Denote the two force vectors by [tex]\vec F_1[/tex] and [tex]\vec F_2[/tex], respectively.
Compute the horizontal and vertical components:
[tex]F_{1,x}=(150\,\mathrm{lb})\cos40^\circ\approx114.91\,\mathrm{lb}[/tex]
[tex]F_{1,y}=(150\,\mathrm{lb})\sin40^\circ\approx96.42\,\mathrm{lb}[/tex]
[tex]F_{2,x}=(100\,\mathrm{lb})\cos170^\circ\approx-98.49\,\mathrm{lb}[/tex]
[tex]F_{2,y}=(100\,\mathrm{lb})\sin170^\circ\approx17.37\,\mathrm{lb}[/tex]
The resultant force [tex]\vec F=\vec F_1+\vec F_2[/tex] has components equal to the sum of the corresponding components of [tex]\vec F_1,\vec F_2[/tex]:
[tex]F_x=F_{1,x}+F_{2,x}\approx16.43\,\mathrm{lb}[/tex]
[tex]F_y=F_{1,y}+F_{2,y}\approx113.78\,\mathrm{lb}[/tex]
The resultant force then has magnitude
[tex]\|\vec F\|=\sqrt{{F_x}^2+{F_y}^2}\approx\boxed{114.96\,\mathrm{lb}}[/tex]
and direction [tex]\theta[/tex], where
[tex]\tan\theta=\dfrac{F_y}{F_x}\implies\boxed{\theta\approx81.79^\circ}[/tex]