$\Sigma F_x=200\cos(36.86)+100=260.0N( \rightarrow)$ $ \Sigma F_y=-200\sin36.86-80+40=-159.97N(\downarrow)$ ***Magnitude of resultant R:-*** ***$R=\sqrt{\Sigma F_x^2+\Sigma F_y^2}=\sqrt{260^2+159.97^2}=305.27N$*** ***Inclination of resultant***$\theta$:- $\theta=\tan^{-1}(\dfrac{\Sigma F_y}{\Sigma F_x})=31.60^0$ $\Sigma M_a=(200\sin(36.86)\times600)+(200\cos36.86\times200)-40000+(80\times300)-(100\times200)+(40\times200)=75987.56$ ***By Varignons Theorem we have:-*** ***$Y=\dfrac{\Sigma M_a}{R}=\dfrac{75987.56}{305.27}=248.91mm$*