# Troublesome Triangles

Geometry Level 4

If $$\displaystyle a\cos{A} + b\cos{B} + c\cos{C} \over \displaystyle a\sin{B} + b\sin{C} + c\sin{A}$$ $$= \displaystyle\frac{a+b+c}{9R}$$. Then the triangle is-

Here $$a , b , c$$ are sides of triangle and $$A , B , C$$ are the angles of Triangle and $$R$$ the the circumradius of $$\vartriangle ABC$$

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