Ответ:
∠A ≈ 60.88°
Дано:
A(2;1;3),
B(7;4;5),
C(4;2;1).
-------------------------
Найти:
∠A - ?
--------------------------
Решение:
[tex]KL=\sqrt{(K_x-L_x)^2+(K_y-L_y)^2+(K_z-L_z)^2}[/tex]
[tex]AB=\sqrt{(2-7)^2+(1-4)^2+(3-5)^2}=\sqrt{38}\\BC=\sqrt{(7-4)^2+(4-2)^2+(5-1)^2}=\sqrt{29}\\AC=\sqrt{(2-4)^2+(1-2)^2+(3-1)^2}= 3[/tex]
[tex]BC^2=AC^2+AB^2-2AC*AB*cos(A)\\cos(A)=\frac{AC^2+AB^2-BC^2}{2AC*AB} \\cos(A)=\frac{3^2+(\sqrt{38})^2-(\sqrt{29})^2 }{2*3*\sqrt{38} } =\frac{3\sqrt{38} }{38} \\[/tex]
∠A = arccos((3√38) / 38) ≈ 60.88°
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Verified answer
Ответ:
∠A ≈ 60.88°
Дано:
A(2;1;3),
B(7;4;5),
C(4;2;1).
-------------------------
Найти:
∠A - ?
--------------------------
Решение:
[tex]KL=\sqrt{(K_x-L_x)^2+(K_y-L_y)^2+(K_z-L_z)^2}[/tex]
[tex]AB=\sqrt{(2-7)^2+(1-4)^2+(3-5)^2}=\sqrt{38}\\BC=\sqrt{(7-4)^2+(4-2)^2+(5-1)^2}=\sqrt{29}\\AC=\sqrt{(2-4)^2+(1-2)^2+(3-1)^2}= 3[/tex]
[tex]BC^2=AC^2+AB^2-2AC*AB*cos(A)\\cos(A)=\frac{AC^2+AB^2-BC^2}{2AC*AB} \\cos(A)=\frac{3^2+(\sqrt{38})^2-(\sqrt{29})^2 }{2*3*\sqrt{38} } =\frac{3\sqrt{38} }{38} \\[/tex]
∠A = arccos((3√38) / 38) ≈ 60.88°