Áp dụng định lý Pytago
\(AC=\sqrt{5^2-3^2}=4cm\)
Ta có:
\(cotB=\dfrac{AB}{AC}=\dfrac{3}{4}\)
\(cotC=\dfrac{AC}{AB}=\dfrac{4}{3}\)
\(\Rightarrow P=\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{25}{12}\)
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Có : \(BC^2=AC^2+AB^2\left(d/l-Pytago\right)\)
\(\Rightarrow AC=\sqrt{BC^2-AB^2}=\sqrt{5^2-3^2}=4\left(cm\right)\)
\(tanB=\dfrac{AC}{AB}=\dfrac{4}{3}\Rightarrow cotB=1:\dfrac{4}{3}=\dfrac{3}{4}\)
\(tanC=\dfrac{AB}{AC}=\dfrac{3}{4}\Rightarrow cotC=1:\dfrac{3}{4}=\dfrac{4}{3}\)
Vậy \(P=cotB+cotC=\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{25}{12}\)
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