If the angles `A, B, C` (in that order) of triangle ABC are in arithmetic progression, and `L=lim_(A rarr C) sqrt(3-4sinAsinC)/|A-C|` then find the value of `100L^2`.

Options

A. 100
B. 2
C. 3
D. 4

Answer

To solve the problem step by step, we will analyze the given information and apply mathematical concepts related to limits and trigonometric identities.

### Step-by-Step Solution:

1. **Understanding the Angles in Arithmetic Progression**:
Given that the angles \( A, B, C \) of triangle \( ABC \) are in arithmetic progression, we can express them as:
\[ A = b - d, \quad B = b, \quad C = b + d \]
where \( b \) is the middle angle and \( d \) is the common difference.

2. **Using the Sum of Angles in a Triangle**:
The sum of the angles in a triangle is \( 180^\circ \):
\[ A + B + C = 180^\circ \implies (b - d) + b + (b + d) = 180^\circ \]
Simplifying this gives:
\[ 3b = 180^\circ \implies b = 60^\circ \]
Therefore, the angles are:
\[ A = 60^\circ - d, \quad B = 60^\circ, \quad C = 60^\circ + d \]

3. **Setting Up the Limit Expression**:
We need to evaluate the limit:
\[ L = \lim_{A \to C} \frac{\sqrt{3 - 4 \sin A \sin C}}{|A - C|} \]
Substituting \( A = 60^\circ - d \) and \( C = 60^\circ + d \):
\[ L = \lim_{d \to 0} \frac{\sqrt{3 - 4 \sin(60^\circ - d) \sin(60^\circ + d)}}{|(60^\circ - d) - (60^\circ + d)|} \]
This simplifies to:
\[ L = \lim_{d \to 0} \frac{\sqrt{3 - 4 \sin(60^\circ - d) \sin(60^\circ + d)}}{| -2d |} = \lim_{d \to 0} \frac{\sqrt{3 - 4 \sin(60^\circ - d) \sin(60^\circ + d)}}{2d} \]

4. **Using the Product-to-Sum Formula**:
We can use the identity:
\[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \]
Thus, we have:
\[ \sin(60^\circ - d) \sin(60^\circ + d) = \frac{1}{2} [\cos(-2d) - \cos(120^\circ)] \]
Since \( \cos(120^\circ) = -\frac{1}{2} \):
\[ \sin(60^\circ - d) \sin(60^\circ + d) = \frac{1}{2} [\cos(2d) + \frac{1}{2}] \]

5. **Substituting Back into the Limit**:
Now substituting this back into our limit expression:
\[ L = \lim_{d \to 0} \frac{\sqrt{3 - 2(\cos(2d) + \frac{1}{2})}}{2d} \]
Simplifying gives:
\[ L = \lim_{d \to 0} \frac{\sqrt{3 - 1 - 2\cos(2d)}}{2d} = \lim_{d \to 0} \frac{\sqrt{2 - 2\cos(2d)}}{2d} \]
Using the identity \( 1 - \cos x = 2 \sin^2(\frac{x}{2}) \):
\[ L = \lim_{d \to 0} \frac{\sqrt{2 \cdot 2 \sin^2(d)}}{2d} = \lim_{d \to 0} \frac{2|\sin(d)|}{2d} = \lim_{d \to 0} \frac{|\sin(d)|}{d} \]
As \( d \to 0 \), \( \frac{\sin(d)}{d} \to 1 \):
\[ L = 1 \]

6. **Calculating \( 100L^2 \)**:
Finally, we calculate:
\[ 100L^2 = 100 \cdot 1^2 = 100 \]

### Final Answer:
The value of \( 100L^2 \) is \( \boxed{100} \).