si acest ex 5 va rog si e​

Răspuns:

10

Explicație pas cu pas:

AC=AB+BC=3i-4j

AC+AC=2AC=6i-8j de modul (lungime) √(6²+(-8)²)=10

6

tii cont ca 2sinαcosα=sin(2α)

deci

4sin(2*π/12)=4*sin(π/6) =4*1/2=2

[tex]\it\ \overrightarrow{AC} =\overrightarrow{AB}+\overrightarrow{BC}=2\overrightarrow{i}- 3\overrightarrow{j}+\overrightarrow{i}-\overrightarrow{j}=3\overrightarrow{i}-4\overrightarrow{j}[/tex]

[tex]\it \overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{BC}= \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{AC}=2\overrightarrow{AC}=2(3\overrightarrow{i}-4\overrightarrow{j})=6\overrightarrow{i}-8\overrightarrow{j}[/tex]

[tex]\it |\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{BC}|=|6\overrightarrow{i}-4\overrightarrow{j}|=\sqrt{6^2+8^2}=\sqrt{100}=10[/tex]