Can someone explain how you do this. i would really appreciate it.

Check the picture below.

now, let's recall that running a segment from a midpoint across to a midpoint in a triangle, creates a midsegment, and a midsegment which is parallel to the "base" of the triangle, is always half of that base.

now, noticing the picture on the top-left triangle, we know those points are midpoints, so those in red are midsegments and therefore half the base, to make it short

IC = HE/2 = 7

JB = IC/2 = 3.5

now onto the top-right triangle, which is the same thing just basing itself on its other end

CG = AH/2 = 10

DF = CG/2 = 5

now, let's go to the  picture on the bottom-center

we  know that DG = 4, and since D and G are midpoints, DG is the midsegment of CEH thus

CH = 2DG = 8

likewise, on the green triangle ACH, the midsegment IB is half of the base CH, we know CH = 8, so IB = 4.

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°