Easier way of finding radius of circle inscribed in a scalene triangle given 2 sides and the included angle I have a scalene triangle with two sides given and the included angle.  I am solving for the radius of the inscribed circle.  See the image below.

I know that I can use the law of Cosines to find the length of the missing side
$c^2=a^2+b^2-2ab cosC$
Then I could use area of a triangle formula
$K=\frac12absin C$
Finally I could find and insert the perimeter and area into the formula below to solve for r.
$r=\frac{2A}p$
My equation for r would be:
$r=\frac{ab \sin{C}}{a+b+\sqrt{a^2+b^2-2ab \cos{C}}}$
Using the given data this would give me:
$r=\frac{119*202*sin(43)}{119+202+\sqrt{119^2+202^2-2*119*202*cos(43)}}=35.5$
Is there a simpler way to do this?
 A: using the cosine rule as you state the third side is $140.73$.
then, since the sides are tangents to the incircle, we may write:
$$
119 = x + y \\
202 = x+z \\
140.73 = y+z
$$
these equations give $x=90.136$
then 
$$
\begin{align}
r &= x \tan 21.5^{\circ}\\
& = 35.5
\end{align}
$$
(apologies if my calculations are in error, but i think the method suggested will work)
A: Since the centre of the incircle sits on the bisectors of the angles, you can decompose the triangle into three pairs of right triangles, by drawing those bisectors and each radius to the $3$ tangent points.
These small triangles each have altitude of the incircle radius and total base (across all six small triangles) equal to the perimeter, giving the formula for the incircle radius:
$$r_{\small I} = \frac{A_T}{s} $$
where $A_T$ is the triangle's area and $s$ is the semiperimeter (half the perimeter).
So you still need the third side as you calculated, to derive the (semi)perimeter, and then you can get the area from Heron's formula, 
$$A_T = \sqrt{s(s-a)(s-b)(s-c)}$$
or, combining these two, 
$$r_{\small I} = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$
In your case $s = (119+202+140.73)/2 = 230.87$ so $r_{\small I} = \sqrt{\frac{(111.87)(28.87)(90.14)}{230.87}} = \sqrt{1261} = 35.51$ .
