| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Exact trigonometric values |
| Difficulty | Moderate -0.3 This is a straightforward application of the cosine rule to find an angle, followed by using the area formula (1/2)ab sin C. Both parts are standard textbook exercises requiring direct formula application with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos BAC = \frac{5^2+9^2-10^2}{2\times5\times9}\) | M1 | 1.1a – Oe. Do not allow for a different angle found |
| \(= \frac{1}{15}\) | A1 [2] | 1.1 – Fraction must be seen in lowest terms. isw \(86.2°\) found |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin BAC = \sqrt{1-\cos^2 BAC}\ \left[= \frac{4\sqrt{14}}{15}\right]\) | B1 | 3.1a – FT their (a) |
| Area \(= \frac{1}{2}\times5\times9\times\sin BAC\) | M1 | 1.1a – Allow if value used for their angle |
| \(= 6\sqrt{14}\ \text{cm}^2\) | A1 [3] | 1.1 – Cao. Must be from exact working. Condone missing units. Use of \(\frac{1}{2}\times5\times9\times\sin86°\) or similar using their value for another angle found: B0 M1 A0 |
## Question 4:
**(a)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos BAC = \frac{5^2+9^2-10^2}{2\times5\times9}$ | M1 | 1.1a – Oe. Do not allow for a different angle found |
| $= \frac{1}{15}$ | A1 [2] | 1.1 – Fraction must be seen in lowest terms. isw $86.2°$ found |
**(b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin BAC = \sqrt{1-\cos^2 BAC}\ \left[= \frac{4\sqrt{14}}{15}\right]$ | B1 | 3.1a – FT their **(a)** |
| Area $= \frac{1}{2}\times5\times9\times\sin BAC$ | M1 | 1.1a – Allow if value used for their angle |
| $= 6\sqrt{14}\ \text{cm}^2$ | A1 [3] | 1.1 – Cao. Must be from exact working. Condone missing units. Use of $\frac{1}{2}\times5\times9\times\sin86°$ or similar using their value for another angle found: B0 M1 A0 |
---4 A triangle ABC has sides $\mathrm { AB } = 5 \mathrm {~cm} , \mathrm { AC } = 9 \mathrm {~cm}$ and $\mathrm { BC } = 10 \mathrm {~cm}$.
\begin{enumerate}[label=(\alph*)]
\item Find the cosine of angle BAC, giving your answer as a fraction in its lowest terms.
\item Find the exact area of the triangle.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2019 Q4 [5]}}