A playground, when viewed from above, is shaped like a quadrilateral, \({\text{ABCD}}\), where \({\text{AB}} = 21.8\,{\text{m}}\) and \({\text{CD}} = 11\,{\text{m}}\) . Three of the internal angles have been measured and angle \({\text{ABC}} = 47^\circ \) , angle \({\text{ACB}} = 63^\circ \) and angle \({\text{CAD}} = 30^\circ \) . This information is represented in the following diagram.

Calculate the distance \({\text{AC}}\).

Calculate angle \({\text{ADC}}\).

There is a tree at \({\text{C}}\), perpendicular to the ground. The angle of elevation to the top of the tree from \({\text{D}}\) is \(35^\circ \).

Calculate the height of the tree.

Chavi estimates that the height of the tree is \(6\,{\text{m}}\).

Calculate the percentage error in Chavi’s estimate.

Chavi is celebrating her birthday with her friends on the playground. Her mother brings a \(2\,\,{\text{litre}}\) bottle of orange juice to share among them. She also brings cone-shaped paper cups.

Each cup has a vertical height of \(10\,{\text{cm}}\) and the top of the cup has a diameter of \(6\,{\text{cm}}\).

Calculate the volume of one paper cup.

Calculate the maximum number of cups that can be completely filled with the \(2\,\,{\text{litre}}\) bottle of orange juice.

\(\frac{{21.8}}{{\sin 63^\circ }} = \frac{{{\text{AC}}}}{{\sin 47^\circ }}\)          (M1)(A1)

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

\(({\text{AC}} = )\,\,17.9\,({\text{m}})\,\,\,(17.8938...\,({\text{m}}))\)           (A1)(G2)

\(\frac{{11}}{{\sin 30}} = \frac{{17.8938...}}{{\sin {\text{ADC}}}}\)          (M1)(A1)(ft)

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

\(\left( {{\text{Angle ADC}} = } \right)\,\,\,54.4^\circ \,\,\,(54.4250...^\circ )\)          (A1)(ft)(G2)

Note: Accept \(54.5\,\,(54.4527...)\) or \(126\,\,(125.547...)\) from using their \(3\) sf answer.
Follow through from part (a). Accept \(125.575...\)

\(11 \times \tan 35^\circ \)  (or equivalent)          (M1)

Note: Award (M1) for correct substitution into trigonometric ratio.

\(7.70\,({\text{m}})\,\,\,(7.70228...\,({\text{m}}))\)          (A1)(G2)

\(\left| {\frac{{6 - 7.70228...}}{{7.70228...}}} \right| \times 100\,\% \)          (M1)

Note: Award (M1) for correct substitution into the percentage error formula.

OR

\(100 - \left| {\frac{{6 \times 100}}{{7.70228...}}} \right|\)          (M1)

Note: Award (M1) for the alternative method.

\(22.1\,(\% )\,\,\,(22.1009...\,(\% ))\)          (A1)(ft)(G2)

Note: Award at most (M1)(A0) for a final answer that is negative. Follow through from part (c).

\(\frac{1}{3}\pi  \times {3^2} \times 10\)          (A1)(M1)

Note: Award (A1) for \(3\) seen, (M1) for their correct substitution into volume of a cone formula.

\(94.2\,{\text{c}}{{\text{m}}^3}\,\,\,(30\pi \,{\text{c}}{{\text{m}}^3},\,\,94.2477...\,{\text{c}}{{\text{m}}^3})\)          (A1)(G3)

Note: The answer is \(94.2\,{\text{c}}{{\text{m}}^3}\), units are required. Award at most (A0)(M1)(A0) if an incorrect value for \(r\) is used.

\(\frac{{2000}}{{94.2477...}}\) OR \(\frac{2}{{0.0942477...}}\)          (M1)(M1)

Note: Award (M1) for correct conversion (litres to \({\text{c}}{{\text{m}}^3}\) or \({\text{c}}{{\text{m}}^3}\) to litres), (M1) for dividing by their part (e) (or their converted part (e)).

