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Date May 2017 Marks available 2 Reference code 17M.2.sl.TZ2.9
Level SL only Paper 2 Time zone TZ2
Command term Find Question number 9 Adapted from N/A

Question

A ship is sailing north from a point A towards point D. Point C is 175 km north of A. Point D is 60 km north of C. There is an island at E. The bearing of E from A is 055°. The bearing of E from C is 134°. This is shown in the following diagram.

M17/5/MATME/SP2/ENG/TZ2/09

Find the bearing of A from E.

[2]
a.

Finds CE.

[5]
b.

Find DE.

[3]
c.

When the ship reaches D, it changes direction and travels directly to the island at 50 km per hour. At the same time as the ship changes direction, a boat starts travelling to the island from a point B. This point B lies on (AC), between A and C, and is the closest point to the island. The ship and the boat arrive at the island at the same time. Find the speed of the boat.

[5]
d.

Markscheme

valid method     (M1)

eg\(\,\,\,\,\,\)\(180 + 55,{\text{ }}360 - 90 - 35\)

235° (accept S55W, W35S)     A1     N2

[2 marks]

a.

valid approach to find \({\rm{A\hat EC}}\) (may be seen in (a))     (M1)

eg\(\,\,\,\,\,\)\({\rm{A\hat EC}} = 180 - 55 - {\rm{A\hat CE}},{\text{ }}134 = {\text{E}} + 55\)

correct working to find \({\rm{A\hat EC}}\) (may be seen in (a))     (A1)

eg\(\,\,\,\,\,\)\(180 - 55 - 46,{\text{ }}134 - 55\), \({\rm{A\hat EC}} = 79^\circ \)

evidence of choosing sine rule (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)\(\frac{a}{{\sin A}} = \frac{b}{{\sin B}}\)

correct substitution into sine rule     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{\text{CE}}}}{{\sin 55^\circ }} = \frac{{175}}{{\sin {\rm{A\hat EC}}}}\)

146.034

\({\text{CE}} = 146{\text{ (km)}}\)     A1     N2

[5 marks]

b.

evidence of choosing cosine rule     (M1)

eg\(\,\,\,\,\,\)\({\text{D}}{{\text{E}}^2} = {\text{D}}{{\text{C}}^2} + {\text{C}}{{\text{E}}^2} - 2 \times {\text{DC}} \times {\text{CE}} \times \cos \theta \)

correct substitution into right-hand side     (A1)

eg\(\,\,\,\,\,\)\({60^2} + {146.034^2} - 2 \times 60 \times 146.034\cos 134\)

192.612

\({\text{DE}} = 193{\text{ (km)}}\)     A1     N2

[3 marks]

c.

valid approach for locating B     (M1)

eg\(\,\,\,\,\,\)BE is perpendicular to ship’s path, angle \({\text{B}} = 90\)

correct working for BE     (A1)

eg\(\,\,\,\,\,\)\(\sin 46^\circ  = \frac{{{\text{BE}}}}{{146.034}},{\text{ BE}} = 146.034\sin 46^\circ ,{\text{ }}105.048\)

valid approach for expressing time     (M1)

eg\(\,\,\,\,\,\)\(t = \frac{d}{s},{\text{ }}t = \frac{d}{r},{\text{ }}t = \frac{{192.612}}{{50}}\)

correct working equating time     (A1)

eg\(\,\,\,\,\,\)\(\frac{{146.034\sin 46^\circ }}{r} = \frac{{192.612}}{{50}},{\text{ }}\frac{s}{{105.048}} = \frac{{50}}{{192.612}}\)

27.2694

27.3 (km per hour)     A1     N3

[5 marks]

d.

Examiners report

[N/A]
a.
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b.
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c.
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d.

Syllabus sections

Topic 3 - Circular functions and trigonometry » 3.6 » The cosine rule.
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