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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.

 May 22, 2019
 #1
avatar+8209 
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 May 22, 2019
 #2
avatar+22358 
+3

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

 

laugh

 May 22, 2019

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