The resultant of two forces P and Q is 1200 N horizontally leftward. Determine the force Q and corresponding angle ? for the system of force s as shown in Fig. 1a.

Given,

$R_x = -1200\ N$

$R_y = 0$

Now,

$\sum F_x = R_x$

$\implies -Q\ sin\theta + 600\ cos\ 60 = -1200$

$\implies Q\ sin\ theta = 600\ cos\ 60 + 1200$

$\implies \therefore Q\ sin \theta = 1500 ......(i)$

Since, $R_y = 0$

$\implies -Q\ cos\theta + 60\ sin\ …