OCR MEI AS Paper 1 2024 June — Question 1 2 marks

Exam BoardOCR MEI
ModuleAS Paper 1 (AS Paper 1)
Year2024
SessionJune
Marks2
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSine and Cosine Rules
TypeBasic sine rule application
DifficultyModerate -0.8 This is a straightforward sine rule application requiring students to set up sin A/13 = sin 15°/10 and solve for A, then recognize the obtuse solution. It's simpler than average as it's a direct single-method question with clear given information, though the obtuse angle requirement adds minimal complexity.
Spec1.05b Sine and cosine rules: including ambiguous case
1 The triangle ABC has an obtuse angle at A . The angle at B is \(15 ^ { \circ }\). The length of AC is 10 cm and the length of BC is 13 cm . Calculate the size of the angle at A .
Question 1:
AnswerMarks Guidance
\(\frac{\sin A}{13} = \frac{\sin 15°}{10}\)M1 Uses the sine rule
\(A = 160.3°\)A1 Do not accept \(\sin^{-1}\left(\frac{13\sin 15°}{10}\right) = 19.7°\) as final answer
[Total: 2]
**Question 1:**

$\frac{\sin A}{13} = \frac{\sin 15°}{10}$ | M1 | Uses the sine rule

$A = 160.3°$ | A1 | Do not accept $\sin^{-1}\left(\frac{13\sin 15°}{10}\right) = 19.7°$ as final answer

**[Total: 2]**

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1 The triangle ABC has an obtuse angle at A . The angle at B is $15 ^ { \circ }$. The length of AC is 10 cm and the length of BC is 13 cm .

Calculate the size of the angle at A .

\hfill \mbox{\textit{OCR MEI AS Paper 1 2024 Q1 [2]}}
This paper (12 questions)
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Q1 2 Q2 2 Q3 3 Q4 4 Q5 6 Q6 6 Q7 6 Q8 10 Q9 7 Q10 8 Q11 6 Q12 10