[tex]\it cosB=\dfrac{3}{5} tgC\ \ \ \ (*)
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cosB=\dfrac{AB}{BC}=\dfrac{4\sqrt2}{BC}
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tgC= \dfrac{AB}{AC} =\dfrac{4\sqrt2}{AC}[/tex]
Relația din enunț devine:
[tex]\it \dfrac{4\sqrt2}{BC} =\dfrac{3}{5}\cdot \dfrac{4\sqrt2}{AC} |_{\cdot\frac{AC}{4\sqrt2} } \Rightarrow \dfrac{AC}{BC} =\dfrac{3}{5} \Rightarrow \left(\dfrac{AC}{BC} \right)^2 =\left(\dfrac{3}{5}\right)^2 \Rightarrow
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\Rightarrow \dfrac{AC^2}{BC^2} =\dfrac{9}{25 } \Rightarrow AC^2= \dfrac{9\cdot BC^2}{25} \ \ \ \ (1)[/tex]
Din teoreme lui Pitagora, rezultă:
[tex]\it AC^2= BC^2-AB^2 = BC^2-(4\sqrt2)^2 =BC^2-32\ \ \ \ (2)
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(1), (2) \Rightarrow BC^2-32=\dfrac{9 BC^2}{25} \Rightarrow 25BC^2-32\cdot25 =9BC^2 \Rightarrow
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\Rightarrow 25BC^2-9BC^2=32\cdot25 \Rightarrow 16BC^2=32\cdot25|_{:16} \Rightarrow BC^2=2\cdot 25[/tex]
[tex]\it \Rightarrow BC=\sqrt{2\cdot25} \Rightarrow BC= 5\sqrt2\ cm[/tex]
