Find the length of AC.

Find the area of triangle ABC.

\(\frac{{{\text{AC}}}}{{\sin 30^\circ }} = \frac{8}{{\sin 40^\circ }}\)     (M1)(A1)

Note: Award (M1) for substitution in the sine rule formula, (A1) for correct substitutions.

AC = 6.22 (m) (6.22289…)     (A1)     (C3)

[3 marks]

Area of triangle \({\text{ABC}} = \frac{1}{2} \times 8 \times 6.22289... \times \sin 110^\circ \)     (M1)(A1)(ft)

Note: Award (M1) for substitution in the correct formula, (A1)(ft) for their correct substitutions. Follow through from their part (a).

Area triangle ABC = 23.4 m² (23.3904...m²)     (A1)(ft)     (C3)

Note: Follow through from a positive answer to their part (a). The answer is 23.4 m², units are required.

[3 marks]

Whilst using the sine rule to find an angle was tested in question 10, here the sine rule was required to be used to find a length. Many scripts showed the correct value of 6.22 m, but a significant number of candidates calculated AB instead of AC and the answer of 11.7 m proved to be a popular, but erroneous, answer. Some candidates turned the problem into two right angled triangles by dropping the perpendicular from C to the line AB. Much working was then required to find AC and again, some of these candidates simply determined the length of AB. Despite a significant number of candidates identifying the incorrect length in part (a), many of these recovered in part (b) to use their value correctly in the formula for the area of a triangle and many correct calculations were seen. Unfortunately, this good work was spoilt as some candidates either missed the units or gave the incorrect units in their final answer and, as a consequence, lost the last mark.

Whilst using the sine rule to find an angle was tested in question 10, here the sine rule was required to be used to find a length. Many scripts showed the correct value of 6.22 m, but a significant number of candidates calculated AB instead of AC and the answer of 11.7 m proved to be a popular, but erroneous, answer. Some candidates turned the problem into two right angled triangles by dropping the perpendicular from C to the line AB. Much working was then required to find AC and again, some of these candidates simply determined the length of AB. Despite a significant number of candidates identifying the incorrect length in part (a), many of these recovered in part (b) to use their value correctly in the formula for the area of a triangle and many correct calculations were seen. Unfortunately, this good work was spoilt as some candidates either missed the units or gave the incorrect units in their final answer and, as a consequence, lost the last mark.