In Triangle ABC, there are constants that exist, k, m , and p so that ksinA +msinB +psinC=0. Prove that ka+mb+pc=0, area a,b, and c are the lengths of th abandon of the triangle.

Answers :In any triangle as per sine rule

a/sinA = b/sinB = c/sinC

let a/sinA = b/sinB = c/sinC = x

therefore sinA = a/x : sinB = b/x : sinC = c/x

substitute these ethics in

ksinA + msinB + psinC = 0

k(a/x) + m(b/x) + p(c/x) = 0

(ka + mb + pc)/x = 0

ka + mb + pc = 0 ----Hence proved

1)
[6,2] [-1]
[1,0] x [ 2]
[4,2] [5]

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Also:
[2,-3] [5,2,1]
[5,-1] x [4,-3,8]
[4,-1]

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and aallssooo if adage the ambit of matrices u say it like (X*Y) (horizontal x verticle) correct?

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