Show that the length of side ST is 1.95 m, correct to 3 significant figures.

Calculate the area of the triangle PTS.

Write down the area of the rectangle STVR.

Calculate the total surface area of the tent, including the base.

Calculate the volume of the tent.

A pole is placed from V to M, the midpoint of PS.

Find in metres,

(i) the height of the tent, TM;

(ii) the length of the pole, VM.

Calculate the angle between VM and the base of the tent.

\({\text{ST}} = \frac{{1.6}}{{\cos 35^\circ }}\)     (M1)(A1)

Note: Award (M1) for correctly substituted trig equation, (A1) for 1.6 seen.

OR

\(\frac{{{\text{ST}}}}{{\sin 35^\circ }} = \frac{{3.2}}{{\sin 110^\circ }}\)     (M1)(A1)

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

ST = 1.95323...     (A1)

= 1.95 (m)     (AG)

Notes: Both unrounded and rounded answer must be seen for final (A1) to be awarded.

\(\frac{1}{2} \times 3.2 \times 1.95323... \times \sin 35^\circ \) or \(\frac{1}{2} \times 1.95323... \times 1.95323... \times \sin 110^\circ \)     (M1)(A1)

Note: Award (M1) for substituted area formula, (A1) for correct substitutions. Do not award follow through marks.

= 1.79 m² (1.79253...m²)     (A1)(G2)

Notes: The answer is 1.79 m², units are required. Accept 1.78955... from using 1.95.

OR

\(\frac{1}{2} \times 3.2 \times 1.12033...\)     (A1) (M1)

Note: Award (A1) for the correct value for TM (1.12033...) OR correct expression for TM (i.e. 1.6tan35°, \(\sqrt {{{(1.95323...)}^2} - {{1.6}^2}} \)), (M1) for correctly substituted formula for triangle area.

= 1.79 m² (1.79253...m²)     (A1)(G2)

Notes: The answer is 1.79 m², units are required. Accept 1.78 m² from using 1.95.

9.18 m² (9.18022 m²)     (A1)(G1)

Notes: The answer is 9.18 m², units are required. Do not penalize if lack of units was already penalized in (b). Do not award follow through marks here. Accept 9.17 m² (9.165 m²) from using 1.95.

\(2 \times 1.79253... + 2 \times 9.18022... + 4.7 \times 3.2\)     (M1)(A1)(ft)

Note: Award (M1) for addition of three products, (A1)(ft) for three correct products.

= 37.0 m² (36.9855...m²)     (A1)(ft)(G2)

Notes: The answer is 37.0 m², units are required. Accept 36.98 m² from using 3sf answers. Follow through from their answers to (b) and (c). Do not penalize if lack of units was penalized earlier in the question.

\(1.79253... \times 4.7\)     (M1)

Note: Award (M1) for their correctly substituted volume formula.

= 8.42 m³ (8.42489...m³)     (A1)(ft)(G2)

Notes: The answer is 8.42 m³, units are required. Accept 8.41 m³ from use of 1.79. An answer of 8.35, from use of TM = 1.11, will receive follow-through marks if working is shown. Follow through from their answer to part (b). Do not penalize if lack of units was penalized earlier in the question.

(i) \({\text{TM}} = 1.6\tan {35^\circ }\)     (M1)

Notes: Award (M1) for their correct substitution in trig ratio.

OR

\({\text{TM}} = \sqrt {{{(1.95323...)}^2} - {{1.6}^2}} \)     (M1)

Note: Award (M1) for correct substitution in Pythagoras’ theorem.

OR

\(\frac{{3.2 \times {\text{TM}}}}{2} = 1.79253...\)     (M1)

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

= 1.12 (m) (1.12033...)     (A1)(ft)(G2)

Notes: Follow through from their answer to (b) if area of triangle is used. Accept 1.11 (1.11467) from use of ST = 1.95.

(ii) \({\text{VM}} = \sqrt {{{1.12033...}^2} + {{4.7}^2}} \)     (M1)

Note: Award (M1) for their correct substitution in Pythagoras’ theorem.

= 4.83 (m) (4.83168 )     (A1)(ft)(G2)

Notes: Follow through from (f)(i).

\({\sin ^{ - 1}}\left( {\frac{{1.12033...}}{{4.83168...}}} \right)\)     (M1)

OR

\({\cos^{ - 1}}\left( {\frac{{4.7}}{{4.83168...}}} \right)\)     (M1)

OR

\({\tan^{ - 1}}\left( {\frac{{1.12033...}}{{4.7}}} \right)\)     (M1)

Note: Award (M1) for correctly substituted trig equation.

OR

\({\cos ^{ - 1}}\left( {\frac{{{{4.7}^2} + {{(4.83168...)}^2} - {{(1.12033...)}^2}}}{{2 \times 4.7 \times 4.83168...}}} \right)\)     (M1)

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

= 13.4° (13.4073...)     (A1)(ft)(G2)

Notes: Accept 13.3°. Follow through from part (f).