Question

The lengths of the sides of a triangle are successive terms of a geometric progression. Let $A$ and $C$ be the smallest and the largest interior angles of the triangle respectively. If the shortest side has length $16$ centimeters and $\frac{\sin A-2\sin B+3\sin C}{\sin C-2\sin B+3\sin A}=\frac{19}{9},$ then the perimeter of the triangle in centimeters is
(correct answer + 3, wrong answer 0)

Answer 1

Let the lengths of the sides of the triangle in centimetres be $16,16r,16r^2$ ( $r>1$ )
Then, $\frac{1-2r+3r^2}{r^2-2r+3}=\frac{19}{9}$
$\Rightarrow 9-18r+27r^2=19r^2-38r+57$
$\Rightarrow 8r^2+20r-48=0$
$\Rightarrow 2r^2+5r-12=0$
$\Rightarrow (r+4)(2r-3)=0$
$\Rightarrow r=\frac{3}{2}$
Hence, the perimeter of the triangle is $16(1+\frac{3}{2}+\frac{9}{4})=76$ cm