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If in a triangle ABC, `sin^4A+sin^4B+sin^4C=sin^2B sin^2C+2 sin^2C sin^2A+2sin^2A sin^2B`, show that, one of the angles of the triangle is `30^@` or `150^@`

Answer» `sin^4A+sin^4B-2sin^2Asin^2B=sin^2Bsin^2C+2sin^2Csin^2Csin^2A-sin^4c`
`(sin^2A-sin^2B)^2=sin^2c(sin^2B+2sin^2A-sin^2(A+B))`
`sin^2(A-B)sin^2C=sin^2C(sin^2B+2sin^2A-sin^2(A+B)`
`sin^2Acos^2B+sin^2Bcos^2A-2sinAcosBsinBcosA= sin^2B+2sin^2A-sin^2Acos^2b-sin^2Bcos^2A-2sinacosBsinBcosA`
`sin^2A[cos^2B+cos^2B-2]=sin^2B[1-cos^2A-cos^2A]`
`2sin^2A[cos^2B-1]=sin^2B[1-2cos^2A]`
`-2sin^2A=1-2cos^2A`
`2(cos^2A-sin2A)=1`
`2A=pi/3,5pi/3`
`A=pi/6,5/6pi`
`A=30^0,150^0`.


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