Answer:
x= 11°
Explanation:
∠BAC= ∠ACD (alt. ∠s, AB//CD)
∠BAC= 33°
∠BAC +∠ABC= ∠BCE (ext. ∠s of triangle ABC)
33° +2x= 5x
5x -2x= 33°
3x= 33°
x= 33° ÷3
x= 11°
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Alternatively,
∠ACB= 180° -5x (adj. ∠s on a str. line)
∠BCD +∠ABC= 180° (int. ∠s, AB//CD)
∠ACB +∠ACD +2x= 180°
180° -5x +33° +2x= 180°
Minus 180° on both sides:
-5x +2x +33°= 0
5x -2x= 33°
3x= 33°
x= 11°