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.
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