find the area of triangle whose vertices are (acosA,asinA), (bcosB,bsinB), (ccosC, csinC)

i) Using e^(ix) = cos(x) + i*sin(x), the 3 vertices are given by: A{cos(a) + i*sin(a)}, B{cos(b) + i*sin(b)}, C{cos(c) + i*sin(C)} ii) So the 3 vertices of the triangle are: (Acosa, Asina), (Bcosb, Bsinb), (Ccosc, Csinc) iii) Area of a triangle ABC whose vertices are A(x₁,y₁), B(x₂, y₂), and C(x₃, y₃) is given by: Area = (1/2)[(x₁)(y₂ - y₃) + (x₂)(y₃ - y₁)) + (x₃)(y₁ - y₂)] NOTE; In applying this, let us take the vertices in anti clock wise direction; else the answer would become negative. So of the above, area is: = (1/2)[(Acosa)(Bsinb - Csinc) + (Bcosb)(Csinc - Asina) + (Ccosc)(Asina - Bsinb)] sq. units.

It can be found out by determinant ie.,Det(acosA asinA 1 bcosB bsinB 1 ccosC csinC 1)HENCE THE ANSWER IS abcosAsinB-accosAsinC-absinAcosB + acsinAcosC + bccosBsinC-bccosCsinB.RIGHT?

It can be found out by determinant ie.,0.5*Det(acosA asinA 1 bcosB bsinB 1 ccosC csinC 1)HENCE THE ANSWER IS 0.5*[abcosAsinB-accosAsinC-absinAcosB + acsinAcosC + bccosBsinC-bccosCsinB].RIGHT?