Suppose an isosceles triangle [tex]$ABC$[/tex] has [tex]$\angle A = \frac{\pi}{6}$[/tex] and [tex]$b = c = 5$[/tex]. What is the length of [tex]$a^2$[/tex]?

A. [tex]$5^2 \sqrt{3}$[/tex]

B. [tex]$5^2(2 + \sqrt{3})$[/tex]

C. [tex]$5^2(\sqrt{3} - 2)$[/tex]

D. [tex]$5^2(2 - \sqrt{3})$[/tex]



Answer :

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.