In a quadrilateral ABCD, ∠A+∠C is 2 times ∠B+∠D. If ∠A= 140° and f∠D = 60°, then ∠B=
Given,
ABCD is a parallelogram
∠A+∠C = 2(∠B+∠D)
∠A = 40°
∵ ∠A+∠B+∠C+∠D = 360° [angle sum property of quadrilateral]
= ∠A+∠C+∠B+∠D = 360°
= 2(∠B+∠D)+ ∠B+∠D = 360°
= 3(∠B+∠D) = 360°
= ∠B+∠D = 
∵∠D= 60° [given]
∴ ∠B = 120°-60° = 60°
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