Question 1150398
<pre>
Given: m∠A = α = 42º
       Area of ∆ACK = 50 c

{{{drawing(400,400,-5,5,-5,5,
locate(-4.2,.3,A), locate(4.1,.2,K),locate(0,0,O), locate(3.0,3,C), 
circle(0,0,4), line(-4,0,4,0), line(-4,0,2.9725793,2.6765224),
red(arc(-4,0,5,-5,0,21)),locate(-3,.65,alpha=42^o),


line(4,0,2.9725793,2.6765224) )}}}

∆ACK is a right triangle because it is inscribed in a semicircle.

Therefore its area is half the product of its legs:

{{{A=expr(1/2)(AC)(CK)}}}

{{{AC/AK=cos(42^o)}}}, {{{AC=AK*cos(42^o)}}},

{{{CK/AK=sin(42^o)}}}, {{{CK=AK*sin(42^o)}}},

{{{AK=diameter=2R}}}

{{{AC=2R*cos(42^o)}}},  {{{CK=2R*sin(42^o)}}},

Substituting in 

{{{A=expr(1/2)(AC)(CK)}}}

{{{50=expr(1/2)(2R*cos(42^o)^"")(2R*sin(42^o)^"")}}}

You finish.  Simplify that and solve for R.

Answer: R ≈ 7.090515777

Edwin</pre>