Exterior angle is sum of interior opposite angles

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For ∆ABC

 ∠1 = ∠ABC + ∠ACB

∠2 = ∠CAB + ∠CBA

∠3 = ∠BAC + ∠BCA

 

Let’s solve some questions     

 

Proof of exterior angle property

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In ∆ABC,

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Also,

  ∠1 + ∠4 = 180°           (Linear Pair)

  (180° − ∠2 − ∠3) + ∠4 = 180°     (From (1 )

  ∠4 = 180° − 180° + ∠2 + ∠3

  ∠4 = 0° + ∠2 + ∠3

  ∠4 = ∠2 + ∠3         

 

∴ Exterior angle is equal to sum of interior opposite angles

 

Find exterior angle ∠ 1

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Here,

  ∠1 = ∠B + ∠C     ( Exterior angle property)

  ∠1 = 45°  + 60° 

  ∠1 = 105° 

        

Find exterior angle

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In ∆ABC,

  ∠BCD = ∠CAB + ∠ABC        (Exterior angle property)

  ∠BCD = 85° + 25°

  ∠BCD = 110° 

 

Find exterior angle

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In ∆PQR,

  ∠SQR = ∠P + ∠R        (Exterior angle property)

  ∠SQR = 30° + 15°

  ∠SQR = 45° 

 

Find exterior angle

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In ∆XYZ,

  ∠XZO = ∠X + ∠Y           (Exterior angle property)

  ∠XZO = 20° + 90°

  ∠XZO = 110° 

 

Find ∠ACB

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In ∆ABC,

  ∠DAC = ∠B + ∠C     (Exterior angle property)

  120° = 40° + ∠C

  120° − 40° = ∠C

  80° = ∠C 

  ∠C = 80°

 

i.e. ∠ACB = 80°   

 

Find ∠PQR

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In ∆PQR,

  ∠PRS = ∠PQR + ∠QPR        (Exterior angle property)

  160° = ∠PQR  + 50°

  160° − 50° = ∠PQR

  110° = ∠PQR

  ∠PQR = 110°   

 

Find ∠XZY

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In ∆XYZ,

  ∠OXZ = ∠XZY + ∠XYZ        (Exterior angle property)

  140° = ∠XZY  + 90°

  140° − 90° = ∠XZY

  50° = ∠XZY

  ∠XZY = 50°      

  1. Chapter 6 Class 7 Triangle and its Properties
  2. Concept wise

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.