Answer:
27°
Step-by-step explanation:
Let line AB be extended to touch line DC at point F, thereby forming triangle BCF.
∠ABC + ∠BCE = 180° (sum of interior angles in a transversal).
∠ABC + 48 = 180
∠ABC = 180 - 48 = 132°
Also in triangle ADF, ∠ADF + ∠DFA + ∠FAD = 180 (sum of angle in a straight line)
85 + ∠DFA + 25 = 180
∠DFA + 110 = 180
∠DFA = 180 - 110 = 70
∠DFA = 70
∠DFA + ∠BFC = 180° (sum of angle on a straight line)
∠BFC + 70 = 180
∠BFC = 180-70 = 110
∠BFC = 110°
∠ABC + ∠CBF = 180° (sum of angle on a straight line)
132 + ∠CBF = 180
∠CBF = 1800 - 132 = 42
∠CBF = 42°
In triangle BCF:
∠BFC + ∠CBF + x = 180° (sum of angle in a triangle)
42 + 110 + x = 180
x + 152 = 180
x = 180 - 152
x = 27°
