Answer
$F=\dfrac{0.6 W \ \cos \theta}{\sin 12^{\circ}}$
Work Step by Step
We simplify the model as: $F=\dfrac{0.6 W(\sin \theta \cos 90^{\circ}+\cos \theta \sin 90^{\circ})}{\sin 12^{\circ}} \\=\dfrac{0.6 W(\sin \theta (0)+\cos \theta (1))}{\sin 12^{\circ}} \\=\dfrac{0.6 W \times (\cos \theta)}{\sin 12^{\circ}}$
