The resultant forces of the two given forces from the reference direction is 17.05 N.
The given parameters;
The x and y components of the Force A is calculated as follows;
[tex]F_x = F cos(\theta)\\\\F_x = 13 \times cos(30) = 11.258 \ N\\\\F_y = Fsin(\theta)\\\\F_y = 13 \times sin(30) = 6.5 \ N[/tex]
The x and y components of the Force B is calculated as follows;
[tex]F_x = 5 \times cos(72) = 1.545 \ N\\\\F_y = 5 \times sin(72) = 4.755 \ N[/tex]
The net force in x and y direction is calculated as;
[tex]\Sigma F_x = 11.258 \ N \ + 1.545 \ N = 12.803 \ N\\\\\Sigma F_y = 6.5 \ N \ + \ 4.755 \ N = 11.255 \ N[/tex]
The resultant forces is calculated as follows;
[tex]F = \sqrt{F_x^2 + F_y^2} \\\\F = \sqrt{12.803^2 +11.255^2} \\\\F = 17.05 \ N \\[/tex]
Thus, the resultant forces of the two given forces from the reference direction is 17.05 N.
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