Question

Cho tam giác ABC biết \(S=b^2-\left(a-c\right)^2\). Tính \(\tan B\)

Answer 1

Ta có : \(S=\dfrac{1}{2}SinB.ac=b^2-a^2-c^2+2ac\)

\(\Rightarrow\dfrac{1}{2}SinB.ac=-\left(a^2+c^2-b^2\right)+2ac\)

Mà \(CosB=\dfrac{a^2+c^2-b^2}{2ac}\)

\(\Rightarrow a^2+c^2-b^2=2ac.CosB\)

\(\Rightarrow\dfrac{1}{2}SinB.ac=2ac-2ac.\cos B\)

\(\Rightarrow SinB=4-4\cos B\)

\(\Rightarrow SinB+4\cos B=4\)

Lại có : \(\sin^2B+\cos^2B=1\)

- Giair hệ ta được : \(\left\{{}\begin{matrix}\left[{}\begin{matrix}\cos B=1\\\cos B=\dfrac{15}{17}\end{matrix}\right.\\\left[{}\begin{matrix}sinB=0\\sinB=\dfrac{8}{17}\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}tanB=0\\tanB=\dfrac{8}{15}\end{matrix}\right.\)

Mà 3 điểm A, B, C là 1 tam giác .

=> TanB = 8/15 .