

correct perpendicular length       (A1)

eg    $\frac{15\sqrt{3}}{3} , 5\sqrt{3}$

$\sin \theta =\frac{\sqrt{3}}{2}$        A1  N2

 

Note: Do not award the final A mark if candidate goes on to state $\sin \theta =\frac{\mathrm{\pi }}{3}$, as this demonstrates a lack of understanding.

 

[3 marks]

attempt to substitute into double-angle formula for cosine       (M1)

$1-2{\left(\frac{\sqrt{3}}{3}\right)}^2, 2{\left(\frac{\sqrt{6}}{3}\right)}^2-1, {\left(\frac{\sqrt{6}}{3}\right)}^2-{\left(\frac{\sqrt{3}}{3}\right)}^2, \cos \left(2\theta \right)=1-2{\left(\frac{\sqrt{3}}{2}\right)}^2, 1-2 {\sin }^2\left(\frac{\sqrt{3}}{3}\right)$

correct working       (A1)

eg  $1-2\times \frac{3}{9}, 2\times \frac{6}{9}-1, \frac{6}{9}-\frac{3}{9}$

$\cos \left(2\times \text{C}\hat{\text{A}}\text{B}\right)=\frac{3}{9} \left(=\frac{1}{3}\right)$          A1  N2

[3 marks]

[N/A]

[N/A]