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Question

If the system of equations xsinα+ysinβ+xsinγ=0, xcosα+ycosβ+zcosγ=0, x+y+z=0 where α,β,γ are angles of a triangle, have a non-trivial solution, then the triangle must be

A
isosceles
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B
equilateral
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C
right- angled
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D
none of these
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Solution

The correct option is A isosceles
Since the given system has a non-trivial solution, therefore,
∣ ∣sinαsinβsinγcosαcosβcosγ111∣ ∣=0
∣ ∣sinαsinβsinαsinγsinαcosαcosβcosαcosγcosα100∣ ∣=0
[Applying C2C2C1,C3C3C1]
∣ ∣ ∣ ∣sinα2cos(β+α2)sin(βα2)2cos(γ+α2)sin(γα2)cosα2sin(β+α2)sin(αβ2)2sin(γ+α2)sin(αγ2)100∣ ∣ ∣ ∣=0
4sin(αβ2)sin(γα2)
[sin(γ+α2)cos(β+α2)sin(β+α2)cos(γ+α2)]=0
4sin(αβ2)sin(βγ2)sin(γα2)=0
sin(αβ2)=0
or sin(βγ2)=0 or sin(γα2)=0
αβ=0 or βγ=0 or γα=0
α=β or β=γ or γ=α
triangle must be isosceles.

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