\(21\)          (A1)(ft)(G2)

Note: The final (A1) is not awarded if the final answer is not an integer. Follow through from part (e), but only if the answer is rounded down.

Question 4: Trigonometry and volumes of 3D solids
This question was done well by most candidates. Trigonometry was a real strength with competent use of the sine rule. A small minority treated CB as parallel to AB and hence used alternate angles. The lack of a diagram in part (c) held some candidates back as they struggled to form the correct trigonometric ratio. Percentage error in part (d) was generally good. Most candidates scored the two marks as their answer to part (c) was followed through in part (d). Some candidates are still giving negative answers to percentage error problems. The common mistake in this part was the use of the new value in the denominator rather than the original value. Part (f) was less successful, in general, with a number of candidates not able to do the conversion.

Question 4: Trigonometry and volumes of 3D solids
This question was done well by most candidates. Trigonometry was a real strength with competent use of the sine rule. A small minority treated CB as parallel to AB and hence used alternate angles. The lack of a diagram in part (c) held some candidates back as they struggled to form the correct trigonometric ratio. Percentage error in part (d) was generally good. Most candidates scored the two marks as their answer to part (c) was followed through in part (d). Some candidates are still giving negative answers to percentage error problems. The common mistake in this part was the use of the new value in the denominator rather than the original value. Part (f) was less successful, in general, with a number of candidates not able to do the conversion.

Question 4: Trigonometry and volumes of 3D solids
This question was done well by most candidates. Trigonometry was a real strength with competent use of the sine rule. A small minority treated CB as parallel to AB and hence used alternate angles. The lack of a diagram in part (c) held some candidates back as they struggled to form the correct trigonometric ratio. Percentage error in part (d) was generally good. Most candidates scored the two marks as their answer to part (c) was followed through in part (d). Some candidates are still giving negative answers to percentage error problems. The common mistake in this part was the use of the new value in the denominator rather than the original value. Part (f) was less successful, in general, with a number of candidates not able to do the conversion.

Question 4: Trigonometry and volumes of 3D solids
This question was done well by most candidates. Trigonometry was a real strength with competent use of the sine rule. A small minority treated CB as parallel to AB and hence used alternate angles. The lack of a diagram in part (c) held some candidates back as they struggled to form the correct trigonometric ratio. Percentage error in part (d) was generally good. Most candidates scored the two marks as their answer to part (c) was followed through in part (d). Some candidates are still giving negative answers to percentage error problems. The common mistake in this part was the use of the new value in the denominator rather than the original value. Part (f) was less successful, in general, with a number of candidates not able to do the conversion.

Question 4: Trigonometry and volumes of 3D solids
This question was done well by most candidates. Trigonometry was a real strength with competent use of the sine rule. A small minority treated CB as parallel to AB and hence used alternate angles. The lack of a diagram in part (c) held some candidates back as they struggled to form the correct trigonometric ratio. Percentage error in part (d) was generally good. Most candidates scored the two marks as their answer to part (c) was followed through in part (d). Some candidates are still giving negative answers to percentage error problems. The common mistake in this part was the use of the new value in the denominator rather than the original value. Part (f) was less successful, in general, with a number of candidates not able to do the conversion.

Question 4: Trigonometry and volumes of 3D solids
This question was done well by most candidates. Trigonometry was a real strength with competent use of the sine rule. A small minority treated CB as parallel to AB and hence used alternate angles. The lack of a diagram in part (c) held some candidates back as they struggled to form the correct trigonometric ratio. Percentage error in part (d) was generally good. Most candidates scored the two marks as their answer to part (c) was followed through in part (d). Some candidates are still giving negative answers to percentage error problems. The common mistake in this part was the use of the new value in the denominator rather than the original value. Part (f) was less successful, in general, with a number of candidates not able to do the conversion.