Posted by **Nick** on Thursday, October 11, 2007 at 9:04pm.

find the area of a circle inscribed in triangle ABC where a=9, b=13, and the measure of angle C=38 degrees

- Pre Calculus/Trig -
**bobpursley**, Thursday, October 11, 2007 at 9:20pm
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.

- Pre Calculus/Trig -
**Reiny**, Thursday, October 11, 2007 at 10:06pm
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)

= 2.393

This is the radius of your circle, so the area is pi(2.393)^2

= 17.99 or 18 units

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