Зная длины всех трех сторон треугольника, мы без проблем найдем все остальные параметры:
\( cos A={AC^2+AB^2-BC^2 \over 2*AC*AB}={\htmlClass{kAC}{8}^2+\htmlClass{kAB}{9}^2-\htmlClass{kBC}{7}^2 \over 2*\htmlClass{kAC}{8}*\htmlClass{kAB}{9}}=\htmlClass{kcosA}{0.667} \)
\( \angle A=arccos( cos A)=arccos( \htmlClass{kcosA}{0.667})=\htmlClass{kA}{48.19}\degree \)
\( cos C={AC^2+BC^2-AB^2 \over 2*AC*BC}={\htmlClass{kAC}{8}^2+\htmlClass{kBC}{7}^2-\htmlClass{kAB}{9}^2 \over 2*\htmlClass{kAC}{8}*\htmlClass{kBC}{7}}=\htmlClass{kcosC}{0.286} \)
\( \angle C=arccos( cos C)=arccos( \htmlClass{kcosC}{0.286})=\htmlClass{kC}{73.398}\degree \)
\(\angle B=180-\angle А- \angleС=180-\htmlClass{kA}{48.19}-\htmlClass{kC}{73.398}=\htmlClass{kB}{58.412}\degree\)
Так же можно найти угол В через косинус В\( cos B={AB^2+BC^2-AC^2 \over 2*AB*BC}={\htmlClass{kAB}{9}^2+\htmlClass{kBC}{7}^2-\htmlClass{kAC}{8}^2 \over 2*\htmlClass{kAB}{9}*\htmlClass{kBC}{7}}=\htmlClass{kcosB}{0.524} \)
\( \angle B=arccos( cos B)=arccos( \htmlClass{kcosB}{0.524})=\htmlClass{kB}{58.412}\degree \)
\(P=AB+BC+AC=\htmlClass{kAB}{9}+\htmlClass{kBC}{7}+\htmlClass{kAC}{8}=\htmlClass{kP}{24}\)
\(S=\htmlClass{fontm}{ \sqrt{{P\over 2}*({P\over 2}-AC)*({P\over 2}-BC)*({P\over 2}-AB)}}=\htmlClass{fontm}{ \sqrt{{\htmlClass{kP}{24}\over 2}*({\htmlClass{kP}{24}\over 2}-\htmlClass{kAC}{8})*({\htmlClass{kP}{24}\over 2}-\htmlClass{kBC}{7})*({\htmlClass{kP}{24}\over 2}-\htmlClass{kAB}{9})}}=\htmlClass{kS}{26.833}\)
\(H={2 * S\over AC}={2 * \htmlClass{kS}{26.833}\over \htmlClass{kAC}{8}}=\htmlClass{kH}{6.708}\)
\( Дано: \triangle ABC;\)
\( AB=\htmlClass{kAB}{9} см;\)
\( BC=\htmlClass{kBC}{7} см;\)
\( AC=\htmlClass{kAC}{8} см;\)
\( Найти: \angle A ?\)
\( Решение:\)
\(\angle A=arccos( cos A)\)
\(cos A={AC^2+AB^2-BC^2 \over 2*AC*AB}={\htmlClass{kAC}{8}^2+\htmlClass{kAB}{9}^2-\htmlClass{kBC}{7}^2 \over 2*\htmlClass{kAC}{8}*\htmlClass{kAB}{9}}=\htmlClass{kcosA}{0.667} \)
\( \angle A=arccos( \htmlClass{kcosA}{0.667})=\htmlClass{kA}{48.19}\degree \)
\( Ответ:\htmlClass{kA}{48.19}\degree\)