Q32 of 153 Page 148

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

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

In ΔABC, side BC is produced to D. If ∠ABC = 40° and ∠ACD = 120°, then ∠A = ?

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 = 40°, ∠ACB = 90° and ∠BED = 100°. Then ∠BDE = ?

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