User interface language: English | Español

Date November 2010 Marks available 3 Reference code 10N.2.sl.TZ0.3
Level SL only Paper 2 Time zone TZ0
Command term Find Question number 3 Adapted from N/A

Question

In the diagram below A, B and C represent three villages and the line segments AB, BC and CA represent the roads joining them. The lengths of AC and CB are 10 km and 8 km respectively and the size of the angle between them is 150°.

Find the length of the road AB.

[3]
a.

Find the size of the angle CAB.

[3]
b.

Village D is halfway between A and B. A new road perpendicular to AB and passing through D is built. Let T be the point where this road cuts AC. This information is shown in the diagram below.

Write down the distance from A to D.

[1]
c.

Show that the distance from D to T is 2.06 km correct to three significant figures.

[2]
d.

A bus starts and ends its journey at A taking the route AD to DT to TA.

Find the total distance for this journey.

[3]
e.

The average speed of the bus while it is moving on the road is 70 km h–1. The bus stops for 5 minutes at each of D and T .

Estimate the time taken by the bus to complete its journey. Give your answer correct to the nearest minute.

[4]
f.

Markscheme

AB2 = 102 + 82 – 2 × 10 × 8 × cos150°     (M1)(A1)

AB = 17.4 km     (A1)(G2)


Note: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer.

[3 marks]

a.

\(\frac{8}{{\sin {\text{C}}\hat {\rm A}{\text{B}}}} = \frac{{17.4}}{{\sin 150^\circ }}\)     (M1)(A1)

\({\text{C}}\hat {\rm A}{\text{B}} = 13.3^\circ \)     (A1)(ft)(G2)


Notes: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer. Follow through from their answer to part (a).

[3 marks]

b.

AD = 8.70 km (8.7 km)     (A1)(ft)


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

[1 mark]

c.

DT = tan (13.29...°) × 8.697... = 2.0550...     (M1)(A1)

= 2.06     (AG)


Notes: Award (M1) for correct substitution in the correct formula, award (A1) for the unrounded answer seen. If 2.06 not seen award at most (M1)(AO).

[2 marks]

d.

\(\sqrt {{{8.70}^2} + {{2.06}^2}}  + 8.70 + 2.06\)     (A1)(M1)

= 19.7 km     (A1)(ft)(G2)


Note: Award (A1) for AT, (M1) for adding the three sides of the triangle ADT, (A1)(ft) for answer. Follow through from their answer to part (c).

[3 marks]

e.

\(\frac{{19.7}}{{70}} \times 60 + 10\)     (M1)(M1)

= 26.9     (A1)(ft)


Note: Award (M1) for time on road in minutes, (M1) for adding 10, (A1)(ft) for unrounded answer. Follow through from their answer to (e).


= 27  (nearest minute)     (A1)(ft)(G3)


Note: Award (A1)(ft) for their unrounded answer given to the nearest minute.

[4 marks]

f.

Examiners report

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

a.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

b.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

c.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

d.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

e.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

f.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.3 » Use of the sine rule: \(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}\).
Show 42 related questions

View options