$BC=\sqrt{5};AC=3;cotC=-2$
Ta có:
`tanC=1/{cot C}=1/{-2}={-1}/2`
`tan^2 C+1={sin^2 C}/{cos^2 C}+1={sin^2 C+cos^2 C}/{cos^2 C}=1/{cos^2 C}`
`=>(-1/ 2)^2+1=1/{cos^2C}`
`=>1/{cos^2 C}=5/ 4`
`=>cos^2 C=4/ 5`
$⇒\left[\begin{array}{l}cosC=\dfrac{2}{\sqrt{5}}\\cosC=\dfrac{-2}{\sqrt{5}}\end{array}\right.$
Vì `0°<\hat{C}<180°`
`=>0<sinC<1; -1<cosC<1`
Mà `cotC={cosC}/{sinC}=-2<0=>cosC<0`
`=>cosC={-2}/{\sqrt{5}}`
$\\$
Áp dụng định lý Cosin ta có:
`AB^2=BC^2+AC^2-2BC.AC.cosC`
`AB^2=5+3^2-2.\sqrt{5}.3.{-2}/{\sqrt{5}}=26`
`=>AB=\sqrt{26}`
Vậy $AB=\sqrt{26}(đ v đ d)$
