# 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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