Diketahui segitiga ABC memiliki sudut BAC = 60° panjang AB = 6cm dan panjang AC = 3cm Panjang BC = ....

 Dengan aturan kosinus:
BC=\sqrt{AB^2+AC^2-2AB.AC.\cos BAC} \\ BC=\sqrt{4^2+3^2-2.4.3.\cos 60} \\ BC=\sqrt{16+9-24\times\frac{1}{2}} \\ BC=\sqrt{25-12} \\ BC=\sqrt{13}
Teorema garis bagi:
\displaystyle \frac{AB}{AC}=\frac{BD}{CD} \\ \\ \frac{4}{3}=\frac{BD}{\sqrt{13}-BD} \\ \\ 4\sqrt{13}-4BD=3BD \\ 7BD=4\sqrt{13} \\$BD=\frac{4}{7}\sqrt{13} \\ $Dan,$ \\ CD=\frac{3}{7}\sqrt{13} \\ \\ $Rumus garis bagi:$ \\ \displaystyle AD^2=AB.AC-BD.CD \\ AD^2=4.3-\frac{4}{7}\sqrt{13}.\frac{3}{7}\sqrt{13} \\ AD^2=12-\frac{12\times 13}{49} \\ \\ AD^2=\frac{12.49-156}{49} \\ AD^2=\frac{588-146}{49}=\frac{442}{49} \\ AD=\sqrt{\frac{442}{49}} \\ \\ AD=\frac{\sqrt{442}}{7}

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