Dua buah vektor gaya besarnya masing-masing \(F_1 = 6\) satuan dan \(F_2 = 3\) satuan, saling membentuk sudut 60°. Tentukan:

(A) Besar \(R\)
\begin{equation*}
\begin{split}
|R| & = \sqrt{(F_1)^2 + (F_2)^2 + 2 \:.\: F_1 \:.\: F_2 \:.\: \cos \theta} \\\\
|R| & = \sqrt{(6)^2 + (3)^2 + 2 \:.\: 6 \:.\: 3 \:.\: \cos 60} \\\\
|R| & = \sqrt{(6)^2 + (3)^2 + 2 \:.\: 6 \:.\: 3 \:.\: \frac 12} \\\\
|R| & = \sqrt{63} \\\\
|R| & = 3 \sqrt{7}
\end{split}
\end{equation*}
(B) Sudut antara \(R\) dan \(F_1\)
\begin{equation*}
\begin{split}
\frac{F_1}{\sin \beta} & = \frac{F_2}{\sin \alpha} = \frac{R}{\sin \theta} \\\\
\frac{F_2}{\sin \alpha} & = \frac{R}{\sin \theta} \\\\
\frac{\cancel {3}}{\sin \alpha} & = \frac{\cancel {3} \sqrt{7}}{\sin 60} \\\\
\sqrt{7} \sin \alpha & = \sin 60 \\\\
\sin \alpha & = \frac {\sin 60}{\sqrt{7}} \\\\
\sin \alpha & = \frac {\frac 12 \sqrt{3}}{\sqrt{7}} \\\\
\sin \alpha & = \frac {\sqrt{3}}{2 \sqrt{7}} \quad {\color {blue} \times \frac {\sqrt{7}}{\sqrt{7}}} \\\\
\sin \alpha & = \frac {\sqrt{21}}{14} \\\\
\alpha & = \sin^{-1} \frac {\sqrt{21}}{14} \\\\
\alpha & = 19^{\text{o}}
\end{split}
\end{equation*}
(C) Sudut antara \(R\) dan \(F_2\)
\(\beta = 60^{\text{o}} - 19^{\text{o}} = 41^{\text{o}}\)