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

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

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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^(@)