Question 894880
A semicircle is inscribed inside a triangle ABC and with its center O 
lying on the side AC and Angle B = 90° and O divides AC such that AO = 15 cm and CO = 20 cm. Find the radius of the circle.
<pre>
  
{{{drawing(400,6200/19,-5+3,33+3,-5,26,
locate(18,13.5,theta),locate(2,1.5,theta),

arc(16,12,24,-24,217,397),locate(22,0,r),locate(22,12,r),
triangle(0,0,16,12,16,0),triangle(16,12,28,12,28,21),rectangle(16,0,28,12),
locate(0,0,C), locate(28,0,B), locate(28,22.5,A),locate(20.5,18,15),
locate(7,8,20),locate(16.25,6.2,r),locate(15,13,O),
locate(16,0,D),locate(28.3,12.4,E))}}}

Angles AOE and OCE are equal in measure.  Both are marked {{{theta}}}.

From the upper triangle,

{{{cos(theta)=r/15}}}
{{{r=15cos(theta)}}}

From the lower triangle,

{{{sin(theta)=r/20}}}
{{{r=20sin(theta)}}}

So setting the two expressions for r equal:

{{{15cos(theta)=20sin(theta)}}}

Divide both sides by {{{20cos(theta)}}}

{{{15cos(theta)/(20cos(theta))=20sin(theta)/(20cos(theta))}}}

{{{15cross(cos(theta))/(20cross(cos(theta)))=cross(20)sin(theta)/(cross(20)cos(theta))}}}

{{{15/20 = sin(theta)/cos(theta)}}}

{{{3/4=tan(theta)}}}

Since tangent = {{{opposite/(adjacent)}}} we draw a right
triangle with opposite=3 and adjacent=4.

hypotenuse = {{{sqrt(3^2+4^2)=sqrt(9+16)=sqrt(25)=5}}}

{{{drawing(100,80,-.5,4.5,-.6,3.4,triangle(0,0,4,0,4,3),
locate(1,.7,theta),locate(2,0,4),locate(4.1,1.5,3),locate(1.9,2.4,5))}}}


And since {{{r=15cos(theta)}}}
          {{{r=15*expr(4/5)}}}
          {{{r=12}}}

Edwin</pre>