A 1350-N uniform boom at phi = 64.5° to the horizontal is supported by a cable at an angle θ = 25.5° to the horizontal as shown in the figure below. The boom is pivoted at the bottom, and an object of weight w = 1850 N hangs from its top.
Here is a link for the diagram: http://www.webassign.net/sercp9/8-p-028.gif

Find the tension in the support cable (kN)
Find the components of the reaction force exerted by the pivot on the boom. (kN)

Hint: start by using sum of moments about the pivot point to find tension in the cable. The rest is then using sum of forces in the x- and y-directions equal 0.

Hint: start by using sum of moments about the pivot point to find tension in the cable. The rest is then using sum of forces in the x- and y-directions equal 0.

so I found the tension in the cable to be 1.45kN
and using the following equation I found Fy to be 2.58kN
Fy(Beam)+Fy(weight)-F(tension)*sin25.5

But now I am a little confused as to how to set up the equation to get Fx. I know that Fx=0 and the components that make up Fx are the x components of the beam, the weight and the tension in the cable, but I don't really know how to approach setting up the equation, and I am unsure of which angle to use as well.

The x-component of the weight and the boom are both zero, because the force of gravity is straight down (in the y-direction only) So you have:

\sum F_x = 0 = F_x(pivot) - F(tension) \cos(25.5)

Thank you so much!