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If the total force is along the line L (in the figure) then the two x components cancel out and you use that property to find FB by setting the components equal to each other. And since the x components canceled out, the total force is just the sum of the
well, i don't know what angles you've got in your figure, but to find FB use: FBsin(thetaB)=FAsin(thetaA) and solve for FB: FB=FAsin(thetaA) / sin(thetaB) and for the magnitude of FA + FB: FAcos(thetaA) + FBcos(thetaB) hope that helps.

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