#b Is angle EFH, which is 90 - angle CFH.

Consider the right triangle CFH, CH is 10 and FH is 20, both given:

sin CFH = 10/20 = 1/2

So, 90 - sin^-1 (1/2) =

For # c and # d.

Angle ACB = BAC = 70 degrees

The length line from the mid-point of AC to the mid-point of GH is the length of the line from the mid-point of AC to B minus the length of the line from the mid-point GH to B:

14sin(70) - 4sin(70) = 9.397

Together with the line joining the mid-points of AC and DF, 10sqrt(3) = 17.321, from #a, and the line joining the mid-points of DF and GH you have a right triangle.

#c tan^-1 (17.321/9.397) =

#d Using the same triangle: (9.387)^2 + (17.321)^2) = 3888.302

sqrt 388.302 =

I hope this helps. I hope it's correct.