Cho tam giác ABC có AB = 6cm, AC = 3cm, góc BAC= 60 độ, M là điểm thỏa mãn

C. $\text{AM = 3 cm}$

Trả lời

Chọn A

Áp dụng định lý coscos ta được:

${\text{\hspace{1em}BC}}^{\text{2}}{\text{ = AB}}^{\text{2}}{\text{ + AC}}^{\text{2}}\text{ - 2AB . AC . cos}\widehat{\text{BAC}}$

$\Rightarrow {\text{BC}}^{\text{2}}{\text{ = 6}}^{\text{2}}{\text{ + 3}}^{\text{2}}{\text{ - 2 . 6 . 3 . cos60}}^{\text{o}}$

$\Rightarrow {\text{BC}}^{\text{2}}\text{ = 27}$

$\Rightarrow \text{BC = 3}\sqrt{\text{3}}\text{ cm}$

$\Rightarrow {\text{AB}}^{\text{2}}{\text{ = BC}}^{\text{2}}{\text{ + AC}}^{\text{2}}$

$\Rightarrow \Delta \text{ABC}$ vuông tại C

Mặt khác:

$\overrightarrow{\text{MB}}\text{ + 2}\overrightarrow{\text{MC}}$

$\Rightarrow \text{CM = }\frac{\text{1}}{\text{3}}\text{BC = }\sqrt{\text{3}}\text{ cm}$

Áp dụng định lý Pytago ta được:

${\text{AM}}^{\text{2}}{\text{ = AC}}^{\text{2}}{\text{ + CM}}^{\text{2}}$

$\Rightarrow \text{AM = }\sqrt{{\text{AC}}^{\text{2}}{\text{ + CM}}^{\text{2}}}\text{ = }\sqrt{\text{9 + 3}}$

$\Rightarrow \text{AM = 2}\sqrt{\text{3}}\text{ cm}$