Problems related to angles made by parallel lines and transversal are presented below along with their detailed solutions.

Problem 1

• Draw BB' parallel to AA' and CC'as shown in the figure below.
  
  
  
• Note that angle ABC is given by
  angle ABC = angle ABB' + angle CBB'
  
• Angle w' and angle ABB' are alternate interior angles and their sizes are equal.
  angle ABB' = angle w'
  
• Angle z' and angle CBB' are alternate interior angles and their sizes are equal.
  angle CBB' = angle z'
  
• Angles w and w' are supplementary which gives
  w' = 180 - w = 180 - 135 = 45 ⁰
  
• Angles z and z' are also supplementary which gives
  z' = 180 - z = 180 - 147 = 33 ⁰
  
• We now substitute angle ABB' by w' and angle CBB' by z' in angle ABC = angle ABB' + angle CBB' found above.
  angle ABC = w' + z' = 45 + 33 = 78 ⁰
  

Problem 2

• Angles A'AC and angle ACC" are alternate interior angles and their sizes are equal.
  angle A'AC = angle ACC"
  
• Angles A'AC and angle ACC" are alternate interior angles and their sizes are equal.
  angle A'AC = angle ACC"
  
• Angles A'AC and angle A"AC are supplementary so that
  angle A"AC = 180 - angle A'AC = 180 - angle ACC"
  
• Rearrange the above to obtain
  angle A"AC + angle ACC" = 180 ⁰
  
• Because AB and CB are bisectors(they divide the angle into two equal angles), angle ABC in triangle ABC is given by
  angle ABC = 180 - (angle A"AC + angle ACC") / 2
  = 180 - 180 / 2 = 90 ⁰

Problem 3

• Angle x and angle ABC are supplementary hence
  angle ABC = 180 - x = 180 - 127 = 53 ⁰
  
• Angle y and angle ACB are supplementary hence
  angle ACB = 180 - y = 180 - 115 = 65 ⁰
  
• Angle ADD' and angle ABC are corresponding angles and their sizes are equal
  angle ADD' = angle ABC = 53 ⁰
  
• Angle AD"D and angle ACB are corresponding angles and their sizes are equal
  angle AD'D = angle ACB = 65 ⁰
  
• Angle DAD' is given by
  angle DAD' = 180 - (angle ADD' + angle AD'D)
  = 180 - 118 = 62 ⁰

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