Trigonometry, SMB-057 Find \(\alpha\), to the nearest degree, such that \(\dfrac{\sin \alpha}{8} = \dfrac{\sin 60^{\circ}}{11} \) (2 marks) Show Answers Only \(\alpha=39^{\circ}\) Show Worked Solution \(\dfrac{\sin \alpha}{8}\) \(= \dfrac{\sin 60^{\circ}}{11} \) \(\sin \alpha\) \(= \dfrac{8 \times \sin 60^{\circ}}{11}\) \(\alpha\) \(=\sin^{-1} (0.6298) \) \(=39^{\circ}\ \text{(nearest degree)} \)