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in Parallelograms by (26.9k points)
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In a quadrilateral ABCD, ∠A+∠C is 2 times ∠B+∠D. If ∠A = 140° and ∠D = 60°, then ∠B = 

A. 60° 

B. 80° 

C. 120° 

D. None of these

1 Answer

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Best answer

Option : (A)

Given that, 

∠A = 140° 

∠D = 60° 

According to question, 

∠A + ∠C = 2 (∠B + ∠D) 

140 + ∠C = 2 (∠B + 60°) 

∠B = \(\frac{1}{2}\)(∠C) + 10° ...(i) 

We know, 

∠A + ∠B + ∠C + ∠D = 360° 

140° + \(\frac{1}{2}\)(∠C) + 10° + ∠C + 60° = 360° 

\(\frac{3}{2}\)∠C = 150° 

∠C = 100° 

∠B = \(\frac{1}{2}\)(100°) + 10° 

= 60°

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