| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Bearings and navigation |
| Difficulty | Moderate -0.3 This is a standard C2 question combining sine/cosine rules with circle sector areas and basic trigonometry. Part (a) is routine bearing/triangle calculation (5 marks), while part (b) involves straightforward sector area formula, right-angled triangle work to find r, and combining areas. All techniques are standard C2 content with no novel insights required, making it slightly easier than average but still requiring careful multi-step work. |
| Spec | 1.05b 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 |
| Answer | Marks |
|---|---|
| 3 (a) | 10.62 + 9.22 − 2×10.6×9.2×cos68° |
| Answer | Marks |
|---|---|
| bearing = 174.9 to 175° | M1 |
| Answer | Marks |
|---|---|
| B1 | Or correct use of Cosine Rule |
| Answer | Marks |
|---|---|
| (i) | 2 |
| Answer | Marks |
|---|---|
| 3 | M1 |
| A1 | 6702.(…) to 2 s.f. or more |
| Answer | Marks |
|---|---|
| (ii) | DC = 80 sin() = 80 3 |
| Answer | Marks |
|---|---|
| 1385.64… to 3s.f. or more | B1 |
| Answer | Marks |
|---|---|
| A1 | both steps required |
| Answer | Marks |
|---|---|
| (iii) | area of ¼ circle = ½ × × (403)2 |
| Answer | Marks |
|---|---|
| = 4300 to 4320 | M1 |
| Answer | Marks |
|---|---|
| A1 | [=3769.9…] |
Question 3:
--- 3 (a) ---
3 (a) | 10.62 + 9.22 − 2×10.6×9.2×cos68°
oo..
QR = 11.1(3…)
sin68 sinQ or sinR oo.
their QR 9.2 10.6
Q = 50.01..° or R = 61.98..°
bearing = 174.9 to 175° | M1
A1
M1
A1
B1 | Or correct use of Cosine Rule
2 s.f. or better
--- 3 (b)
(i) ---
3 (b)
(i) | 2
(A) ½ × 8 2 ×
3
= 6400
3 | M1
A1 | 6702.(…) to 2 s.f. or more
--- 3 (b)
(ii) ---
3 (b)
(ii) | DC = 80 sin() = 80 3
3 2
Area = ½×their DA×40√3
or ½×40√3×80×sin(their DCA)
oo..
area of triangle = 8003 or
1385.64… to 3s.f. or more | B1
M1
A1 | both steps required
ss..o.
--- 3 (b)
(iii) ---
3 (b)
(iii) | area of ¼ circle = ½ × × (403)2
oo..
“6702” + “1385.6” – “3769.9”
= 4300 to 4320 | M1
M1
A1 | [=3769.9…]
i.e. their(b) (i) + their (b) (ii) – their
¼ circle o.e.
933⅓π + 8003\begin{enumerate}[label=(\alph*)]
\item A boat travels from P to Q and then to R. As shown in Fig. 11.1, Q is 10.6 km from P on a bearing of 045°. R is 9.2 km from P on a bearing of 113°, so that angle QPR is 68°.
\includegraphics{figure_4}
Calculate the distance and bearing of R from Q. [5]
\item Fig. 11.2 shows the cross-section, EBC, of the rudder of a boat.
\includegraphics{figure_5}
BC is an arc of a circle with centre A and radius 80 cm. Angle CAB = $\frac{2\pi}{3}$ radians.
EC is an arc of a circle with centre D and radius $r$ cm. Angle CDE is a right angle.
\begin{enumerate}[label=(\roman*)]
\item Calculate the area of sector ABC. [2]
\item Show that $r = 40\sqrt{3}$ and calculate the area of triangle CDA. [3]
\item Hence calculate the area of cross-section of the rudder. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 Q3 [13]}}