Higher June 2017 Paper 2 Q22
22.
Diagram not drawn to scale
Calculate the area of triangle \(BCD\). [9]
| Scheme | Marks | Comments |
|---|---|---|
| \(DB^2 = 4.2^2 + 2.6^2 - 2 \times 4.2 \times 2.6 \times \cos 55.5\)(°) | M1 | |
| \(DB^2 = 12.02968\ldots\) | m1 | Accept rounded or truncated |
| \(DB = 3.468\ldots\) (cm), or \(DB = \sqrt{12.02968\ldots}\) | A1 | Accept answers in the inclusive range 3.468 to 3.5 (cm) for DB |
| \(\sin ADB = \dfrac{2.6 \times \sin 55.5}{3.468(\ldots)}\) | M2 | FT ‘their DB’ provided not a value given in the question M1 for \(\dfrac{\sin ADB}{2.6} = \dfrac{\sin A}{DB}\) or equivalent |
| Angle \(ADB\) 38(.15…°) | A1 | Accept answers in the range 37.7(°) to 38.2(°) inclusive (from max DB = 3.5 cm to min DB 3.468…cm) |
| Angle \(BDC\) (= 82° – \(ADB\) =) 43.8(…°) | A1 | Accept answers in the range 43.8(°) to 44.3(°) inclusive (from DB = 3.468…cm to 3.5 cm) FT 82 – ‘their ADB’ provided previous M1 or M2 awarded |
| Area \(BDC\) = \(\tfrac12 \times 6.4 \times 3.468\ldots \times \sin 43.8\)(°) | M1 | FT ‘their DB’ and ‘their angle BDC’ provided at least 2 method marks (M2 or M1, M1 or M1 and m1) previously awarded |
| Answer in the inclusive range 7.68 to 7.82(…) (cm2) | A1 | CAO |
| (9) |