we know that
in the right triangle FGJ
[tex]cos(FGJ)=\frac{GJ}{FG}= \frac{GJ}{3x-8}[/tex] ----> equation A
in the right triangle GJH
[tex]cos(JGH)=\frac{GJ}{GH}= \frac{GJ}{16}[/tex] ----> equation B
m∠(FGJ)=m∠(JGH) ------> because GJ is an angle bisector of ∠G
so
equate equation A and equation B
[tex]\frac{GJ}{3x-8}=\frac{GJ}{16}\\ 3x-8=16\\3x=24 \\x=8\ units[/tex]
that means
[tex]FG=3*8-8=16\ units[/tex]
[tex]FG=GH[/tex]-------> the triangle FGH is an isosceles triangle
hence
[tex]FJ=JH\\FJ=8\ units[/tex]
[tex]JH=8\ units[/tex]
[tex]FH=FJ+JH=8+8=16\ units[/tex]
therefore
the answer is
[tex]FH=16\ units[/tex]
