Pls ASsist Soon!!!!!

See original post:  https://web2.0calc.com/questions/ahellelepoeowoe-gelp

There's some number of triangles satisfying
What is the sum of all the possible values of BC? If there are no possible values, answer with 0.

$\mathbf{\text{cos-rule}}:$

$\begin{array}{|rcll|} \hline a^2 &=& b^2+c^2-2bc\cos(A) \\ \mathbf{\cos(A)} &=& \mathbf{\dfrac{b^2+c^2-a^2}{2bc}} \\ \hline \end{array}$

$\mathbf{\text{projection-rule}}:$

$\begin{array}{|rcll|} \hline b &=& c\cos(A)+a\cos(C) \quad | \quad \mathbf{\cos(A)=\dfrac{b^2+c^2-a^2}{2bc}} \\ b &=& c\left(\dfrac{b^2+c^2-a^2}{2bc} \right)+a\cos(C) \\ b &=& \dfrac{b^2+c^2-a^2}{2b } +a\cos(C)\quad | \quad *2b \\ 2b^2 &=& b^2+c^2-a^2 +2ab\cos(C) \\ b^2 &=& c^2-a^2 +2ab\cos(C) \\ \mathbf{a^2 -2ab\cos(C) +b^2-c^2} &=& \mathbf{0} \\\\ a &=& \dfrac{2b\cos(C)\pm \sqrt{4b^2\cos^2(C)-4*(b^2-c^2)} }{2} \\ a &=& b\cos(C) \pm \dfrac{\sqrt{4b^2\cos^2(C)-4*(b^2-c^2)} }{2} \\ a_1+a_2 &=& 2b\cos(C) \quad | \quad b=10,\quad C = \dfrac{\pi}{6} \\ &=& 2*10\cos(\dfrac{\pi}{6}) \quad | \quad \cos(\dfrac{\pi}{6}) = \dfrac{\sqrt{3}}{2} \\ &=& 2*10*\dfrac{\sqrt{3}}{2} \\ &=& 10 \sqrt{3} \\ \mathbf{a_1+a_2} &=& \mathbf{17.3205080757} \\ \hline \end{array}$

The sum of all the possible values of BC is 17.3205080757