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

Explanation

Since `DeltaABC` is equilateral,
`therefore angleACB=60^(@)`
`therefore angleACE 180^(@)-angleACB`
`=180^(@)-60^(@)=120^(@)`
`120^(@)xx(pi)/(180^(@))=(2pi)/(3).`
`therefore CE =BC (given )=AC [therefore Delta` ABC is equilateral]
`therefore angleCAE=angleAEC..............................(1)`
Again, `angleACE+angleCAE+angleAEC=180^(@)`
or, `120^(@)+angleAEC+angleAEC=180^(@)[by(1)]`
or, `120^(@)+2angleAEC=180^(@) or , 2angleAEC=180^(@)-120^(@)`
or `2angleAEC=60^(@)`
or, `angleAEC=(60^(@))/(2)=30^(@)=30xx(pi)/(180^(@))=(pi)/(6)`.
`therefore angleACE=(2pi)/(3),angleAEC=(pi)/(6) and angle(CAE=angleAEC=(pi)/(6)[by (1)]`
Hence the required angles are `(2pi)/(3), (pi)/(6) "and" (pi)/(6)`.