## Engineering Mechanics: Statics & Dynamics (14th Edition)

$78.7^{\circ};50.9KN$
We sum the forces as follows: $\Sigma F_y=0$ $\implies 100 sin\theta-T_A-\frac{12}{13}T_A=0$......eq(1) and $\Sigma F_x=0$ $\implies -100cos\theta+\frac{5}{13}T_A=0$......eq(2) Solving these two equations, we obtain: $\theta=78.7^{\circ}$ and $T_A=50.9KN$