OCR MEI Paper 3 2021 November — Question 11 5 marks

Exam BoardOCR MEI
ModulePaper 3 (Paper 3)
Year2021
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSine and Cosine Rules
TypePoint on side of triangle
DifficultyChallenging +1.2 This question requires applying the cosine rule twice (once in triangle BCD to find angle BDC, then in triangle ABD to find AB) and using the supplementary angle relationship. While it involves multiple steps and careful angle work, it follows a standard problem-solving pattern for sine/cosine rule questions with all necessary information given. The 'exact answer' requirement adds minor complexity but the algebraic manipulation is straightforward.
Spec1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)

11 In this question you must show detailed reasoning. The diagram shows triangle ABC , with \(\mathrm { BC } = 8 \mathrm {~cm}\) and angle \(\mathrm { BAC } = 45 ^ { \circ }\).
The point D on AC is such that \(\mathrm { DC } = 5 \mathrm {~cm}\) and \(\mathrm { BD } = 7 \mathrm {~cm}\). \includegraphics[max width=\textwidth, alt={}, center]{a0d9573f-8273-4562-a2d3-07f15d9da1af-7_684_553_1119_258} Determine the exact length of AB .

Question 11:
AnswerMarks Guidance
Answer/WorkingMark Guidance
In triangle BDC, \(\cos D = \frac{7^2 + 5^2 - 8^2}{2 \times 7 \times 5}\)M1 3.1a — Use of cosine rule in triangle BDC (for any angle). Or \(\cos C = \frac{5^2+8^2-7^2}{2\times5\times8}\)
\(\cos D = \frac{1}{7}\)A1 1.1 — Or \(\cos C = \frac{1}{2}\), or \(C = 60°\)
\(\sin D = \sqrt{1 - \frac{1}{49}}\)
\(\sin D = \frac{\sqrt{48}}{7}\)M1 1.1 — Approximate values must not be seen to earn M1 i.e. must be exact. Or \(\sin C = \frac{\sqrt{3}}{2}\)
\(\frac{AB}{\sin D} = \frac{7}{\sin 45°} \Rightarrow \frac{7AB}{\sqrt{48}} = \frac{7\times 2}{\sqrt{2}}\)M1 3.1a — Use of sin rule in triangle ABD. Exact values must be seen to earn M1. Or use of sin rule in triangle ABC: \(\frac{2AB}{\sqrt{3}} = \frac{8\times2}{\sqrt{2}}\)
\(AB = 4\sqrt{6}\) [cm] oeA1 2.2a — Must be exact answer
[5]
## Question 11:

| Answer/Working | Mark | Guidance |
|---|---|---|
| In triangle BDC, $\cos D = \frac{7^2 + 5^2 - 8^2}{2 \times 7 \times 5}$ | M1 | 3.1a — Use of cosine rule in triangle BDC (for any angle). Or $\cos C = \frac{5^2+8^2-7^2}{2\times5\times8}$ |
| $\cos D = \frac{1}{7}$ | A1 | 1.1 — Or $\cos C = \frac{1}{2}$, or $C = 60°$ |
| $\sin D = \sqrt{1 - \frac{1}{49}}$ | | |
| $\sin D = \frac{\sqrt{48}}{7}$ | M1 | 1.1 — Approximate values must not be seen to earn M1 i.e. must be exact. Or $\sin C = \frac{\sqrt{3}}{2}$ |
| $\frac{AB}{\sin D} = \frac{7}{\sin 45°} \Rightarrow \frac{7AB}{\sqrt{48}} = \frac{7\times 2}{\sqrt{2}}$ | M1 | 3.1a — Use of sin rule in triangle ABD. Exact values **must** be seen to earn M1. Or use of sin rule in triangle ABC: $\frac{2AB}{\sqrt{3}} = \frac{8\times2}{\sqrt{2}}$ |
| $AB = 4\sqrt{6}$ [cm] oe | A1 | 2.2a — Must be exact answer |
| | **[5]** | |

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11 In this question you must show detailed reasoning.
The diagram shows triangle ABC , with $\mathrm { BC } = 8 \mathrm {~cm}$ and angle $\mathrm { BAC } = 45 ^ { \circ }$.\\
The point D on AC is such that $\mathrm { DC } = 5 \mathrm {~cm}$ and $\mathrm { BD } = 7 \mathrm {~cm}$.\\
\includegraphics[max width=\textwidth, alt={}, center]{a0d9573f-8273-4562-a2d3-07f15d9da1af-7_684_553_1119_258}

Determine the exact length of AB .

\hfill \mbox{\textit{OCR MEI Paper 3 2021 Q11 [5]}}