$\beta = 180^\circ - 100^\circ - (10^\circ + \alpha)$
$\beta = 70^\circ - \alpha$
$\dfrac{R_2}{\sin 100^\circ} = \dfrac{W}{\sin \beta}$
$\dfrac{R_2}{\sin 100^\circ} = \dfrac{W}{\sin (70^\circ - \alpha)}$
$R_2 = \dfrac{W \sin 100^\circ}{\sin (70^\circ - \alpha)}$
$\theta = 180^\circ - 80^\circ - (80^\circ - \alpha)$
$\theta = 20^\circ + \alpha$
$\dfrac{R_2}{\sin 80^\circ} = \dfrac{P}{\sin \theta}$
$R_2 = \dfrac{P \sin 80^\circ}{\sin \theta}$
$\dfrac{W \sin 100^\circ}{\sin (70^\circ - \alpha)} = \dfrac{W \sin 80^\circ}{\sin (20^\circ + \alpha)}$
$\dfrac{\sin 100^\circ}{\sin (70^\circ - \alpha)} = \dfrac{\sin 80^\circ}{\sin (20^\circ + \alpha)}$
$\sin 100^\circ \sin (20^\circ + \alpha) = \sin 80^\circ \sin (70^\circ - \alpha)$
$\sin 100^\circ (\sin 20^\circ \cos \alpha + \cos 20^\circ \sin \alpha) = \sin 80^\circ (\sin 70^\circ cos \alpha - \cos 70^\circ \sin \alpha)$
$\sin 100^\circ \sin 20^\circ cos \alpha + \sin 100^\circ \cos 20^\circ \sin \alpha = \sin 80^\circ \sin 70^\circ \cos \alpha - \sin 80^\circ \cos 70^\circ \sin \alpha$
$\sin 100^\circ \cos 20^\circ \sin \alpha + \sin 80^\circ \cos 70^\circ \sin \alpha = \sin 80^\circ \sin 70^\circ \cos \alpha - \sin 100^\circ \sin 20^\circ \cos \alpha$
$(\sin 100^\circ \cos 20^\circ + \sin 80^\circ \cos 70^\circ) \sin \alpha = (\sin 80^\circ \sin 70^\circ - \sin 100^\circ \sin 20^\circ) \cos \alpha$
$\dfrac{\sin \alpha}{\cos \alpha} = \dfrac{\sin 80^\circ \sin 70^\circ - \sin 100^\circ \sin 20^\circ}{\sin 100^\circ \cos 20^\circ + \sin 80^\circ \cos 70^\circ}$
$\tan \alpha = \dfrac{\sin 80^\circ \sin 70^\circ - \sin 100^\circ \sin 20^\circ}{\sin 100^\circ \cos 20^\circ + \sin 80^\circ \cos 70^\circ}$
$\tan \alpha = 0.4663076582$
$\alpha = 25^\circ$ answer
