Q33 of 153 Page 148

In the given figure, ∠BAC = 40°, ∠ACB = 90° and ∠BED = 100°. Then ∠BDE = ?

In triangle AEF,

BED = EFA + EAF

EFA = 100 – 40 = 60^o

CFD = EFA (vertical opposite angles)

= 60^o

In triangle CFD, we have

CFD + FCD + CDF = 180^o

CDF = 180^o – 90^o – 60^o

= 30^o

So, BDE = 30^o

Side BC of ΔABC has been produced to D on left hand side and to E on right hand side such that ∠ABD = 125° and ∠ACE = 130°. Then ∠A = ?

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

In the given figure, BO and CO are the bisectors of ∠B and ∠C respectively. If ∠A = 50°, then ∠BOC = ?

In the given figure, AB || CD. If ∠EAB = 50° and ∠ECD = 60°, then ∠AEB = ?