OCR MEI C2 — Question 5 12 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSine and Cosine Rules
TypeBearings and navigation
DifficultyModerate -0.3 This is a standard C2 trigonometry question involving the cosine rule, sine rule, bearings, and arc length - all routine applications of formulae. Part (i) requires calculating triangle side lengths and converting between angles and bearings, while part (ii) applies the arc length formula. These are textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta

  1. The course for a yacht race is a triangle, as shown in Fig. 11.1. The yachts start at A, then travel to B, then to C and finally back to A. \includegraphics{figure_7}
    1. Calculate the total length of the course for this race. [4]
    2. Given that the bearing of the first stage, AB, is 175°, calculate the bearing of the second stage, BC. [4]
  2. Fig. 11.2 shows the course of another yacht race. The course follows the arc of a circle from P to Q, then a straight line back to P. The circle has radius 120 m and centre O; angle POQ = 136°. \includegraphics{figure_8} Calculate the total length of the course for this race. [4]

Question 5:
AnswerMarks
5iA
iB
AnswerMarks
iiBC2 = 3482 + 3022 − 2 × 348 ×
302 × cos 72°
BC = 383.86…
1033.86…[m] or ft 650 + their BC
sinB sin72
=
302 their BC
B = 48.4..
355 − their B o.e.
answer in range 306 to307
224
Arc length PQ = ×2π×120
360
oo..e. or 469.1... to 3 sf or mo
QP = 222.5…to 3 sf or more
AnswerMarks
answer in range 690 to 692 [m]M2
A1
1
M1
A1
M1
A1
M2
B1
AnswerMarks
A1M1 for recognisable attempt at
Cosine Rule
to 3 sf or more
accept to 3 sf or more
Cosine Rule acceptable or Sine Rule
to find C
or 247 + their C
136
M1 for ×2π×120
AnswerMarks
3604
4
4
Question 5:
5 | iA
iB
ii | BC2 = 3482 + 3022 − 2 × 348 ×
302 × cos 72°
BC = 383.86…
1033.86…[m] or ft 650 + their BC
sinB sin72
=
302 their BC
B = 48.4..
355 − their B o.e.
answer in range 306 to307
224
Arc length PQ = ×2π×120
360
oo..e. or 469.1... to 3 sf or mo
QP = 222.5…to 3 sf or more
answer in range 690 to 692 [m] | M2
A1
1
M1
A1
M1
A1
M2
B1
A1 | M1 for recognisable attempt at
Cosine Rule
to 3 sf or more
accept to 3 sf or more
Cosine Rule acceptable or Sine Rule
to find C
or 247 + their C
136
M1 for ×2π×120
360 | 4
4
4
\begin{enumerate}[label=(\roman*)]
\item The course for a yacht race is a triangle, as shown in Fig. 11.1. The yachts start at A, then travel to B, then to C and finally back to A.

\includegraphics{figure_7}

\begin{enumerate}[label=(\Alph*)]
\item Calculate the total length of the course for this race. [4]
\item Given that the bearing of the first stage, AB, is 175°, calculate the bearing of the second stage, BC. [4]
\end{enumerate}

\item Fig. 11.2 shows the course of another yacht race. The course follows the arc of a circle from P to Q, then a straight line back to P. The circle has radius 120 m and centre O; angle POQ = 136°.

\includegraphics{figure_8}

Calculate the total length of the course for this race. [4]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C2  Q5 [12]}}