The first force has magnitude 60 N and makes an angle of -30° relative to the positive horizontal axis (i.e. to the immediate right), while the second force has magnitude 150 N and direction +30° from the horizontal.
In component form, we have
[tex]\vec F_1 = F_{1,x}\,\vec\imath + F_{1,y}\,\vec\jmath \\\\ \vec F_1 = (60\,\mathrm N)\cos(-30^\circ)\,\vec\imath + (60 \,\mathrm N)\sin(-30^\circ)\,\vec\jmath \\\\ \vec F_1 \approx (52.0\,\mathrm N)\,\vec\imath - (30\,\mathrm N)\,\vec\jmath[/tex]
[tex]\vec F_2 = F_{2,x}\,\vec\imath + F_{2,y}\,\vec\jmath \\\\ \vec F_2 = (150\,\mathrm N)\cos(30^\circ)\,\vec\imath + (150\,\mathrm N)\sin(30^\circ)\,\vec\jmath \\\\ \vec F_2 \approx (130\,\mathrm N)\,\vec\imath + (75\,\mathrm N)\,\vec\jmath[/tex]
The resultant force is the vector
[tex]\vec F = \vec F_1 + \vec F_2 \\\\ \vec F \approx (182\,\mathrm N)\,\vec\imath + (45\,\mathrm N)\,\vec\jmath[/tex]
Its magnitude is
[tex]R = \|\vec F\| \\\\ R \approx \sqrt{(182\,\mathrm N)^2 + (45\,\mathrm N)^2} \\\\ R = 187.35\,\mathrm N \approx \boxed{190\,\mathrm N}[/tex]
Its direction [tex]\theta[/tex] is such that
[tex]\tan(\theta) \approx \dfrac{45\,\mathrm N}{182\,\mathrm N}[/tex]
Because the components of the resultant force are both positive, that means the angle [tex]\vec F[/tex] makes with the horizontal is between 0° and 90°. So
[tex]\theta \approx \tan^{-1}\left(\dfrac{45\,\mathrm N}{182\,\mathrm N}\right) \approx \boxed{14^\circ}[/tex]
