the side of BC of  triangle abc has been produced to a point D , such that angle ACD=140 degree. If angle B =2/5 angle A, then find angleA .

∠ACD = 140⁰  (Given)

∠ABC = (2/5) ∠ BAC

∠ACD = ∠ABC + ∠ BAC  (Exterior angle property)

140⁰ = (2/5) ∠ BAC + ∠ BAC 

140⁰ = (7/5) ∠ BAC 

∠ BAC = 20 × 5

∠ BAC = 100⁰