Required number of triangles =total number of triangles-number of triangles having two sides common -number of triangles having one side common
$\Rightarrow 8C_3-8-8\times 4$
$\Rightarrow \large\frac{8!}{3!5!}$$-8-8\times 4$
$\Rightarrow \large\frac{8\times 7\times 6\times 5!}{3\times 2\times5!}$$-8-32$
$\Rightarrow 56-8-32$
$\Rightarrow 56-40$
$\Rightarrow 16$
Hence (A) is the correct answer.