The net force
"\\bf F=F_1+F_2+F_3."In projections on the axes
"F_x=F_1 \\sin20\\degree-F_2+F_3 \\cos30\\degree\\\\\nF_y=F_1 \\cos20\\degree+0-F_3 \\sin30\\degree"Hence
"F_x=75 \\sin20\\degree-30+100 \\cos30\\degree=82.3 N,\\\\\nF_y=75 \\cos20\\degree-100 \\sin30\\degree=20.5 N."The magnitude of a net force
"F=\\sqrt{F_x^2+F_y^2}=\\sqrt{82.3^2+20.5^2}=84.8 N."The direction angle
"\\theta =\\tan^{-1}\\frac{F_y}{F_x }=\\tan^{-1}\\frac{20.5}{82.3}=14\\degree."Answer: "84.8\\rm N[N14\\degree E]."
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