To find the length of [tex]\( a^2 \)[/tex] in the given isosceles triangle [tex]\( \triangle ABC \)[/tex] with angle [tex]\( A = \frac{\pi}{6} \)[/tex] and sides [tex]\( b = 5 \)[/tex] and [tex]\( c = 5 \)[/tex], we can use the Law of Cosines.
The Law of Cosines states:
[tex]\[ a^2 = b^2 + c^2 - 2bc \cos(A) \][/tex]
Here, because [tex]\( b = c = 5 \)[/tex] and [tex]\( A = \frac{\pi}{6} \)[/tex], we can substitute in these values:
[tex]\[ a^2 = 5^2 + 5^2 - 2 \cdot 5 \cdot 5 \cdot \cos \left( \frac{\pi}{6} \right) \][/tex]
First, calculate [tex]\( \cos \left( \frac{\pi}{6} \right) \)[/tex]. We know:
[tex]\[ \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \][/tex]
Now, substitute [tex]\( \cos \left( \frac{\pi}{6} \right) \)[/tex] back into the formula:
[tex]\[ a^2 = 25 + 25 - 2 \cdot 5 \cdot 5 \cdot \frac{\sqrt{3}}{2} \][/tex]
[tex]\[ a^2 = 25 + 25 - 25 \sqrt{3} \][/tex]
[tex]\[ a^2 = 50 - 25 \sqrt{3} \][/tex]
This result needs to be simplified further to match one of the given options:
[tex]\[ a^2 = 25(2 - \sqrt{3}) \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{25(2 - \sqrt{3})} \][/tex]
Given the options in the problem, this corresponds to:
[tex]\[ \boxed{5^2(2-\sqrt{3})} \][/tex]
which is option D.
