Answer:   ∠A=48°,∠B=48°,∠C=84°.
Step by-step explanation:
 Given:  AD and BE are the angle bisectors  of ∠A and ∠B
i.e ∠6=∠7    ( ∵ Angles formed after  AD bisected ∠A)
∠4=∠5       ( ∵ Angles formed after  BE bisected ∠B)
Also,  DE║AB 
⇒ ∠2=∠7   (∵ Alternate interior angles)
    ∠3=∠6   (∵ Alternate interior angles)
And ∠ADE : ∠ADB =∠2:∠3= 2:9 =2x : 9x     ..(1)
To Find:  ∠A,∠B,∠C.
Solution:  ∠2=∠7 (∵ Given)    ...(2)
 ∠2=∠4  (∵ angles on the same segment)    ...(3)
∠4=∠5 =∠B/2 (∵ Given)    ...(4)
∴ In Δ ABD
 ∠3+∠4+∠5+∠7 = 180 (∵ Sum of interior angles of a triangle)
From equation 2,3,4,5, Put values
9x+2x+2x+2x =180°
⇒15x = 180° 
⇒x=12°
Putting values in equation (4) ⇒ ∠ B =2*(2*12) = 48°
 Also, <u>∠B=∠A=48°</u>
Now,in Δ ABC 
 ∠C+∠B+∠A= 180°
⇒48°+48°+∠C= 180°
<u><em>⇒∠C=84°</em></u>