PARALLEL LINES TRANSVERSALS TRIANGLES

Problem 1 :

Solution :

AB ∥ CE

a^º = 60^º (corresponding angle)

∠ABD = ∠CDE = 80 (Corresponding angles)

b^º = 80^º (Vertically opposite angles)

Problem 2 :

Solution :

CF ∥ AB

a^º = 43^º (alternate angles)

DA ∥ CB

b^º = a^º = 43^º (Corresponding angles)

Problem 3 :

Solution :

FG ∥ DE

∠DAC + ∠CAE = 180

110 + ∠CAE = 180

 ∠CAE = 180 - 110

 ∠CAE = 70 = b (alternate interior angles)

∠ABC + ∠CBE = 180

∠ABC + 130 = 180

 ∠ABC = 180 - 130

 ∠ABC = 50 = a(alternate interior angles)

a + b + c = 180

50 + 70 + c = 180

c + 120 = 180

c = 180 -120

c = 60

Problem 4 :

Solution :

AC ∥ BE

a^º = 60^º (corresponding angle)

Problem 5 :

Solution :

AC ∥ BE

∠ACB = ∠CBE 

50^º =  a^º (alternate interior angles)

Problem 6 :

Solution :

FG ∥ HD

a = 40 (alternate interior angles)

FG ∥ EB

∠CBA = ∠GCB

b = 70 (Alternate interior angles)

a + b = c + 70

40 + 70 = c + 70

c = 40

Problem 7 :

Solution :

∠CBE = ∠BED

a = ∠BED = 40 (Alternate interior angles)

a + b = 180 (supplementary)

40 + b = 180

b = 180 - 40

b = 140

c + 40 = 180 (co interior angles)

c = 180 - 40

c = 140