8. The lengths of the sides of a triangle are successive terms of a geometric progression. Let $A$ and $C$ be the smallest and largest interior angles of the triangle respectively. If the shortest side has length $16 \mathrm{~cm}$ and
$$
\frac{\sin A-2 \sin B+3 \sin C}{\sin C-2 \sin B+3 \sin A}=\frac{19}{9},
$$
find the perimeter of the triangle in centimetres.
8. Answer: 76
Solution. Let the lengths of the sides of the triangle in centimetres be $16,16 r$ and $16 r^{2}$ (where $r>1$ ). Then $\frac{1-2 r+3 r^{2}}{r^{2}-2 r+3}=\frac{19}{9}$ so that $r=\frac{3}{2}$. Hence, the perimeter of the triangle $=16\left(1+\frac{3}{2}+\frac{9}{4}\right)=76 \mathrm{~cm}$