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, ∠B=∠A=48°
Now,in Δ ABC
∠C+∠B+∠A= 180°
⇒48°+48°+∠C= 180°
⇒∠C=84°
