Work Shown:
Since the diameter AB cuts chord EF in half, we can prove AB is perpendicular to EF. So we have a right angle forming at ACF.
angle ACF = 90 degrees
angle ACF = 2(7x-4)
2(7x-4) = 90
14x-8 = 90
14x-8+8 = 90+8 .... add 8 to both sides
14x = 98
14x/14 = 98/14 ... divide both sides by 14
x = 7
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If x = 7, then
2(7x-4) = 2*(7*7-4) = 2*(49-4) = 2*45 = 90
this confirms the answer
