Cho tam giác ABC có AB = 3; AC = 5; $\widehat{BAC}=60°.$Gọi M là điểm thuộc đoạn BC sao cho BM = 2MC . Tính độ dài đoạn AM.

Áp dụng định lý cos, ta có:

$\text{BC = }\sqrt{{\text{AB}}^{\text{2}}{\text{ + AC}}^{\text{2}}\text{ - 2AB . AC . cosBAC}}\text{ = }\sqrt{\text{19}}$

$\Rightarrow \text{cosB = }\frac{{\text{AB}}^{\text{2}}{\text{ + BC}}^{\text{2}}{\text{ - AC}}^{\text{2}}}{\text{2AB . BC}}\text{ = }\frac{\sqrt{\text{19}}}{\text{38}}$

$\text{BM = 2MC}\Rightarrow \text{BM = }\frac{\text{2}}{\text{3}}\text{BC = }\frac{\text{2}\sqrt{\text{19}}}{\text{3}}$

$\Rightarrow \text{AM = }\sqrt{{\text{AB}}^{\text{2}}{\text{ + BM}}^{\text{2}}\text{ - 2AB . BM . cosB}}\text{ = }\frac{\sqrt{\text{139}}}{\text{3}}$.