Question

Tema: Vektor
Jumlah Soal: 1

Answer 1

Jawab:

A

Penjelasan dengan langkah-langkah:

[tex]\displaystyle \frac{\sqrt{\overrightarrow{a^2}+\overrightarrow{b^2}-2\overrightarrow{a}\overrightarrow{b}\cos \theta}}{\sqrt{\overrightarrow{a^2}+\overrightarrow{b^2}+2\overrightarrow{a}\overrightarrow{b}\cos \theta}}=\frac{\sqrt{2}}{2}\\\frac{\overrightarrow{a^2}+\overrightarrow{b^2}-2\overrightarrow{a}\overrightarrow{b}\cos \theta}{\overrightarrow{a^2}+\overrightarrow{b^2}+2\overrightarrow{a}\overrightarrow{b}\cos \theta}=\frac{2}{4}[/tex]

[tex]\displaystyle \frac{\overrightarrow{a^2}+\overrightarrow{a^2}-2\overrightarrow{a}\overrightarrow{a}\cos \theta}{\overrightarrow{a^2}+\overrightarrow{a^2}+2\overrightarrow{a}\overrightarrow{a}\cos \theta}=\frac{1}{2}\\\frac{2\overrightarrow{a^2}-2\overrightarrow{a^2}\cos \theta}{2\overrightarrow{a^2}+2\overrightarrow{a^2}\cos \theta}=\frac{1}{2}[/tex]

[tex]\displaystyle \frac{2\overrightarrow{a^2}(1-\cos \theta)}{2\overrightarrow{a^2}(1+\cos \theta)}=\frac{1}{2}\\2(1-\cos \theta)=1+\cos \theta\\1=3\cos \theta\\\cos \theta=\frac{1}{3}[/tex]