| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Ambiguous case (two solutions) |
| Difficulty | Standard +0.8 This question requires applying the cosine rule in part (a), then recognizing and solving the ambiguous case of the sine rule in part (b) where two valid triangles exist. The ambiguous case requires students to identify both acute and obtuse angle solutions, which goes beyond routine application and tests deeper understanding of trigonometric problem-solving. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(BC^2 = 6^2 + 15^2 - 2 \times 6 \times 15 \times \cos 30°\) | M1 (AO 1.1a) | Attempt use of cosine rule. Allow either omission of 2, or \(+\) not \(-\), but no other errors. Allow other fully complete methods, such as basic trigonometry, possibly combined with Pythagoras. |
| \(BC = 10.3\) cm | A1 (AO 1.1) | Obtain 10.3 cm, or better. If \(> 3\) sf then allow 10.25, or answers that round to 10.25. Condone no units. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{\sin 30}{4} = \dfrac{\sin D}{6}\) | M1 (AO 1.1) | Attempt use of sine rule. Correct equation seen, with fractions either way up. Could also be implied by method e.g. \(\sin^{-1}(0.75)\) is M1, but just \(0.75\) is M0. Allow other fully complete methods. |
| \(D = 48.6°\) | A1 (AO 1.1) | Obtain \(D = 48.6°\), or better. \(D = 48.590377\ldots\) Allow \(D = 0.848\) radians. |
| or \(D = 131°\) | A1FT (AO 3.1a) | Obtain \(D = 131°\), or better. FT their first angle as long as \(< 150°\). A0 if additional angles given as well. Allow \(D = 2.29\) radians (could be FT on incorrect acute angle in radians, as long as \(D < 2.618\)). |
## Question 1:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $BC^2 = 6^2 + 15^2 - 2 \times 6 \times 15 \times \cos 30°$ | M1 (AO 1.1a) | Attempt use of cosine rule. Allow either omission of 2, or $+$ not $-$, but no other errors. Allow other fully complete methods, such as basic trigonometry, possibly combined with Pythagoras. |
| $BC = 10.3$ cm | A1 (AO 1.1) | Obtain 10.3 cm, or better. If $> 3$ sf then allow 10.25, or answers that round to 10.25. Condone no units. |
**[2 marks]**
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{\sin 30}{4} = \dfrac{\sin D}{6}$ | M1 (AO 1.1) | Attempt use of sine rule. Correct equation seen, with fractions either way up. Could also be implied by method e.g. $\sin^{-1}(0.75)$ is M1, but just $0.75$ is M0. Allow other fully complete methods. |
| $D = 48.6°$ | A1 (AO 1.1) | Obtain $D = 48.6°$, or better. $D = 48.590377\ldots$ Allow $D = 0.848$ radians. |
| or $D = 131°$ | A1FT (AO 3.1a) | Obtain $D = 131°$, or better. FT their first angle as long as $< 150°$. A0 if additional angles given as well. Allow $D = 2.29$ radians (could be FT on incorrect acute angle in radians, as long as $D < 2.618$). |
**[3 marks]**
---1 In the triangle $A B C$, the length $A B = 6 \mathrm {~cm}$, the length $A C = 15 \mathrm {~cm}$ and the angle $B A C = 30 ^ { \circ }$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the length $B C$.\\
$D$ is the point on $A C$ such that the length $B D = 4 \mathrm {~cm}$.
\item Calculate the possible values of the angle $A D B$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2023 Q1 [5]}}