Question

20. In the diagram, ACJL, BCEF, DEGH and IGJK are straight lines. Calculate ∆a + ∆b + ∆c + ∆d + ∆e + ∆f+ ∆g + ∆h.​

Answer 1

∠a + ∠b + ∠c + ∠d + ∠e + ∠f  + ∠g + ∠h = 360°

 

Explanation

The sum of angles in triangles is 180°. Hence, based on the given picture, we get:

• ∠a + ∠b = 180° – ∠BCA
  ∠BCA and ∠ECJ are vertical/opposite angles, so ∠BCA = ∠ECJ. Therefore:
  ⇒ ∠a + ∠b = 180° – ∠ECJ   ...(i)
• ∠c + ∠d = 180° – ∠DEF
  ∠DEF and ∠CEG are vertical/opposite angles, so ∠DEF = ∠CEG. Therefore:
  ⇒ ∠c + ∠d = 180° – ∠CEG    ....(ii)
• ∠e + ∠f = 180° – ∠IGH
  ∠IGH and ∠EGJ are vertical/opposite angles, so ∠IGH = ∠EGJ. Therefore:
  ⇒ ∠e + ∠f = 180° – ∠EGJ   ....(iii)
• ∠g + ∠h = 180° – ∠KJL
  ∠KJL and ∠CJG are vertical/opposite angles, so ∠KJL = ∠CJG. Therefore:
  ⇒ ∠g + ∠h = 180° – ∠CJG    ...(iv)

Then, we calculate the sum of (i), (ii), (iii), and (iv).

∠a + ∠b = 180° – ∠ECJ
∠c + ∠d = 180° – ∠CEG
∠e + ∠f = 180° – ∠EGJ
∠g + ∠h = 180° – ∠CJG
---------------------------------- +
∠a + ∠b + ∠c + ∠d + ∠e + ∠f  + ∠g + ∠h
= 720° – ∠ECJ – ∠CEG – ∠EGJ –∠CJG
⇒ ∠a + ∠b + ∠c + ∠d + ∠e + ∠f  + ∠g + ∠h
    = 720° – (∠ECJ + ∠CEG + ∠EGJ + ∠CJG)

We know that ∠ECJ, ∠CEG, ∠EGJ, and ∠CJG are angles inside CEGJ. Hence:
∠ECJ + ∠CEG + ∠EGJ + ∠CJG = 360°

So:
∠a + ∠b + ∠c + ∠d + ∠e + ∠f  + ∠g + ∠h = 720° – 360°
∴  ∠a + ∠b + ∠c + ∠d + ∠e + ∠f  + ∠g + ∠h = 360°  (answer)