Question

Answer 1

MC=2MB

mà MC+MB=BC=6

nên \(MC=\frac23\cdot6=4;MB=4\cdot\frac12=2\)

Xét ΔACB có \(cosC=\frac{CA^2+CB^2-AB^2}{2\cdot CA\cdot CB}=\frac{6^2+\left(2\sqrt7\right)^2-4^2}{2\cdot6\cdot2\sqrt7}\)

\(=\frac{36+28-16}{24\sqrt7}=\frac{48}{24\sqrt7}=\frac{2}{\sqrt7}\)

xét ΔCAM có \(cosC=\frac{CA^2+CM^2-AM^2}{2\cdot CA\cdot CM}\)

=>\(\frac{\left(2\sqrt7\right)^2+4^2-AM^2}{2\cdot2\sqrt7\cdot4}=\frac{2}{\sqrt7}\)

=>\(28+16-AM^2=\frac{2}{\sqrt7}\cdot4\cdot\sqrt7\cdot4=2\cdot4\cdot4=2\cdot16=32\)

=>\(AM^2=44-32=12\)

=>\(AM=2\sqrt3\)