OCR C2 — Question 7 10 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSine and Cosine Rules
TypeTriangle with circular sector
DifficultyStandard +0.3 This is a straightforward multi-part question testing sine rule, triangle area formula, and sector area. All parts follow standard procedures with no novel insight required. Part (ii) is given as 'show that', making it easier. The combination of triangle and sectors is routine for C2 level, making this slightly easier than average.
Spec1.05b 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

7. \includegraphics[max width=\textwidth, alt={}, center]{de1a3480-0d83-43c2-a5a2-2f117b8a50fd-3_376_892_221_427} The diagram shows a design painted on the wall at a karting track. The sign consists of triangle \(A B C\) and two circular sectors of radius 2 metres and 1 metre with centres \(A\) and \(B\) respectively. Given that \(A B = 7 \mathrm {~m} , A C = 3 \mathrm {~m}\) and \(\angle A C B = 2.2\) radians,
  1. find the size of \(\angle A B C\) in radians to 3 significant figures,
  2. show that \(\angle B A C = 0.588\) radians to 3 significant figures,
  3. find the area of triangle \(A B C\),
  4. find the area of the wall covered by the design.

AnswerMarks Guidance
(i) \(\frac{\sin \theta}{3} = \frac{\sin 2.2}{7}\)M1
\(\sin B = \frac{7}{3} \sin 2.2\)
\(\angle ABC = 0.354\) (3sf)A1
(ii) \(\angle BAC = \pi - (2.2 + 0.3538) = 0.588\) (3sf)M1 A1
(iii) \(= \frac{1}{2} \times 3 \times 7 \times \sin 0.5878 = 5.82 \text{ m}^2\) (3sf)M1 A1
(iv) \(= 5.822 + [\frac{1}{2} \times 2^2 \times (2\pi - 0.5878)] + [\frac{1}{2} \times 1^2 \times (2\pi - 0.3538)]\)M3
\(= 20.2 \text{ m}^2\) (3sf)A1 (10)
**(i)** $\frac{\sin \theta}{3} = \frac{\sin 2.2}{7}$ | M1 |
$\sin B = \frac{7}{3} \sin 2.2$ | |
$\angle ABC = 0.354$ (3sf) | A1 |

**(ii)** $\angle BAC = \pi - (2.2 + 0.3538) = 0.588$ (3sf) | M1 A1 |

**(iii)** $= \frac{1}{2} \times 3 \times 7 \times \sin 0.5878 = 5.82 \text{ m}^2$ (3sf) | M1 A1 |

**(iv)** $= 5.822 + [\frac{1}{2} \times 2^2 \times (2\pi - 0.5878)] + [\frac{1}{2} \times 1^2 \times (2\pi - 0.3538)]$ | M3 |
$= 20.2 \text{ m}^2$ (3sf) | A1 | (10)

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7.\\
\includegraphics[max width=\textwidth, alt={}, center]{de1a3480-0d83-43c2-a5a2-2f117b8a50fd-3_376_892_221_427}

The diagram shows a design painted on the wall at a karting track. The sign consists of triangle $A B C$ and two circular sectors of radius 2 metres and 1 metre with centres $A$ and $B$ respectively.

Given that $A B = 7 \mathrm {~m} , A C = 3 \mathrm {~m}$ and $\angle A C B = 2.2$ radians,\\
(i) find the size of $\angle A B C$ in radians to 3 significant figures,\\
(ii) show that $\angle B A C = 0.588$ radians to 3 significant figures,\\
(iii) find the area of triangle $A B C$,\\
(iv) find the area of the wall covered by the design.\\

\hfill \mbox{\textit{OCR C2  Q7 [10]}}