Higher November 2024 Paper 3 Q17
17 Here is triangle \(ABC\).
Work out the area of triangle \(ABC\).
Give your answer correct to 3 significant figures. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 56.0 | P1 | for a start to the process by correctly substituting into the cosine rule to find an angle eg, \(18.2^2 = 14.6^2 + 7.9^2 - 2 \times 14.6 \times 7.9 \times \cos A\) |
| P1 | for rearranging to find \(\cos A\), eg \(\cos A = \dfrac{14.6^2 + 7.9^2 - 18.2^2}{2 \times 14.6 \times 7.9}\ (= -0.2413\ldots)\) or \(A = 103.965\ldots\) | |
| P1 | for process to find the area of triangle \(ABC\) eg Area \(= \frac{1}{2} \times 7.9 \times 14.6 \times \sin(\text{``}103.965\ldots\text{''})\) or Area \(= \frac{1}{2} \times 7.9 \times 14.6 \times \sin[A]\) Area \(= \frac{1}{2} \times 14.6 \times 18.2 \times \sin(\text{``}24.912\ldots\text{''})\) or Area \(= \frac{1}{2} \times 14.6 \times 18.2 \times \sin[B]\) Area \(= \frac{1}{2} \times 7.9 \times 18.2 \times \sin(\text{``}51.1222\ldots\text{''})\) or Area \(= \frac{1}{2} \times 7.9 \times 18.2 \times \sin[C]\) | |
| A1 | for answer in the range 55.96 to 56.0 |
Additional guidance
\(\cos B = 0.9069\ldots\)
\(B = 24.912\ldots\)
\(\cos C = 0.6276\ldots\)
\(C = 51.1222\ldots\)
[\(A\)], [\(B\)], [\(C\)] must be a numerical value and clearly identified by labelling or on the diagram with no contradiction
If an answer is given in the range in working and then rounded incorrectly award full marks