C2 January 2011 Q2
2. In the triangle \(ABC\), \(AB = 11\) cm, \(BC = 7\) cm and \(CA = 8\) cm.
| Scheme | Marks |
|---|---|
| \(11^2 = 8^2 + 7^2 - \left(2 \times 8 \times 7\cos C\right)\) | M1 |
| \(\cos C = \dfrac{8^2 + 7^2 - 11^2}{2 \times 8 \times 7}\) (or equivalent) | A1 |
| \(\left\{\hat{C} = 1.64228\ldots\right\} \Rightarrow \hat{C} = \text{awrt } 1.64\) | A1 cso |
| (3) |
Notes
M1 is also scored for \(8^2 = 7^2 + 11^2 - \left(2 \times 7 \times 11\cos C\right)\) or \(7^2 = 8^2 + 11^2 - \left(2 \times 8 \times 11\cos C\right)\) or \(\cos C = \dfrac{7^2 + 11^2 - 8^2}{2 \times 7 \times 11}\) or \(\cos C = \dfrac{8^2 + 11^2 - 7^2}{2 \times 8 \times 11}\)
1st A1: Rearranged correctly to make \(\cos C = \ldots\) and numerically correct (possibly unsimplified). Award A1 for any of \(\cos C = \dfrac{8^2 + 7^2 - 11^2}{2 \times 8 \times 7}\) or \(\cos C = \dfrac{-8}{112}\) or \(\cos C = -\dfrac{1}{14}\) or \(\cos C = \text{awrt } -0.071\).
SC: Also allow 1st A1 for \(112\cos C = -8\) or equivalent.
Also note that the 1st A1 can be implied for \(\hat{C} = \text{awrt } 1.64\) or \(\hat{C} = \text{awrt } 94.1^\circ\).
Special Case: \(\cos C = \dfrac{1}{14}\) or \(\cos C = \dfrac{11^2 - 8^2 - 7^2}{2 \times 8 \times 7}\) scores a SC: M1A0A0.
2nd A1: for awrt 1.64 cao
Note that \(A = 0.6876\ldots^c\ \left(\text{or } 39.401\ldots^\circ\right)\), \(B = 0.8116\ldots^c\ \left(\text{or } 46.503\ldots^\circ\right)\)
| Scheme | Marks |
|---|---|
| Use of Area \(\Delta ABC = \dfrac{1}{2}ab\sin(\text{their } C)\), where \(a, b\) are any of 7, 8 or 11. | M1 |
| \(= \dfrac{1}{2}(7 \times 8)\sin C\) using the value of their \(C\) from part (a). | A1 ft |
| \(\{= 27.92848\ldots \text{ or } 27.93297\ldots\} = \text{awrt } 27.9\) (from angle of either \(1.64^c\) or \(94.1^\circ\)) | A1 cso |
| (3) | |
| [6] |
Notes
M1: alternative methods must be fully correct to score the M1.
For any (or both) of the M1 or the 1st A1; their \(C\) can either be in degrees or radians.
Candidates who use \(\cos C = \dfrac{1}{14}\) to give \(C = 1.499\ldots\), can achieve the correct answer of awrt 27.9 in part (b). These candidates will score M1A1A0cso, in part (b).
Finding \(C = 1.499\ldots\) in part (a) and achieving awrt 27.9 with no working scores M1A1A0.
Otherwise with no working in part (b), awrt 27.9 scores M1A1A1.
Special Case: If the candidate gives awrt 27.9 from any of the below then award M1A1A1.
\(\dfrac{1}{2}(7 \times 11)\sin(0.8116^c \text{ or } 46.503^\circ) = \text{awrt } 27.9\), \(\dfrac{1}{2}(8 \times 11)\sin(0.6876\ldots^c \text{ or } 39.401\ldots^\circ) = \text{awrt } 27.9\).
Alternative: Hero’s Formula: \(A = \sqrt{13(13 - 11)(13 - 8)(13 - 7)} = \text{awrt } 27.9\), where M1 is attempt to apply \(A = \sqrt{s(s - 11)(s - 8)(s - 7)}\) and the first A1 is for the correct application of the formula.