Question

tam giác ABC có BC=\(\sqrt{5}\), AC=3 và cotC=2. tính cạnh AB.

Answer 1

cot C=2

=>\(tanC=\dfrac{1}{cotC}=\dfrac{1}{2}\)

\(1+tan^2C=\dfrac{1}{cos^2C}\)

=>\(cos^2C=1+\dfrac{1}{4}=\dfrac{5}{4}\)

=>\(cosC=\dfrac{2}{\sqrt{5}}\) hoặc \(cosC=-\dfrac{2}{\sqrt{5}}\)

TH1: \(cosC=\dfrac{2}{\sqrt{5}}\)

=>\(\dfrac{BC^2+AC^2-AB^2}{2\cdot BC\cdot AB}=\dfrac{2}{\sqrt{5}}\)

=>\(\dfrac{5+9-AB^2}{6\sqrt{5}}=\dfrac{2}{\sqrt{5}}\)

=>\(14-AB^2=12\)

=>AB^2=2

=>\(AB=\sqrt{2}\)

TH2: \(cosC=-\dfrac{2}{\sqrt{5}}\)

=>\(\dfrac{5+9-AB^2}{6\sqrt{5}}=-\dfrac{2}{\sqrt{5}}\)

=>\(14-AB^2=\dfrac{-2}{\sqrt{5}}\cdot6\sqrt{5}=-12\)

=>AB^2=26

=>\(AB=\sqrt{26}\)