Bài 2:
a) Ta có công thức: \(S_{ABC}=\dfrac{1}{2}AB.AC.sin\widehat{A}\Rightarrow sin\widehat{A}=\dfrac{2S}{AB.AC}=\dfrac{2S}{bc}\).
Tương tự: \(sin\widehat{B}=\dfrac{2S}{ca};sin\widehat{C}=\dfrac{2S}{ab}.\)
\(\Rightarrow sin\widehat{B}+sin\widehat{C}=\dfrac{2S}{ca}+\dfrac{2S}{ab}=\dfrac{2S}{a}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{2S}{a}.\dfrac{b+c}{bc}=\dfrac{2S}{a}.\dfrac{2a}{bc}=\dfrac{4S}{bc}\)
Mà \(2sin\widehat{A}=2.\dfrac{2S}{bc}=\dfrac{4S}{bc}\Rightarrow2sin\widehat{A}=sin\widehat{B}+sin\widehat{C}\).
b) Ta có: \(S=\dfrac{1}{2}h_a.BC\Rightarrow h_a=\dfrac{2S}{BC}=\dfrac{2S}{a}\)
Tương tự: \(h_b=\dfrac{2S}{b};h_c=\dfrac{2S}{c}\)
\(\Rightarrow\dfrac{1}{h_b}+\dfrac{1}{h_c}=\dfrac{b}{2S}+\dfrac{c}{2S}=\dfrac{b+c}{2S}=\dfrac{2a}{2S}=\dfrac{a}{S}\)
Mà \(\dfrac{2}{h_a}=2.\dfrac{a}{2S}=\dfrac{a}{S}\Rightarrow\dfrac{2}{h_a}=\dfrac{1}{h_b}+\dfrac{1}{h_c}\)
