Can you find the resultant force between these two vectors? Determine the magnitude of the resultant force on an object if force $A$ is pulling the object with $150$ lbs of force and force $B$ is pulling with $300$ lbs, and the angle between the two forces is $110^\circ$.
 A: 1) Draw a diagram with coordinate axes:

2) Determine the components of the two forces:

$${F_A}_x = F_A = 150 lbs \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space {F_A}_y = 0$$
$$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space{F_B}_x = -F_B\cos70° = -103 lbs  \space\space\space\space\space\space\space\space\space\space{F_B}_y = F_B\sin70° = 282 lbs$$
3) Find the components of the resultant force
$$R_x = {F_A}_x + {F_B}_x = 47 lbs$$
$$R_y = {F_A}_y + {F_B}_y = 282lbs$$
4) Apply Pythagoras' theorem to get the magnitude of the resultant:
$$R = \sqrt {R_x^2 + R_y^2} = 286lbs$$
If you want to know more, here's a good article which explains each step in more detail: How to find the resultant force acting on an object
A: 
use this formula of triangle  $a^2=b^2+c^2-2bc\cdot \cos A $ where a,b,c are sides of triangle.
so resultant force $$\vec R^2={150}^2+{300}^2-2\cdot150\cdot 300 \cos 70^\circ$$
$$\vec R=\sqrt {22500+90000-30781.81}$$
$$\vec R=285.86\,lbs$$
There was typo in the formula which I have now corrected. 
a^2=b^2+c^2 - 2bc * cos A
A: If we assume $f_A$ points in the positive $x$ direction and $f_B$ is located in quadrant II, we have,
$$f_A = 
\left[ \begin{array}{c}
150\\
0 \end{array} \right]$$
and 
$$f_B = 
\left[ \begin{array}{c}
300\cos(110) \\
300\sin(110) \end{array} \right]$$
The resultant is 
$$f_R = f_A + f_B = 
\left[ \begin{array}{c}
150 + 300\cos(110) \\
300\sin(110) \end{array} \right]$$
then
$$\|f_R\| = \sqrt{(150 + 300\cos(110))^2 + (300\sin(110))^2}=285.86$$
Note:  Notice that $\cos(110)$ is negative.
