To solve for [tex]\(a^2\)[/tex] in the given isosceles triangle [tex]\(ABC\)[/tex] where angle [tex]\(A = \frac{\pi}{6}\)[/tex] and sides [tex]\(b = c = 5\)[/tex], we can utilize the Law of Cosines.
The Law of Cosines states that for any triangle with sides [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex], and an angle [tex]\(A\)[/tex] opposite to side [tex]\(a\)[/tex], the following relationship holds:
[tex]\[
a^2 = b^2 + c^2 - 2bc \cos(A)
\][/tex]
Given:
[tex]\[
A = \frac{\pi}{6}, \quad b = 5, \quad c = 5
\][/tex]
First, let's calculate [tex]\(\cos\left(\frac{\pi}{6}\right)\)[/tex]. We know that:
[tex]\[
\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}
\][/tex]
Now, substituting the values into the Law of Cosines formula:
[tex]\[
a^2 = 5^2 + 5^2 - 2 \cdot 5 \cdot 5 \cdot \cos\left(\frac{\pi}{6}\right)
\][/tex]
Substitute [tex]\(\cos\left(\frac{\pi}{6}\right)\)[/tex]:
[tex]\[
a^2 = 25 + 25 - 2 \cdot 5 \cdot 5 \cdot \frac{\sqrt{3}}{2}
\][/tex]
Simplify the expression step-by-step:
[tex]\[
a^2 = 25 + 25 - 25\sqrt{3}
\][/tex]
[tex]\[
a^2 = 50 - 25\sqrt{3}
\][/tex]
Therefore, the length of [tex]\(a^2\)[/tex] is:
[tex]\[
a^2 = 25(2 - \sqrt{3})
\][/tex]
Comparing this result with the given choices, we find:
A. [tex]\(5^2(2-\sqrt{3})\)[/tex]
Thus, the correct answer is [tex]\(A\)[/tex].
