Triangle's possibility

Using Cosine Rule, we have:

$BC^2 = AB^2 + AC^2 - 2\dot{}AB\dot{}AC\dot{}\cos\theta \\ \Rightarrow AB^2 + AC^2 - 112 \cos 0^\circ < BC^2 < AB^2 + AC^2 - 112 \cos 180^\circ \\ \sqrt{AB^2 + AC^2 - 112} < BC < \sqrt{AB^2 + AC^2 + 112}$

For the $8$ possible cases of $AB$ and $AC$ , we have:

$\begin{cases} (AB,AC) = (1,56); (56,1) & \Rightarrow 55 < BC < 57 & \Rightarrow \begin{cases} \text{acceptable } BC = 56 \\ \text{acceptable triangles} = 2 \end{cases} \\ (AB,AC) = (2,28); (28,2) & \Rightarrow 26 < BC < 30 & \Rightarrow \begin{cases} \text{acceptable } BC = 27,28,29 \\ \text{acceptable triangles} = 6 \end{cases} \\ (AB,AC) = (4,14); (14,4) & \Rightarrow 10 < BC < 18 & \Rightarrow \begin{cases} \text{acceptable } BC = 11,12,13,...17 \\ \text{acceptable triangles} = 14 \end{cases} \\ (AB,AC) = (7,8); (8,7) & \Rightarrow 1 < BC < 15 & \Rightarrow \begin{cases} \text{acceptable } BC = 2,3,4,...14 \\ \text{acceptable triangles} = 26 \end{cases} \end{cases}$

$\Rightarrow$ The total number of acceptable triangles is $=2+6+14+26 = \boxed{48}$ .

The possible values of $AB ,AC$ are $(1,56),(2,28),(4,14),(7,8),(8,7),(14,4),(28,2),(56,1)$

Case 1: $1+56>x gives x<57$

$x+1>56 gives x>55$

So the sides are $(56,1,56)$

Case 2: $2+28>x$ gives $x<30$

$x+2>28$ gives $x>26$

So The sides are $(28,2,27),(28,2,28),(28,2,29)$

Case 3: $4+14>x$ gives $x<18$

$x+4>14$ gives $x>10$

Just as above there are $7$ values satisfying $10<x<18$

Case 4: $8+7>x$ gives $x<15$

$x+7>8$ gives $x>1$

There are $13$ values for $1<x<15$

From the above cases 24 triplets of sides are possible,but the values of sides $AB$ and $AC$ can be exchanged, so there are $24*2=\boxed{48}$ such triplets.