Given,

In a parallelogram ABCD,

∠DAB = 75^∘

∠DBC = 60^∘

∠A + ∠B = 180^∘ (supplementary angles)

∠B = 180^∘ – 75^∘ = 105^∘

∠DBA +∠DBC = 105^∘

∠DBA = 105^∘ – 60^∘ = 45^∘

In a triangle ABD

∠DAB + ∠DBA + ∠ADB = 180^∘

75^∘ + 45^∘ + ∠ADB = 180^∘

∠ADB = 180^∘ -120^∘ = 60^∘

∠A = ∠C = 75^∘ (opposite angles of parallelogram)

∠C + ∠D = 180^∘ (supplementary angles of parallelogram)

∠D = 180^∘ – 75^∘ = 105^∘

∠ADB + ∠CDB = 105^∘

∠CDB = 105^∘ - ∠ADB

∠CDB = 105^∘ – 60^∘ = 45^∘