1. find the reactions at the supports a.

Explanation:

Step1: Set up moment - equilibrium about point A

Let the vertical reaction at A be $A_y$ and the horizontal reaction at A be $A_x$. The tension in the cable is $T$. Taking moments about point A, $\sum M_A=0$. The weight of the load is $F = 50kN$. The moment due to the load is $M_{load}=50\times(2 + 2+1)$ and the moment due to the vertical component of the tension $T$ is $T\sin40^{\circ}\times2$. So, $50\times(2 + 2 + 1)-T\sin40^{\circ}\times2=0$. $250-2T\sin40^{\circ}=0$ $T=\frac{250}{2\sin40^{\circ}}=\frac{250}{2\times0.6428}\approx194.5kN$

Step2: Set up force - equilibrium equations

In the x - direction, $\sum F_x = 0$. So, $A_x=T\cos40^{\circ}$. Substituting $T\approx194.5kN$, we get $A_x = 194.5\times\cos40^{\circ}=194.5\times0.766\approx149.0kN$. In the y - direction, $\sum F_y=0$. So, $A_y + T\sin40^{\circ}-50 = 0$. Substituting $T\approx194.5kN$, we have $A_y=50 - T\sin40^{\circ}=50-194.5\times0.6428=50 - 125=- 75kN$ (the negative sign indicates the direction is opposite to the assumed positive direction).

Answer:

The horizontal reaction at A, $A_x\approx149.0kN$ and the vertical reaction at A, $A_y\approx - 75kN$