Part (a)
$\tan 30^\circ = \dfrac{1000}{P}$
$P = 1732.05 \, \text{ lb}$ answer
$\sin 30^\circ = \dfrac{1000}{R}$
$R = 2000 \, \text{ lb}$ answer
Part (b)
$\dfrac{P}{\sin 60^\circ} = \dfrac{1000}{\sin (30^\circ + \theta)}$
$P = \dfrac{1000 \sin 60^\circ}{\sin (30^\circ + \theta)}$
$\dfrac{dP}{d\theta} = \dfrac{-1000 \sin 60^\circ \cos (30^\circ + \theta)}{\sin^2 (30^\circ + \theta)} = 0$
$-1000 \sin 60^\circ \cos (30^\circ + \theta) = 0$
$\cos (30^\circ + \theta) = 0$
$30^\circ + \theta = 90^\circ$
$\theta = 60^\circ$ answer
$P_{min} = \dfrac{1000 \sin 60^\circ}{\sin (30^\circ + 60^\circ)}$
$P_{min} = 866.02 \, \text{ lb}$ answer
$\alpha = 180^\circ - 60^\circ - (30^\circ + \theta)$
$\alpha = 180^\circ - 60^\circ - (30^\circ + 60^\circ)$
$\alpha = 30^\circ$
$\dfrac{R}{\sin \alpha} = \dfrac{1000}{\sin (30^\circ + \theta)}$
$\dfrac{R}{\sin 30^\circ} = \dfrac{1000}{\sin (30^\circ + 60^\circ)}$
$R = 500 \, \text{ lb}$ answer
