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?
Best Answer
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)