In `DeltaABC, angle C = 2 angle A, and AC = 2BC`, then the value of `(a^(2) + b^(2)+ c^(2))/(R^(2))` (where R is circumradius of triangle) is ______

Answer

8

Explanation

`A + B + C = pi`
Given `C = 2A`
`rArr B = pi - 3A`
As `0 lt C lt pi rArr 0 lt 2A lt pi rArr 0 lt A lt (pi)/(2)`
By sine rule, `(a)/(sin A) = (b)/(sin B)`
or `(a)/(sin A) = (2a)/(sin (pi -3A)) or (a)/(1) = (2a)/(3-4 sin^(2)A)`
or `3 -4 sin^(2) A = 2`
or `sin^(2)A = (1)/(4) or sin A = (1)/(2) or A = (pi)/(6) or (5pi)/(6)`
But `0 lt A lt (pi)/(2)`
`rArr A = (pi)/(6), angle B = (pi)/(2), and angleC = (pi)/(3)`
`rArr a^(2) + b^(2) + c^(2) =4R^(2) [sin^(2)A + sin^(2)B + sin^(2)C]`
`=4R^(2) [sin^(2).(pi)/(6) + sin^(2).(pi)/(2) + sin^(2).(pi)/(3)]`
`= 4R^(2) [(1)/(4) + 1 + (3)/(4)] = 8R^(2)`
or `(a^(2) +b^(2) + c^(2))/(R^(2)) = 8`