Higher June 2018 Paper 4 Q18
18 Calculate the area of this triangle.
Not to scale
[6]| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 11.1 or 11.14 or 11.13[6…] or accept 11 with supporting working. | 6 | M3 for correct cos rule with cos as subject e.g. [cos = ] \(\dfrac{6.4^2 + 5.8^2 - 3.9^2}{2 \times 6.4 \times 5.8}\) or M2 for the above (M3) formula with one error or for \(3.9^2 = 6.4^2 + 5.8^2 - 2 \times 6.4 \times 5.8 \times \cos[.]\) or M1 for this (M2) formula with one error AND M2 for \(\frac{1}{2} \times 6.4 \times 5.8 \times \sin(\textit{their}\,36.87\ldots)\) or M1 for the use of this formula with one error | accept any correct method and they can find any angle, see additional guidance for the other angles this angle (opposite 3.9) is 36.87… which implies M3 |