Higher November 2022 Paper 2 Q19
19.
Diagram not drawn to scale
In triangle \(ABC\), \(AB\) = 7 cm, \(AC\) = 9 cm and \(A\hat{B}C\) = 61°.
\(ACD\) is a straight line.
Find \(B\hat{C}D\). [4]
| Scheme | Marks | Comments |
|---|---|---|
| An explicit calculation for \(\sin BCA\) e.g. \(\sin BCA = 7 \times \dfrac{\sin 61}{9}\) | M2 | M1 for any correct implicit form e.g. \(\dfrac{\sin BCA}{7} = \dfrac{\sin 61}{9}\) |
| \(BCA =\) 42.86… si | A1 | Accept 43 |
| (\(B\hat{C}D\) = 180 − 42.86... = ) Answer in range 137.1° to 137.2° inclusive | B1 | FT ‘their 42.8 to 43’ providing at least M1 previously awarded. Accept 137 following complete working; degree symbol may be omitted |
| (4) |