Angles and forces 
Two forces of 40 pounds and 28 pounds act on an object. The angle between the two forces is 65 degrees. Find the magnitude of the resultant force to the nearest pound. Using this answer, find the measure of the angle formed between the resultant and the smaller force, to the nearest degree.

I saw this problem in a text while I was reading and was wondering how to do this problem. I was trying to teach myself about this topic and I would like to see how a problem like this would be solved. Can someone please show me.
 A: Hint 1: Using properties of the dot product, we get
$$
\begin{align}
(f_1+f_2)\cdot(f_1+f_2)
&=f_1\cdot f_1+2f_1\cdot f_2+f_2\cdot f_2\\
&=40^2+2\cdot40\cdot28\cos(65^\circ)+28^2
\end{align}
$$
Hint 2: The Law of Sines says
$$
\frac{\sin{\theta}}{40}=\frac{\sin(65^\circ-\theta)}{28}
$$
Hint 3: if we let $f$ be the magnitude of the resulting force, then we can also use the Law of Cosines to get
$$
40^2=28^2+f^2-2\cdot28f\cos(\theta)
$$

$\hspace{5.5cm}$
A: Given: 
$\theta=65^\circ$
$f_1=40lb$
$f_2=28lb$
use law of cosines SSA
$R^2=F_1+F_2-2F_1*F_2COS(65^\circ)$
find R by taking the square root
resulting angle is $(180-65)^\circ$
A: If you've learned to represent forces by vectors, you can place, say, the 40-pound force on the positive $ \ x-$ axis, so that it is represented as $ \ 40 \ \hat{i} \ $ .  The 28-pound force can then be set up as pointing in the direction 65º counterclockwise relative to the positive $ \ x-$ axis; it will then have an $ \ x-$ component of $ \ 28 \ \cos 65º \ \hat{i} \ $ and a $ \ y-$ component of  $ \ 28 \ \sin 65º \ \hat{j} \ $ .
We then add the $ \ x- $ and $ \ y-$ components separately to find the resultant force represented as $$ \ ( \ 40 \ + \ 28 \ \cos 65º \ ) \hat{i} \ + \ 28 \ \sin 65º \ \hat{j} \ \approx \ 51.8 \hat{i} \ + \ 25.4 \ \hat{j} \ \ \text{lbs} \ \ . $$
These two components form the legs of a right triangle.  The magnitude is found from the hypotenuse of that triangle, so we apply the Pythagorean Theorem.  The angle $ \ \theta \ $ that the result makes to the positive $ \ x-$ axis is given by 
$$ \tan \theta \ = \ \frac{\text{y-component}}{\text{x-component}} \ \approx \ \frac{25.4}{51.8} \ \ . $$
The size of the requested angle is the difference of that angle $ \ \theta \ $ from 65º . 

