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

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

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

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

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

$=15 + 40 \times 0.8 - 25 \times 0.7660$

$= 27.8489 \rightarrow\ (1)$

• $Along\ the\ Y-axis\ we\ resolve\ the\ forces.$

$R_y = 15 + 40\sin\theta + 25\cos50$

$=15 + 40 \times 0.6 + 25 \times 0.6428$

$= 90.0697\rightarrow\ (2)$

• $From\ equation\ (1)\ and\ (2)\ we\ get$
• $Resultant\ force\ R = \sqrt{R_x^2+R_y^2}$

$ = \sqrt{(27.8489)^2+(90.0697)^2}$

$= 94.2768 N$

• $Inclination\ of\ the\ resultant\ \theta$

$\theta = tan^{-1}\left( \dfrac {R_y} {R_x}\right)$

$= tan^{-1}\left( \dfrac {90.0697} {27.8489}\right)$

$= 72.8188\ ^\circ$

• $Thus, Resultant\ force\ = 94.2768N\ and\ \angle 72.8188\ ^\circ$