найти косинусы глов треугольника с вершинами А(2.8) В(-1.5) С(3.1)
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Ответил Матов
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найдем сторону с начало
![AB= sqrt{(-1-2)^2+(5-8)^2}=sqrt{9+9}=3sqrt{2}\
BC=sqrt{(3+1)^2+(1-5)^2}=sqrt{16+16}=4sqrt{2}\
AC=sqrt{(3-2)^2+(1-8)^2}=sqrt{1+49}= 5sqrt{2}\
\
cosBAC=frac{32-50-18}{-60}= frac{2}{5}\
cosACB=frac{18-32-50}{-80}=frac{4}{5}\
cosABC=frac{50-18-32}{-48}=frac{4}{3}\](https://tex.z-dn.net/?f=AB%3D+sqrt%7B%28-1-2%29%5E2%2B%285-8%29%5E2%7D%3Dsqrt%7B9%2B9%7D%3D3sqrt%7B2%7D%5C%0ABC%3Dsqrt%7B%283%2B1%29%5E2%2B%281-5%29%5E2%7D%3Dsqrt%7B16%2B16%7D%3D4sqrt%7B2%7D%5C%0AAC%3Dsqrt%7B%283-2%29%5E2%2B%281-8%29%5E2%7D%3Dsqrt%7B1%2B49%7D%3D+5sqrt%7B2%7D%5C%0A%5C%0AcosBAC%3Dfrac%7B32-50-18%7D%7B-60%7D%3D+frac%7B2%7D%7B5%7D%5C%0AcosACB%3Dfrac%7B18-32-50%7D%7B-80%7D%3Dfrac%7B4%7D%7B5%7D%5C%0AcosABC%3Dfrac%7B50-18-32%7D%7B-48%7D%3Dfrac%7B4%7D%7B3%7D%5C "AB= sqrt{(-1-2)^2+(5-8)^2}=sqrt{9+9}=3sqrt{2}\
BC=sqrt{(3+1)^2+(1-5)^2}=sqrt{16+16}=4sqrt{2}\
AC=sqrt{(3-2)^2+(1-8)^2}=sqrt{1+49}= 5sqrt{2}\
\
cosBAC=frac{32-50-18}{-60}= frac{2}{5}\
cosACB=frac{18-32-50}{-80}=frac{4}{5}\
cosABC=frac{50-18-32}{-48}=frac{4}{3}\")
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