m∠ABX = 112
m∠CBX=68
1)
According to the segment addition postulate, in those 2 angles summed we have ∠ABC.
2) ABC is equal to 180º
∠ABX +∠CBX =∠ABC
Plugging into them
14x +70+20x+8=180º
34x +78 =180
34x +78-78=180-78
34x=102 : 34
x= 3
So m∠ABX = 14(3) +70 = 112 and m∠CBX =180-112=68
