Write down the length of XM.

Use your graphic display calculator to find the standard deviation of the number of trees.

Calculate the length of VM.

Calculate the angle between VM and ABCD.

Calculate the length of the third side of the triangle. Give your answer correct to the nearest 10 m.

Calculate the area enclosed by the path that goes around the forest.

Inside the forest a second path forms the three sides of another triangle named ABC. Angle BAC is 53°, AC is 180 m and BC is 230 m.

Calculate the size of angle ACB.

UP applies in this question

(UP)     XM = 5 cm     (A1)

[1 mark]

16.8     (G1)

UP applies in this question

VM² = 5² + 8²     (M1)

Note: Award (M1) for correct use of Pythagoras Theorem.

(UP)     VM = \(\sqrt{89}\) = 9.43 cm     (A1)(ft)(G2)

[2 marks]

\(\tan {\text{VMX}} = \frac{8}{5}\)     (M1)

Note: Other trigonometric ratios may be used.

\({\rm{V\hat MX}} = 58.0^\circ \)     (A1)(ft)(G2)

[2 marks]

UP applies in this question

l² = 290² + 550² − 2 × 290 × 550 × cos115°     (M1)(A1)

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

l = 722     (A1)(G2)

(UP)     = 720 m     (A1)

Note: If 720 m seen without working award (G3).

The final (A1) is awarded for the correct rounding of their answer.

[4 marks]

UP applies in this question

\({\text{Area}} = \frac{1}{2} \times 290 \times 550 \times \sin 115\)     (M1)(A1)

Note: Award (M1) for substituted correct formula (A1) for correct substitution.

(UP)     = \(72\,300{\text{ }}{{\text{m}}^2}\)     (A1)(G2)

[3 marks]

\(\frac{{180}}{{\sin {\text{B}}}} = \frac{{230}}{{\sin 53}}\)     (M1)(A1)

Note: Award (M1) for substituted sine rule formula, (A1) for correct substitution.

B = 38.7°     (A1)(G2)

\({\operatorname{A\hat CB}} = 180 - (53^\circ  + 38.7^\circ )\)

\( = 88.3^\circ \)     (A1)(ft)

 [4 marks]

This part proved accessible to the great majority of candidates. The common errors were (1) the inversion of the tangent ratio (2) the omission of the units and (3) the incorrect rounding of the answer; with 58° being all too commonly seen.

A straightforward question that saw many fine attempts. Given its nature – where much of the work was done on the GDC – it must be emphasised to candidates that incorrect entry of data into the calculator will result in considerable penalties; they must check their data entry most carefully.

The use of the inappropriate standard deviation was seen, but infrequently.

This part proved accessible to the great majority of candidates. The common errors were (1) the inversion of the tangent ratio (2) the omission of the units and (3) the incorrect rounding of the answer; with 58° being all too commonly seen.

This part proved accessible to the great majority of candidates. The common errors were (1) the inversion of the tangent ratio (2) the omission of the units and (3) the incorrect rounding of the answer; with 58° being all too commonly seen.

Again, this part proved accessible to the majority with a large number of candidates attaining full marks. However, there were also a number of candidates who seemed not to have been prepared in the use of trigonometry in non right-angled triangles. Also, failing to round the answer in (a) to the nearest 10m was a common omission.

Again, this part proved accessible to the majority with a large number of candidates attaining full marks. However, there were also a number of candidates who seemed not to have been prepared in the use of trigonometry in non right-angled triangles. Also, failing to round the answer in (a) to the nearest 10 m was a common omission.

Again, this part proved accessible to the majority with a large number of candidates attaining full marks. However, there were also a number of candidates who seemed not to have been prepared in the use of trigonometry in non right-angled triangles. Also, failing to round the answer in (a) to the nearest 10 m was a common omission.