- A50°
- B75°
- C105°
- D55°
Step-by-Step Solution
∠ACD = ∠ACE + ∠ECD = 105°, so ∠ACB = 180° − 105° = 75°. Also ∠A = 50° (alternate) and ∠B = 55° (corresponding), and 50° + 55° + 75° = 180°.
Answer: B) 75°
Through C a line CE is drawn parallel to BA, as shown. ∠ACE = 50° and ∠ECD = 55°. Find ∠ACB.
Class 9Exterior Angle of a TriangleGeometry — Lines and Angles
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