posted by Nick .
find the area of a circle inscribed in triangle ABC where a=9, b=13, and the measure of angle C=38 degrees
You can get c with the law of cosines.
Then, draw the figure. You have a multitude of similar triangles. The idea is to use them to find the measure of radius, the distance on many of those triangles.
I've tried that multiple times, but I was told the answer should be more than one. I get something around .07
As suggested by bobpursley, use the cosine law to get c, I got 8.1
Now use the Sine Law to find angle A, which I got to be 43.16º
Let P be the centre of the incsribed triangle. A property of that circle is that its centre is the intersection of the angle bisectors of the triangle.
Consider triangle APC. Angle C is 19º, AC = 13 and angle A is 21.58º
(If you have a triangle with its base known and the two angles on that base A and C are known, then the height is given by
Height = base/(cot A + cot C)
in this case
height = 13/(cot19º + cot21.58º)
This is the radius of your circle, so the area is pi(2.393)^2
= 17.99 or 18 units