Ответ:
По теореме косинусов
[tex]\bf AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot cos\angle{B}\ \ \ \ \Rightarrow \\\\\\cos\angle{B}=\dfrac{AB^2+BC^2-AC^2}{2\cdot AB\cdot BC}=\dfrac{4+16-4\cdot 3}{2\cdot 2\cdot 4}=\dfrac{8}{16}=\dfrac{1}{2}\ \ \ \ \Rightarrow \\\\\\\angle{B}=60^\circ[/tex]
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Ответ:
По теореме косинусов
[tex]\bf AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot cos\angle{B}\ \ \ \ \Rightarrow \\\\\\cos\angle{B}=\dfrac{AB^2+BC^2-AC^2}{2\cdot AB\cdot BC}=\dfrac{4+16-4\cdot 3}{2\cdot 2\cdot 4}=\dfrac{8}{16}=\dfrac{1}{2}\ \ \ \ \Rightarrow \\\\\\\angle{B}=60^\circ[/tex]