Question

Cho tam giác $A B C$, chứng minh rằng:

a) $\cot A=\dfrac{b^{2}+c^{2}-a^{2}}{4 S}$.

b) $\cot A+\cot B+\cot C \geq \sqrt{3}$.

Answer 1

`Answer:`

a) \(a^2=b^2+c^2-2bc\cos A\)

\(2S=bc.\sin A\)

\(\Rightarrow2bc=\frac{4S}{\sin A}\)

\(\Rightarrow a^2=b^2+c^2-\frac{4S\cos A}{\sin A}=b^2+c^2-4S\cot A\)

\(\Rightarrow\cot A=\frac{b^2+c^2-a^2}{4S}\)

Answer 2

A