An exterior angle of a triangle is 115° and one of the interior opposite angles is 35°. Find the other two angles.

In Δ ABC, let ∠ A = 35° and exterior ∠ ACD = 115°.
Exterior ∠ ACD = ∠ A +∠ B
⇒ 115° = 35° + ∠ B
⇒ ∠ B = 115° - 35° = 80°
In Δ ABC, ∠ A + ∠ B + ∠ C = 180° [Sum of the three angles of a triangle is 180° ]
⇒ 35° + 80° + ∠ C = 180°
⇒ ∠ C = 180° - (80° +35° )
        = 180° - 115°
        = 65°.
∴ The other two angles are ∠ B = 80° and ∠ C = 65°.