Use your brains -2

soln

Let the center of the circumcircle be $O$ . Then $AO=BO=CO=5$ . Since $\triangle ABC$ is an isosceles triangle, it is noted that $\angle BAO = \angle CAO$ and let it be $\theta$ .

Using Cosine Rule, we have:

$5^2 = 6^2 + 5^2 - 2(6)(5)\cos {\theta}\quad \Rightarrow 60 \cos {\theta} = 36$

$\Rightarrow \cos {\theta} = \dfrac {36}{60} = \dfrac {3}{5}\quad \Rightarrow \sin {\theta} = \dfrac {4}{5}$

We note that $BC = 2(6)\sin {\theta} = 12 \times \dfrac {4}{5} = \boxed {\frac {48}{5}}$

We know that R=(abc/4 * area) or 5=[36x/4 * {(x/4) * sqrt(4 * 6^2)-x^2)}] or 5=36x/x * sqrt(144-x^2) or sqrt(144-x^2)=36/5 or 144-x^2=1296/25 now solving it we get x=48/5

i used the formula abc/4 area of triangle=radius 36x/4 [(144+x)/2(x^2/4)]^1/2 36/[(144-x^2)(x^2)]^1/2 so 144-x x=36 36/5*5 x^2=2304/25 x=48/5