Using ‘distance = speed $\times$ time’, show that the distance from ${\text{A}}$ to ${\text{B}}$ is $1220$ metres correct to 3 significant figures.

The distance from ${\text{B}}$ to ${\text{C}}$ is $850$ metres. Running this part of the course takes Sarah $5$ minutes and $3$ seconds.

Calculate the speed, in ${\text{m}}{{\text{s}}^{ - 1}}$, that Sarah runs from ${\text{B}}$ to ${\text{C}}$.

The distance from ${\text{B}}$ to ${\text{C}}$ is $850$ metres. Running this part of the course takes Sarah $5$ minutes and $3$ seconds.

Calculate the distance, in metres, from ${\mathbf{C}}$ to ${\mathbf{A}}$.

The distance from ${\text{B}}$ to ${\text{C}}$ is $850$ metres. Running this part of the course takes Sarah $5$ minutes and $3$ seconds.

Calculate the total distance, in metres, of the cross-country running course.

The distance from ${\text{B}}$ to ${\text{C}}$ is $850$ metres. Running this part of the course takes Sarah $5$ minutes and $3$ seconds.

Find the size of angle ${\text{BCA}}$.

The distance from ${\text{B}}$ to ${\text{C}}$ is $850$ metres. Running this part of the course takes Sarah $5$ minutes and $3$ seconds.

Calculate the area of the cross-country course bounded by the lines ${\text{AB}}$, ${\text{BC}}$ and ${\text{CA}}$.

$3.8 \times 320$     (A1)

Note: Award (A1) for $320$ or equivalent seen.

$= 1216$     (A1)

$= 1220{\text{ (m)}}$     (AG)

Note: Both unrounded and rounded answer must be seen for the final (A1) to be awarded.

[2 marks]

$\frac{{850}}{{303}}{\text{ (m}}{{\text{s}}^{ - 1}}){\text{ (2.81, 2.80528}} \ldots {\text{)}}$     (A1)(G1)

[1 mark]

${\text{A}}{{\text{C}}^2} = {1220^2} + {850^2} - 2(1220)(850)\cos 110^\circ$     (M1)(A1)

Note: Award (M1) for substitution into cosine rule formula, (A1) for correct substitutions.

${\text{AC}} = 1710{\text{ (m) (1708.87}} \ldots {\text{)}}$     (A1)(G2)

Notes: Accept $1705{\text{ }} (1705.33…)$.

[3 marks]

$1220 + 850 + {\text{1708.87}} \ldots$     (M1)

$= {\text{3780 (m) (3778.87}} \ldots {\text{)}}$     (A1)(ft)(G1)

Notes: Award (M1) for adding the three sides. Follow through from their answer to part (c). Accept $3771{\text{ }} (3771.33…)$.

[2 marks]

$\frac{{\sin C}}{{1220}} = \frac{{\sin 110^\circ }}{{{\text{1708.87}} \ldots }}$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into sine rule formula, (A1)(ft) for correct substitutions. Follow through from their part (c).

$C = 42.1^\circ {\text{ (42.1339}} \ldots {\text{)}}$     (A1)(ft)(G2)

Notes: Accept $41.9^{\circ}, 42.0^{\circ}, 42.2^{\circ}, 42.3^{\circ}$.

OR

$\cos C = \frac{{{\text{1708.87}}{ \ldots ^2} + {{850}^2} - {{1220}^2}}}{{2 \times {\text{1708.87}} \ldots  \times 850}}$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into cosine rule formula, (A1)(ft) for correct substitutions. Follow through from their part (c).

$C = 42.1^\circ {\text{ (42.1339}} \ldots {\text{)}}$     (A1)(ft)(G2)

Notes: Accept $41.2^{\circ}, 41.8^{\circ}, 42.4^{\circ}$.

[3 marks]

$\frac{1}{2} \times 1220 \times 850 \times \sin 110^\circ$     (M1)(A1)(ft)

OR

$\frac{1}{2} \times {\text{1708.87}} \ldots  \times 850 \times \sin {\text{42.1339}} \ldots ^\circ$     (M1)(A1)(ft)

OR

$\frac{1}{2} \times 1220 \times {\text{1708.87}} \ldots  \times \sin {\text{27.8661}} \ldots ^\circ$     (M1)(A1)(ft)

Note: Award (M1) for substitution into area formula, (A1)(ft) for correct substitution.

$= 487\,000{\text{ }}{{\text{m}}^2}{\text{ (487}}\,{\text{230}} \ldots {\text{ }}{{\text{m}}^2})$     (A1)(ft)(G2)

Notes: The answer is $487\,000{\text{ }}{{\text{m}}^2}$, units are required.

     Accept $486\,000{\text{ }}{{\text{m}}^2}{\text{ (485}}\,{\text{633}} \ldots {\text{ }}{{\text{m}}^2})$.

     If workings are not shown and units omitted, award (G1) for $487\,000{\text{ or }}486\,000$.

     Follow through from parts (c) and (e).

[3 marks]