Answer
$x\approx11.15$
Work Step by Step
$A=31^{\circ},s=11,r=12$
Formula for Arc-length,
$s=r.C$
$11=12.C$
$C=\frac{11}{12}radian=52.52^{\circ}$
Therefore, $D=180^{\circ}-(31^{\circ}+52.52^{\circ})=96.48^{\circ}$
Using laws of sines,
$\frac{r}{sinA}=\frac{r+x}{sinD}$
$\frac{12}{sin31^{\circ}}=\frac{12+x}{sin96.48^{\circ}}$
$12+x=12\times\frac{sin96.48^{\circ}}{sin31^{\circ}}$
$12+x=23.15$
Hence, $x=11.15$
