(i) Write down the length of OM .

(ii) Find the length of VM .

Find the area of triangle VBC .

Calculate the volume of the pyramid.

Show that the angle between the line VM and the base of the pyramid is 52° correct to 2 significant figures.

Ahmed is at point P , a distance x metres from M on horizontal ground, as shown in the following diagram. The size of angle VPM is 27° . Q is a point on MP .

Write down the size of angle VMP .

Ahmed is at point P , a distance x metres from M on horizontal ground, as shown in the following diagram. The size of angle VPM is 27° . Q is a point on MP .

Using your value of VM from part (a)(ii), find the value of x.

Ahmed is at point P , a distance x metres from M on horizontal ground, as shown in the following diagram. The size of angle VPM is 27° . Q is a point on MP .

Ahmed walks 50 m from P to Q.

Find the length of QV, the distance from Ahmed to the vertex of the pyramid.

(i) 115.2 (m)     (A1)

Note: Accept 115 (m)

(ii) \(\sqrt{(146.5^2 + 115.2^2)}\)     (M1)

Note: Award (M1) for correct substitution.

186 (m) (186.368…)     (A1)(ft)(G2)

Note: Follow through from part (a)(i).

[3 marks]

\(\frac{1}{2} \times 230.4 \times 186.368...\)     (M1)

Note: Award (M1) for correct substitution in area of the triangle formula.

21500 m² (21469.6…m²)     (A1)(ft)(G2)

Notes: The final answer is 21500 m²; units are required. Accept 21400 m² for use of 186 m and/or 115 m.

[2 marks]

\(\frac{1}{3} \times 230.4^2 \times 146.5\)     (M1)

Note: Award (M1) for correct substitution in volume formula.

2590000 m³ (2592276.48 m³)     (A1)(G2)

Note: The final answer is 2590000 m³; units are required but do not penalise missing or incorrect units if this has already been penalised in part (b).

 [2 marks]

\(\tan^{-1}\left( {\frac{{146.5}}{{115.2}}} \right)\)     (M1)

Notes: Award (M1) for correct substituted trig ratio. Accept alternate correct trig ratios.

= 51.8203...= 52°     (A1)(AG)

Notes: Both the unrounded answer and the final answer must be seen for the (A1) to be awarded. Accept 51.96° = 52°, 51.9° = 52° or 51.7° = 52°

128°     (A1)

[1 mark]

\(\frac{{186.368}}{{\sin27}} = \frac{{x}}{{\sin25}}\)     (A1)(M1)(A1)(ft)

Notes: Award (A1)(ft) for their angle MVP seen, follow through from their part (e). Award (M1) for substitution into sine formula, (A1) for correct substitutions. Follow through from their VM and their angle VMP.

x = 173 (m) (173.490...)     (A1)(ft)(G3)

Note: Accept 174 from use of 186.4.

[4 marks]

VQ² = (186.368...)² + (123.490...)² − 2 × (186.368...) × (123.490...) × cos128     (A1)(ft)(M1)(A1)(ft)

Notes: Award (A1)(ft) for 123.490...(123) seen, follow through from their x (PM) in part (f), (M1) for substitution into cosine formula, (A1)(ft) for correct substitutions. Follow through from their VM and their angle VMP.

OR

173.490... − 50 = 123.490... (123)     (A1)(ft)

115.2 + 123.490... = 238.690...     (A1)(ft)

\(\text{VQ} = \sqrt{(146.5^2 + 238.690...^2)}\)     (M1)

VQ = 280 (m) (280.062...)     (A1)(ft)(G3)

Note: Accept 279 (m) from use of 3 significant figure answers.

[4 marks]

(a) This part was very well done on the whole.

(b) Amazingly badly done. Many candidates used 146.4 for the height and others tried unsuccessfully to find slant heights and angles to that they could use the area of a triangle formula \(\frac{{1}}{{2}}ab \sin C\).

(c) This was fairly well done.

(d) Quite a few candidates managed to show this although they did not always put down the unrounded answer and so lost the last mark. Some even tried to use 52° to verify its value.

.

(e) Very well done on the whole – even if part (d) was wrong.

(f) This was well done by those who attempted it. Not all candidates used VM to find x and so lost one mark. There were quite a few different methods of finding the answer.

(g) Again this was well done by those who attempted it. Again there were many different ways to reach the correct answer.