In `Delta"ABC,AC=BC and BC is extended upto the point D. If angleACD"=144^(@)," then determine the circular value of each of the angles of" Delta`ABC.

Explanation

In `Delta`ABC,AC=BC,
`therefore angleABC=angleBAC`
`angleACD-144^(@), therefore angle ACB=180^(@)-angleACD`
`=180^(@)-144^(@)=36^(@)`
Again , `angleABC+angleBAC+angleACB=180^(@)`
or,`angleABC+angleABC+36^(@)=180^(@) " " [therefore angleBAC=angleABC]`
or, `2angleABC=180^(@)-36^(@)`
or, `2angleABC=144^(@)`
or, `angleABC=(144^(@))/(2)=72^(@)`
`therefore angleBAC=angleABC=72^(@)`
Hence threee of the triangle `DeltaABC "are" 72^(@),72^(@),36^(@)`
or, `72^(@)xx(pi)/180^(@), 72^(@)xx(pi)/(180^(@)), 36^(@)xx(pi)/(180^(@))`
or, `(2pi)/(5),(2pi)/(5),(pi)/(5).`