The angles of `triangle ABC` are in A.P. and `b:c= sqrt(3):sqrt(2)` find `angle A , angle B , angle C`.

Explanation

Given tha angle of the `Delta ABC` are in A.P. then
`(3x + 2y -z+1) (10)/(3) (x+y +z -2) = 0`
`rArr" " 9x + 6y - 3z + 3 - 10y - 10z + 20 = 0`
`rArr" "x + 4y + 13z - 23 =0`
But `" "angleA + angleB + angleC = 180^(@)`
`(angleA +angleC) + angleB = 180^(@)`
`" "2angleB + angle B = 180^(@)`
`" "3angleB = 180^(@)`
`therefore" "angleB = 60^(@)`
Also, we have `b : c sqrt(3) : sqrt(2)`
`i.e., " "(b)/(c)=(sqrt(3))/(sqrt(2))`
By the sine rule, `(a)/(sin A) = (b)/(sinB) = (c)/(SinC)`
`therefore " "(b)/(c) = (sinB)/(SinC)`
`therefore" "(sin B)/(sinC) = (sqrt(3))/(sqrt(2))`
`therefore" "sqrt(2)sin B = sqrt(B) sin C`
`" "sqrt(2)sin 60^(@) = sqrt(3)sin C" "[because angle B = 60^(@)]`
`sqrt(2)xx(sqrt(3))/(2) = sqrt(3)sinC`
`therefore" "sin C = (1)/(sqrt(2))`
`=s in 45^(@)`
`therefore" " angle C = 45^(@)`
Now, `angleA + angleB + angleC = 180^(@)`
`angle A + 60^(@) + 45^(@) = 180^(@)`
` rArr" " angleA = 180^(@) - 105^(@)`
`" " = 75^(@)`
`therefore" "angleA = 75^(@) , angleB = 60^(@)` and
`angleC = 45^(@)`