Xét `\triangle ABC` vuông tại `A` có:
`@cot B=[AB]/[AC]=>5/8=5/[AC]=>AC=8(cm)
`@AB^2+AC^2=BC^2=>5^2+8^2=BC^2=>BC=\sqrt{89}(cm)`
cotB = \(\dfrac{AB}{AC}= \dfrac{5}{8}\)
=> \(\dfrac{5}{AC}= \dfrac{5}{8}\)
=> AC = 8 cm
áp dụng định lí pi - ta - go :
BC2 = AB2 + AC2
=> BC2 = 52 + 82
=> BC = √89
\(CotB=\dfrac{5}{8}\Rightarrow\dfrac{BA}{AC}=\dfrac{5}{8}\Rightarrow AC=\dfrac{8.5}{5}=8\left(cm\right)\)
\(\Delta ABC\perp A\)
\(BC^2=BA^2+AC^2\\ \Rightarrow BC^2=5^2+8^2\\ \Rightarrow BC=\sqrt{89}\)