What are the angles of a triangle which shares points with a regular pentagon adjacent to a square? [Image] My cousin had a geometry homework question, which asked to find the angles of a triangle inside the following shape
That was fairly simple. (The angles are 45, 54, and 81 degrees because both are regular polygons.)
I was wondering what would happen if the triangle instead was this
What would its angles be? I only know that angle ACG is 117 degrees; the other two seem to require some abstruse trigonometry work to calculate.

Draw a perpendicular line $AH$ to $FG$, let $AC=x$.
Then $AH=x+{1\over2}x\cdot tan(72)=({1\over2}tan(72)+1)x$.
Hence $AGF=arctan({({1\over2}tan(72)+1)x\over{1\over2}x})=78.86$.
$AGE=90-78.86=11.14$, $AGC=11.14+9=20.14$ and $GAC=180-117-20.14-=42.86$
An alternative way will be use pythagorean theorem to get $AG=\sqrt{({1\over2}tan(72)+1)^2x^2+({1\over2}x)^2}$ and use since law where $AGC=arcsin({AC\over({AG\over sin(117)})})$ you still get $AGC=20.14$ here.
I am sorry this is different from the comment, I must have entered something wrong into wolfram alpha.
