Cho tam giác ABC có hb + hc = 2ha. Chứng minh rằng: 1/sin B + 1/sin C
Cho tam giác ABC có hb + hc = 2ha. Chứng minh rằng:
\(\frac{1}{{\sin B}} + \frac{1}{{\sin C}} = \frac{2}{{\sin {\rm{A}}}}\).
Ta có \({S_{ABC}} = \frac{1}{2}{h_a}a = \frac{1}{2}{h_b}b = \frac{1}{2}{h_c}c\)
Suy ra \[{h_b} = \frac{{2{{\rm{S}}_{ABC}}}}{b},{h_c} = \frac{{2{{\rm{S}}_{ABC}}}}{c},{h_a} = \frac{{2{{\rm{S}}_{ABC}}}}{a}\]
Ta có hb + hc = 2ha
\( \Leftrightarrow \frac{{2{{\rm{S}}_{ABC}}}}{b} + \frac{{2{{\rm{S}}_{ABC}}}}{c} = \frac{{{\rm{4}}{{\rm{S}}_{ABC}}}}{a}\)
\( \Leftrightarrow \frac{1}{b} + \frac{1}{c} = \frac{2}{a}\)
Áp dụng định lí sin ta có
\(\frac{1}{{\sin {\rm{A}}}} + \frac{1}{{\sin B}} = \frac{{2R}}{b} + \frac{{2{\rm{R}}}}{c} = 2{\rm{R}}\left( {\frac{1}{b} + \frac{1}{c}} \right) = 2{\rm{R}}.\frac{2}{a} = \frac{2}{{\sin {\rm{A}}}}\)
Vậy \(\frac{1}{{\sin B}} + \frac{1}{{\sin C}} = \frac{2}{{\sin {\rm{A}}}}\).