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]