If α, β, γ are the angles of a triangle and the system of equationscos⁡$$($$α−β$$)x+$$$$cos$$⁡$$($$β−γ$$)y+$$$$cos$$⁡$$($$γ−α$$)z=0cos$$⁡$$($$α$$+$$β$$)x+$$$$cos$$⁡$$($$β$$+$$γ$$)y+$$$$cos$$⁡$$($$γ$$+$$α$$)z=0sin$$⁡$$($$α$$+$$β$$)x+$$$$sin$$⁡$$($$β$$+$$γ$$)y+$$$$sin$$⁡$$($$γ$$+$$α$$)z=0has$$ $$non-trivial$$ solutions, then triangle is necessarily

Question

If α, β, γ are the angles of a triangle and the system of equationscos⁡$$($$α−β$$)x+$$$$cos$$⁡$$($$β−γ$$)y+$$$$cos$$⁡$$($$γ−α$$)z=0cos$$⁡$$($$α$$+$$β$$)x+$$$$cos$$⁡$$($$β$$+$$γ$$)y+$$$$cos$$⁡$$($$γ$$+$$α$$)z=0sin$$⁡$$($$α$$+$$β$$)x+$$$$sin$$⁡$$($$β$$+$$γ$$)y+$$$$sin$$⁡$$($$γ$$+$$α$$)z=0has$$ $$non-trivial$$ solutions, then triangle is necessarily

Infinity Learn · Accepted Answer

Let Δ$$=$$$$cos$$⁡$$($$α−β$$)$$$$cos$$⁡$$($$β−γ$$)$$$$cos$$⁡$$($$γ−α$$)$$$$cos$$⁡$$($$α$$+$$β$$)$$$$cos$$⁡$$($$β$$+$$γ$$)$$$$cos$$⁡$$($$γ$$+$$α$$)$$$$sin$$⁡$$($$α$$+$$β$$)$$$$sin$$⁡$$($$β$$+$$γ$$)$$$$sin$$⁡$$($$γ$$+$$α$$)It$$ is clear that either α$$=$$β or β$$=$$γ or γ$$=$$α s sufficient to make ∆$$=0$$. It is not necessary that triangle is equilateral.Also, isosceles triangle can be obtuse one.

If α, β, γ are the angles of a triangle and the system of equationscos⁡(α−β)x+cos⁡(β−γ)y+cos⁡(γ−α)z=0cos⁡(α+β)x+cos⁡(β+γ)y+cos⁡(γ+α)z=0sin⁡(α+β)x+sin⁡(β+γ)y+sin⁡(γ+α)z=0has non-trivial solutions, then triangle is necessarily

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