Higher June 2017 Unit 2 Q13
13.
Diagram not drawn to scale
By first calculating the size of \(B\hat{A}C\), calculate the area of triangle ABC.
You must show all your working. [5]
From the formula list: In any triangle \(ABC\): Sine rule \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\); Area of triangle \(= \dfrac{1}{2}ab\sin C\)
| Scheme | Marks | Comments |
|---|---|---|
| (sin BAC=) \(\dfrac{6.4 \times \text{sin } 46^\circ}{5.3}\) | M2 | M1 for \(\dfrac{\text{sin BAC}}{6.4} = \dfrac{\text{sin } 46^\circ}{5.3}\) or equivalent |
| 60.3(006…°) | A1 | Allow 60(°) from correct working. |
| Area = ½ × 5.3 × 6.4 × sin (180° - 46° - 60.3(006..°) | M1 | FT 'their derived 60.3(006….°)' |
| = 16.2(78….cm2) or 16.3(cm2) | A1 | Accept 16(cm2) from correct working. SC1 for 11.78(cm2). |