Write down the size of angle AQB.

Calculate the length of AQ.

Calculate the length of AC.

Show that the length of CQ is 50.37 m, correct to 4 significant figures.

Find the size of the angle AQC.

Calculate the total area of the glass needed to construct

(i) the two rectangular faces of the greenhouse;

(ii) the two triangular faces of the greenhouse.

The cost of one square metre of glass used to construct the greenhouse is 4.80 USD.

Calculate the cost of glass to make the greenhouse. Give your answer correct to the nearest 100 USD.

110°     (A1)

\(\frac{{AQ}}{{\sin 35^\circ }} = \frac{{10}}{{\sin 110^\circ }}\)     (M1)(A1)

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

OR

\(AQ = \frac{5}{{\cos 35^\circ }}\)     (A1)(M1)

Note: Award (A1) for 5 seen, (M1) for correctly substituted trigonometric ratio.

\(AQ = 6.10\) (6.10387...)     (A1)(ft)(G2)

Notes: Follow through from their answer to part (a).

\(AC^2 = 10^2 + 50^2\)     (M1)

Note: Award (M1) for correctly substituted Pythagoras formula.

\(AC = 51.0 (\sqrt{2600}, 50.9901...)\)     (A1)(G2)

\(QC^2 = (6.10387...)^2 + (50)^2\)     (M1)

Note: Award (M1) for correctly substituted Pythagoras formula.

\(QC = 50.3711...\)     (A1)

\(= 50.37\)     (AG)

Note: Both the unrounded and rounded answers must be seen to award (A1).

   If 6.10 is used then 50.3707... is the unrounded answer.

   For an incorrect follow through from part (b) award a maximum of (M1)(A0) – the given answer must be reached to award the final (A1)(AG).

\(\cos AQC = \frac{{{{(6.10387...)}^2} + {{(50.3711...)}^2} - {{(50.9901...)}^2}}}{{2(6.10387...)(50.3711...)}}\)     (M1)(A1)(ft)

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

= 92.4°   (\({92.3753...^\circ }\))     (A1)(ft)(G2)

Notes: Follow through from their answers to parts (b), (c) and (d). Accept 92.2 if the 3 sf answers to parts (b), (c) and (d) are used.

     Accept 92.5° (\({92.4858...^\circ }\)) if the 3 sf answers to parts (b), (c) and 4 sf answers to part (d) used.

(i) \(2(50 \times 6.10387...)\)     (M1)

Note: Award (M1) for their correctly substituted rectangular area formula, the area of one rectangle is not sufficient.

= 610 m² (610.387...)     (A1)(ft)(G2)

Notes: Follow through from their answer to part (b).

    The answer is 610 m². The units are required.

(ii) Area of triangular face \( = \frac{1}{2} \times 10 \times 6.10387... \times \sin 35^\circ \)     (M1)(A1)(ft)

OR

Area of triangular face \( = \frac{1}{2} \times 6.10387... \times 6.10387... \times \sin 110^\circ \)     (M1)(A1)(ft)

\(= 17.5051...\)

Note: Award (M1) for substituted triangle area formula, (A1)(ft) for correct substitutions.

OR

(Height of triangle) \( = {(6.10387...)^2} - {5^2}\)

\(= 3.50103...\)

Area of triangular face \( = \frac{1}{2} \times 10 \times their{\text{ }} height\)

\(= 17.5051...\)

Note: Award (M1) for substituted triangle area formula, (A1)(ft) for correctly substituted area formula. If 6.1 is used, the height is 3.49428... and the area of both triangular faces 34.9 m²

Area of both triangular faces = 35.0 m² (35.0103...)     (A1)(ft)(G2)

Notes: The answer is 35.0 m². The units are required. Do not penalize if already penalized in part (f)(i). Follow through from their part (b).

(610.387... + 35.0103...) × 4.80     (M1)

= 3097.90...     (A1)(ft)

Notes: Follow through from their answers to parts (f)(i) and (f)(ii).

    Accept 3096 if the 3 sf answers to part (f) are used.

= 3100     (A1)(ft)(G2)

Notes: Follow through from their unrounded answer, irrespective of whether it is correct. Award (M1)(A2) if working is shown and 3100 seen without the unrounded answer being given.

Most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates were able to recognize sine rule, substitute correctly and reach the required result.

Most candidates were able to recognize sine rule, substitute correctly and reach the required result. The use of Pythagoras’ theorem was also successful, the major source of error being the lack of unrounded and rounded answers in part (d).

Again, most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates were able to recognize sine rule, substitute correctly and reach the required result. Part (e) was less well answered, due in part to the triangle being in three dimensions. However, all three sides had either been asked for in previous parts or given and all that was required was a sketch of a triangle with the vertices labelled; such a diagram was never on any script and this technique should be encouraged.

Again, most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.