Answer:

Applying Lamis Theorem

$\dfrac{T_{BC}}{sin 90^{\circ}}=\dfrac{R_A}{Sin150^{\circ}}=\dfrac{3}{Sin120^{\circ}}$

$\dfrac{T_{BC}}{sin 90^{\circ}}=\dfrac{3}{Sin120^{\circ}}$

$Tension\ in \ cable\ BC\ (T_{BC})=\dfrac{3}{Sin120^{\circ}}\times Sin150^{\circ}=2\sqrt{3} \ KN$

$\dfrac{R_A}{Sin150^{\circ}}=\dfrac{3}{Sin120^{\circ}}$

$Hinge \ reaction\ (R_A)=\dfrac{3}{Sin120^{\circ}}\times Sin150^{\circ}=\sqrt{3}\ \ KN$