Question

Answer 1

Đặt \(A=2R\cdot sinB\cdot sinC\)

Xét ΔABC có

\(\dfrac{AB}{sinC}=\dfrac{AC}{sinB}=\dfrac{BC}{sinA}=2R\)

=>\(sinC=\dfrac{AB}{2R};sinB=\dfrac{AC}{2R};sinA=\dfrac{BC}{2R}\)

\(A=2R\cdot sinB\cdot sinC=2R\cdot\dfrac{AB}{2R}\cdot\dfrac{AC}{2R}=\dfrac{AB\cdot AC}{2R}\)

\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinA=\dfrac{1}{2}\cdot h_A\cdot BC\)

=>\(AB\cdot AC=h_A\cdot\dfrac{BC}{sinA}\)

\(A=\dfrac{AB\cdot AC}{2R}=\dfrac{h_A\cdot BC}{sinA}:2R=\dfrac{h_A\cdot2R\cdot sinA}{2R\cdot sinA}\)

\(=h_A\)