In Δ ABC, let A = x°
By Angle-sum property,
A + B + C = 180°
But, it is given that C = 130°
So, x + B + 130 = 180
B = 180 - 130 - x
B = 50 - x
Since AD and BD are internal bisectors of A and B,
∠ DAB = x/2 and
∠ DBA = [tex]\frac{50-x}{2}[/tex]
[tex]=25-\frac{x}{2}[/tex]
In Δ ADB, by angle-sum property,
∠ DBA + ∠ DAB +∠ ADB = 180°
[tex]=(25-\frac{x}{2} )+\frac{x}{2} +[/tex] + ∠ ADB = 180°
25 + ∠ ADB = 180°
∠ ADB = 180 - 25 = 155°
Hence, ∠ ADB = 155°.
