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Find \(BC\).

 

https://latex.artofproblemsolving.com/f/e/4/fe45662f573f87e30f9f30d0b1ef1f94b4c22fc2.png

 May 27, 2019

Best Answer 

 #1
avatar+8884 
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By the Law of Cosines:

 

\(a^2 = c^2 + b^2 - 2cb \cos A\\~\\ (\overline{\text{BC}})^2 \,=\,(7\sqrt2)^2+(6)^2-2(7\sqrt2)(6)\cos(45°)\\~\\ (\overline{\text{BC}})^2 \,=\,98+36-(84\sqrt2)(\frac{\sqrt2}{2})\\~\\ (\overline{\text{BC}})^2 \,=\,98+36-84\\~\\ (\overline{\text{BC}})^2 \,=\,50\\~\\ \overline{\text{BC}} \,=\,\sqrt{50}\\~\\ \overline{\text{BC}} \,=\,5\sqrt{2}\)

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 May 27, 2019
 #1
avatar+8884 
+3
Best Answer

 

By the Law of Cosines:

 

\(a^2 = c^2 + b^2 - 2cb \cos A\\~\\ (\overline{\text{BC}})^2 \,=\,(7\sqrt2)^2+(6)^2-2(7\sqrt2)(6)\cos(45°)\\~\\ (\overline{\text{BC}})^2 \,=\,98+36-(84\sqrt2)(\frac{\sqrt2}{2})\\~\\ (\overline{\text{BC}})^2 \,=\,98+36-84\\~\\ (\overline{\text{BC}})^2 \,=\,50\\~\\ \overline{\text{BC}} \,=\,\sqrt{50}\\~\\ \overline{\text{BC}} \,=\,5\sqrt{2}\)

hectictar May 27, 2019

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