$a = 16 \cos 30^\circ = 13.86 \text{ m}$
$b = 16 \sin 30^\circ = 8 \text{ m}$
$c = a \tan 37^\circ = 13.86 \tan 37^\circ = 10.44 \text{ m}$
Tension on cable CD
$\Sigma M_A = 0$
$(T \sin 53^\circ)(8 + 10.44) = 8000(13.86) + 12\,000(4)$
$T = 10\,788.47 \, \text{ lb}$ answer
Reaction at A
$\Sigma M_D = 0$
$A_x(8 + 10.44) = 8000(13.86) + 12\,000(4)$
$A_x = 8616.05 \, \text{ lb}$
$\Sigma F_V = 0$
$A_y + T \cos 53^\circ = 8000 + 12\,000$
$A_y + 10\,788.47 \cos 53^\circ = 8000 + 12\,000$
$A_y = 13\,507.34 \, \text{ lb}$
$R_A = \sqrt{{A_y}^2 + {A_x}^2}$
$R_A = \sqrt{8616.05^2 + 13\,507.34^2}$
$R_A = 16\,021.38 \, \text{ lb}$
$\tan \theta_{Ax} = \dfrac{A_y}{A_x} = \dfrac{13\,507.34}{8616.05}$
$\theta_{Ax} = 57.47^\circ$
Thus, RA = 16 021.38 lb at θAx = 57.47° with the horizontal. answer
Force on member CD
$\tan \beta = \dfrac{8}{13.86 - 8}$
$\beta = 53.78^\circ$
$\Sigma M_A = 0$
$(F_{BC} \sin \beta)(8) = 12\,000(4)$
$(F_{BC} \sin 53.78^\circ)(8) = 12\,000(4)$
$F_{BC} = 7437.21 \, \text{ lb tension}$ answer
