$Let\ '\theta'\ be\ the\ inclination\ of\ 40N\ force.$

$From\ fig,\sin \theta= \dfrac {3}{5}=0.6, \cos \theta=\dfrac {4}{5}=0.8$

$Let\ the\ resultant\ be\ 'R' acting\ at\ horizontal\ with\ an\ angle\ '\theta'$

$Along\ the\ X-axis\ we\ resolve\ the\ forces.$

$R_x = 15 + 40\cos\theta - 25\sin50$

$=15 + 40 \times 0.8 - 25 \times …