ABC is a non-degenerate triangle such that 2\sin \angle B \cdot \cos \angle C + \sin \angle C = \sin \angle A + \sin \angle B, what is the value of \lfloor 100 \frac {AC}{AB} \rfloor ?

To find the value of \lfloor 100 \frac {AC}{AB} \rfloor, we need to solve the given equation 2\sin \angle B \cdot \cos \angle C + \sin \angle C = \sin \angle A + \sin \angle B in terms of \frac{AC}{AB}.

Let's start by manipulating the equation.

2\sin \angle B \cos \angle C + \sin \angle C = \sin \angle A + \sin \angle B

Using the trigonometric identity \sin (A + B) = \sin A \cos B + \cos A \sin B, we can rewrite the equation as:

2\sin \angle B \cos \angle C + \sin \angle C = \sin \angle B \cos \angle A + \cos \angle B \sin \angle A

Next, we can rearrange the terms to get:

2\sin \angle B \cos \angle C - \cos \angle B \sin \angle A = \sin \angle B \cos \angle A - \sin \angle C

Now, factor out the common trigonometric factors from both sides of the equation:

\sin \angle B (2\cos \angle C - \cos \angle A) = \sin \angle C - \sin \angle A

Recall the trigonometric identity \sin (A - B) = \sin A \cos B - \cos A \sin B, we can rewrite the equation as:

\sin \angle B (2\cos \angle C - \cos \angle A) = -2\sin \left(\frac{\angle A - \angle C}{2}\right) \cos \left(\frac{\angle A + \angle C}{2}\right)

So, we have:

\frac{\sin \angle B}{\sin \left(\frac{\angle A - \angle C}{2}\right)} = -\frac{2\cos \left(\frac{\angle A + \angle C}{2}\right)}{2\cos \angle C - \cos \angle A}

Now, recall the trigonometric identity \sin 2\theta = 2\sin \theta \cos \theta, we can substitute in the equation:

\frac{\sin \angle B}{\sin \left(\frac{\angle A - \angle C}{2}\right)} = -\frac{\sin (\angle A + \angle C)}{2\cos \angle C - \cos \angle A}

Since ABC is a non-degenerate triangle, we know that \angle A, \angle B, and \angle C are non-zero angles.

Now, consider the ratio \frac{AC}{AB} in the triangle ABC:

\frac{AC}{AB} = \frac{\sin \angle A}{\sin \angle B}

Dividing both sides of the equation by \sin \angle B, we get:

\frac{AC}{AB} = \frac{\sin \angle A}{\sin \angle B} = \frac{\sin (\angle A + \angle C)}{\sin \left(\frac{\angle A - \angle C}{2}\right)} \cdot \frac{2\cos \angle C - \cos \angle A}{2\cos \angle C - \cos \angle A}

Simplifying the equation, we have:

\frac{AC}{AB} = \frac{\sin (\angle A + \angle C)}{\sin \left(\frac{\angle A - \angle C}{2}\right)} \cdot \frac{2\cos \angle C - \cos \angle A}{2\cos \angle C - \cos \angle A} = -1

Therefore, we have found that \frac{AC}{AB} = -1.

Now we can substitute this value into \lfloor 100 \frac {AC}{AB} \rfloor:

\lfloor 100 \frac {AC}{AB} \rfloor = \lfloor 100(-1) \rfloor = \lfloor -100 \rfloor = -100.

So, the value of \lfloor 100 \frac {AC}{AB} \rfloor is -100.