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2015-09-21T13:11:27+05:30
I think you forgot to mention that AE and BC are parallel. I am assuming they are parallel.

If AE and BC are parallel, 
∠DAE = ∠ABC
⇒ x+19 = y - 11
⇒ x = y - 11 - 19
⇒ x = y - 30

∠BAC, ∠ABC and ∠ACB are the three angles of a triangle. Sum of these angles is 180°.
(2x+y)°+ (y-11)°+ (x+11)° = 180°
⇒ 2x+y + y - 11 + x + 11 = 180
⇒ 3x + 2y = 180
⇒ 3(y-30) + 2y = 180
⇒ 5y - 90 = 180
⇒ 5y = 270
⇒ y = 270/5 = 54°
x = y - 30 = 24°

∠DAE = 24+19 = 43°
∠BAC = 2×24+54=102°
∠ABC = 54-11 = 43°
∠ACB = 24+11 = 35°
2 5 2
2015-09-21T13:16:21+05:30

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The sum of angles in triangle ABC = 180 deg =>
       angle BAC + angle ABC  + angle ACB = 180°
          2x+y  +  y - 11 +  x + 11  = 180°
           3x + 2y = 180°   --- (1)
  
Angle DAE = angle ABC      as AE parallel to BC.  DAB is the transversal line.
          x+19 = y - 11
          y - x = 30°      --- (2)

(1) + 3 * (2)   =>   5 y = 270°      =>    y = 54°
         using  (2)  we get  :              x = 24°
 
Now the angles are
            ∠DAE = x+19° = 43°
            ∠BAC =  2x+y = 102°   
            ∠ ABC  = y - 11°  = 43°
            ∠ACB  = x + 11° = 35°
1 5 1
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