Question

In Fig,

ABCD is a parallelogram in which ∠DAB =75° and ∠DBC = 60°. 

Compute ∠CDB and ∠ADB.

Answer 1

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^∘