Question

in the given fig. DE || BC, angle ADE =60° and angle BAC=40°, then angle BCA​

Answer 1

Answer:

By converse of Thale’s theorem DE II BC ∠ADE = ∠ABC = 70° Given ∠BAC = 50° ∠ABC + ∠BAC + ∠BCA = 180° (Angle sum prop of triangles) 70° + 50° + ∠BCA = 180° ∠BCA = 180° - 120° = 60°