Math, asked by smile1234, 1 year ago

solve no 4 ..........

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Answered by KVaishu

4.(i) Let's draw an imaginary line through E parallel to AB.
Let the point on that imaginary line (to the right of point E) be F.
Then,
∠ABE = ∠BEF = 35° (∵ Alternate interior angles)
∠CDE =∠DEF = 65° (∵ Alternate interior angles)
    x = ∠BEF + ∠DEF
= 35° + 65°
= 100°
∴ x = 100°
(ii) Let's draw an imaginary line through O parallel to AB.
  Let the point on that imaginary line (to the left of point O) be P.
Then,
∠ABO + ∠BOP = 180° (∵ co-interior angles)
  55° + ∠BOP = 180°
∠BOP = 180° - 55°
  ∠BOP = 125°
  ∠CDO + ∠DOP = 180° (∵ co-interior angles)
25° + ∠DOP = 180°
  ∠DOP = 180° - 25°
∠DOP = 155°
  Now, x = ∠BOP + ∠DOP
   x = 125° + 155°
  x = 280°
  ∴ x = 280°
(iii) Let's draw an imaginary line through E parallel to AB.
  Let the point on that imaginary line (to the right of point E) be F.
Then,
  ∠BAE + ∠AEF = 180° (∵ co-interior angles)
  116° + ∠AEF = 180°
∠AEF = 180° - 116°
  ∠AEF = 64°
  ∠DCE + ∠CEF = 180° (∵ co-interior angles)
124° + ∠CEF = 180°
  ∠CEF = 180° - 124°
∠CEF = 56°
  Now, x = ∠AEF + ∠CEF
   x = 64° + 56°
  x = 120°
  ∴ x = 120°

Hope my answer helps you.. : )

KVaishu: If it is really helpful to you, then please grade it as the Brainliest answer

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