Question 8
The lengths of the sides of a triangle are successive terms of a geometric progression. Let $A$ and $C$ be its smallest and largest interior angles respectively. The shortest side is $16$ cm and
$$\frac{\sin A-2\sin B+3\sin C}{\sin C-2\sin B+3\sin A}=\frac{19}{9}.$$
Find the perimeter in centimetres.
$$\frac{\sin A-2\sin B+3\sin C}{\sin C-2\sin B+3\sin A}=\frac{19}{9}.$$
Find the perimeter in centimetres.