Q. 325.0( 1 Vote )

# In the given figure, ∠BAC = 30°, ∠ABC = 50° and ∠CDE = 40°. Then ∠AED = ?

A. 120°

B. 100°

C. 80°

D. 110°

Answer :

ACB + ABC + BAC =180

ACB = 180 – 50 – 30 = 100^{o}(Sum of angles of triangle is 180)

ACB + ACD = 180 (linear pair of angles)

ACD = 180 – 100 = 80^{o}

In triangle ECD,

ECD + CDE + DEC = 180

DEC = 180 – 80 – 40

= 60^{o}

DEC + AED = 180^{o}(linear pair of angles)

AED = 180^{o} – 60^{o}

= 120^{o}

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