Calculate the size of angle ADB.

Find the area of triangle ADB.

Calculate the size of angle BCD.

Show that the triangle ABC is not right angled.

$\cos {\text{ADB}} = \frac{{{{12}^2} + {{20}^2} - {{28}^2}}}{{2(12)(20)}}$     (M1)(A1)

Notes: Award (M1) for substituted cosine rule formula, (A1) for correct substitutions.

\(\angle {\text{ADB}} = 120\)     (A1)(G2)

[3 marks]

${\text{Area}} = \frac{{(12)(20)\sin {{120}^ \circ }}}{2}$     (M1)(A1)(ft)

Notes: Award (M1) for substituted area formula, (A1)(ft) for their correct substitutions.

$= 104{\text{ c}}{{\text{m}}^2}$ ($103.923 \ldots {\text{ c}}{{\text{m}}^2}$)     (A1)(ft)(G2)

Note: The final answer is $104{\text{ c}}{{\text{m}}^2}$ , the units are required. Accept $100{\text{ c}}{{\text{m}}^2}$ .

[3 marks]

$\frac{{\sin {\text{BCD}}}}{{12}} = \frac{{\sin {{60}^ \circ }}}{{13}}$     (A1)(ft)(M1)(A1)

Note: Award (A1)(ft) for their 60 seen, (M1) for substituted sine rule formula, (A1) for correct substitutions.

${\text{BCD}} = {53.1^ \circ }$ ($53.0736 \ldots$)     (A1)(G3)

Note: Accept $53$, do not accept $50$ or $53.0$.

[4 marks]

Using triangle ABC

$\frac{{\sin {\text{BAC}}}}{{13}} = \frac{{\sin {{53.1}^ \circ }}}{{28}}$     (M1)(A1)(ft)

OR

Using triangle ABD

$\frac{{\sin {\text{BAD}}}}{{12}} = \frac{{\sin {{120}^ \circ }}}{{28}}$     (M1)(A1)(ft)

Note: Award (M1) for substituted sine rule formula (one of the above), (A1)(ft) for their correct substitutions. Follow through from (a) or (c) as appropriate.

${\text{BAC}} = {\text{BAD}} = {21.8^ \circ }$ ($21.7867 \ldots$)     (A1)(ft)(G2)

Notes: Accept $22$, do not accept $20$ or $21.7$. Accept equivalent methods, for example cosine rule.

${180^ \circ } - ({53.1^ \circ } + {21.8^ \circ }) \ne {90^ \circ }$, hence triangle ABC is not right angled     (R1)(AG)

OR

$\frac{{{\text{CD}}}}{{\sin {{66.9}^ \circ }}} = \frac{{13}}{{\sin {{60}^ \circ }}}$     (M1)(A1)(ft)

Note: Award (M1) for substituted sine rule formula, (A1)(ft) for their correct substitutions. Follow through from (a) and (c).

${\text{CD}} = 13.8{\text{ }}(13.8075 \ldots )$     (A1)(ft)

${13^3} + {28^2} \ne {33.8^2}$, hence triangle ABC is not right angled.     (R1)(ft)(AG)

Note: The complete statement is required for the final (R1) to be awarded.

[4 marks]

The vast majority of candidates scored very well on this question. Those who did not attempted it using the trigonometry associated with right angled triangles. There were few problems with the use of radians and part (d), which was expected to prove challenging, was successfully overcome by more than half of the candidature. Problems arose mainly because of a lack of clarity in identifying the correct triangle.

The vast majority of candidates scored very well on this question. Those who did not attempted it using the trigonometry associated with right angled triangles. There were few problems with the use of radians and part (d), which was expected to prove challenging, was successfully overcome by more than half of the candidature. Problems arose mainly because of a lack of clarity in identifying the correct triangle.

The vast majority of candidates scored very well on this question. Those who did not attempted it using the trigonometry associated with right angled triangles. There were few problems with the use of radians and part (d), which was expected to prove challenging, was successfully overcome by more than half of the candidature. Problems arose mainly because of a lack of clarity in identifying the correct triangle.

The vast majority of candidates scored very well on this question. Those who did not attempted it using the trigonometry associated with right angled triangles. There were few problems with the use of radians and part (d), which was expected to prove challenging, was successfully overcome by more than half of the candidature. Problems arose mainly because of a lack of clarity in identifying the correct triangle.