First we find [tex]\angle ACB[/tex]
[tex]\angle ACB = 180 - (4x - 100) = 180 - 4x + 100 = 280 - 4x[/tex]
Then we find a equation for all the interior angles
[tex]180 = \angle ACB + \angle BAC + \angle ABC\\= 280 - 4x + x + 20 + x\\300 - 2x[/tex]
We solve for x
[tex]180 = 300 - 2x\\180 - 300 = -2x\\-120 = -2x\\2x = 120\\x = 60\\[/tex]
Now we find the exterior angle
[tex]\angle DCB = 4x - 100 = 4 \cdot 60 - 100 = 240 - 100 = 140[/tex]
