Q12 of 25 Page 276

The base BC of the equilateral triangle ABC is extended upto the point E so that CE=BC. By joining A,E, let us determine the circular values of the angles of ΔABC


∠ABC = ∠BAC = ∠BCA = 60°




∠ACE + ∠ACB = 180°


⇒ ∠ACE = 180° - 60°


⇒ ∠ACE = 120°


∵ BC = CE and BC = AC


⇒ AC = AE


⇒ ∠CAE = ∠AEC = x


∠CAE + ∠AEC + ∠ACE = 180°


⇒ 2x + 120° = 180°


⇒ x = 30°





And ∠ACE = 120°




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