В треугольнике \(\displaystyle ABC\) известно, что \(\displaystyle AB=8 {\small,}\) \(\displaystyle BC=10 {\small,}\) \(\displaystyle AC=12 {\small.}\) Найдите \(\displaystyle \cos \angle ABC {\small.}\)
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Теорема косинусов
В треугольнике \(\displaystyle ABC\) \(\displaystyle \color {#339900}{AC}^2=\red{AB}^2+\color {blue}{BC}^2-2\cdot \red{AB} \cdot \color {blue}{BC}\cdot \cos \angle \color {#339900}{B}{\small .}\) |  |
\(\displaystyle {AC}^2={AB}^2+{BC}^2-2\cdot {AB} \cdot {BC}\cdot \cos \angle {ABC}{\small .}\)
Так как \(\displaystyle AB=8 {\small,}\) \(\displaystyle BC=10 {\small,}\) \(\displaystyle AC=12 {\small ,}\) то
\(\displaystyle {12}^2={8}^2+{10}^2-2\cdot {8} \cdot {10}\cdot \cos \angle ABC {\small .}\)
Значит,
\(\displaystyle 144=64+100-160\cdot \cos \angle ABC {\small ,}\)
\(\displaystyle 144=164-160\cdot \cos \angle ABC {\small ,}\)
\(\displaystyle 160\cdot \cos \angle ABC =20{\small ,}\)
\(\displaystyle \cos \angle ABC =0{,}125{\small .}\)
Ответ: \(\displaystyle 0{,}125{\small.}\)