Chapter 3 - Equilibrium of a Particle - Section 3.3 - Coplanar Force Systems - Problems - Page 99: 7

$T=30.5lb$ $\theta=20^{\circ}$

First we will sum the forces along rope D $(\searrow +) \sum \overrightarrow{F}_{x} =0 $ $60\cos10^{\circ}-T-T\cos\theta =0$ Now we will sum the forces perpendicularly to rope D $(\nearrow +) \sum \overrightarrow{F}_{y} =0 $ $T\sin\theta-60\sin10^{\circ}$ Solving the system of two equations, we obtain : $T=30.5lb$ $\theta=20^{\circ}$