Q22 of 77 Page 16

In a quadrilateral ABCD, ∠A+∠C is 2 times ∠B+∠D. If ∠A= 140° and f∠D = 60°, then ∠B=

Given that,

∠A = 140°


∠D = 60°


According to question,


∠A + ∠C = 2 (∠B + ∠D)


140 + ∠C = 2 (∠B + 60°)


∠B = (∠C) + 10° (i)


We know,


∠A + ∠B + ∠C + ∠D= 360°


140° + (∠C) + 10° + ∠C + 60° = 360°


∠C = 150°


∠C = 100°


∠B = (100°) + 10°


= 60°

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