Triangle and circle

This one is a little difficult, Solveit

Look at the following illustration........

On CB, draw BE = BF = 5

And  arccos(CBA)  = 5/18 

And we can bisect this angle and form kite BFDE........where  triangles BFD and BDE are congruent right triangles by  SAS.......note that DF = DE   and  DEB and DFB are right angles, so any inscribed circle will have a radius that is tangent to CB  and AB at  E and F

And (1/2) of  angle CBA  = (1/2)arccos(5/18)

The x coordinate for the center of the circle = 0....   and the y coordinate lies on the bisector (BD) of base angle CBA  at  point D and is given by :

tan [ (1/2)arccos(5/18)]  = y / 5

5 tan [ (1/2)arccos(5/18)]  =  y  =  DF   = 3.75904705778

And this is the approximate radius of the inscribed circle......and DF  = DE  = 3.75904705778

Ho? Haha.. Sorry . :/

i found sqrt(299/9)

Ho Ho where u have been ?)

Haha, Missed me?!

yea too much

Haha, no need to cry, I'm here!

Ha :P

^-^.

Ive got a headache. :(

god bless you or how you saying be healthy

Water & Captin Crunch

Water & Captin Crunch bless you :)

haha :)

Thanks CPhill ! :)

Nice one Chris  :))

Thanks, Melody and Solveit......

i found way esear in the internet they devided (area)triangle into three triangles and their height will be radius of circles

r*18/2+r*18/2+r*10/2=sqrt(299)*10/2

23r=sqrt(299)*10/2

r=3.7590470577805614

Thanks, Solveit......that IS easier.......!!!

That's the thing about some problems - the  simplest way is usually the best, but there may be other [more complex] methods, too......looks like I chose one of those.......!!!!

don t worry you are not alone my teacher did the same :)