a body is acted on by forces F1=16N,180°,F2= 50 N ,040°,F3=28,150°,calculate the magnitude (correct to one decimal)and the direction(to the nearest degree)of the resultant of the three forces

I don't know the answer help me solve it

$R_x = \Sigma F_x$

$R_x = 16 \cos 180^\circ + 50 \cos 040^\circ + 28 \cos 150^\circ$

$R_y = \Sigma F_y$

$R_x = 16 \sin 180^\circ + 50 \sin 040^\circ + 28 \sin 150^\circ$

$R = \sqrt{{R_x}^2 + {R_y}^2}$

Direction of resultant: $\tan \theta_x = \dfrac{R_y}{R_x}$

Just follow the solution above with your calculator.

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I don't know the answer help me solve it

$R_x = \Sigma F_x$

$R_x = 16 \cos 180^\circ + 50 \cos 040^\circ + 28 \cos 150^\circ$

$R_y = \Sigma F_y$

$R_x = 16 \sin 180^\circ + 50 \sin 040^\circ + 28 \sin 150^\circ$

$R = \sqrt{{R_x}^2 + {R_y}^2}$

Direction of resultant:

$\tan \theta_x = \dfrac{R_y}{R_x}$

Just follow the solution above with your calculator.