Question

знайти косинус Кута А в ∆АВС, якщо А(-2;3),В(2;-1)С(1;6).​

Answer 1

Объяснение:

$A(-2;3)\ \ \ \ B(2;-1)\ \ \ \ \ C(1;6)\ \ \ \ cosA=?\\|AB|=\sqrt{(-2-2)^2+(3-(-1))^2} =\sqrt{(-4)^2+4^2} =\sqrt{16+16}=\sqrt{32}=4\sqrt{2}.\\ |BC|=\sqrt{(2-1)^2+(-1-6)^2}=\sqrt{1^2+(-7)^2}=\sqrt{1+49}=\sqrt{50} =5\sqrt{2}.\\| AC| =\sqrt{(-2-1)^2+(3-6)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}.\\ |BC|^2=|AB|^2+|AC|^2-2*|AB|*|AC|*cosA$

$cosA=\frac{|AB|^2+|AC|^2-|BC|^2}{2*|AB|*|AC|} =\frac{(4\sqrt{2})^2+(3\sqrt{2})^2 -(5\sqrt{2})^2 }{2*4\sqrt{2} *3\sqrt{2} } =\frac{32+18-50}{48} =\frac{0}{48}=0.$

Ответ: cosA=0.