Question

Answer 1

a.

Do ABC vuông cân \(\Rightarrow\left\{{}\begin{matrix}\widehat{ABC}=45^0\\BC=AB\sqrt[]{2}=a\sqrt{2}\end{matrix}\right.\)

\(\Rightarrow BM=\dfrac{2}{3}BC=\dfrac{2a\sqrt{2}}{3}\)

Áp dụng định lý cosin cho tam giác ABM:

\(AM=\sqrt{AB^2+BM^2-2AB.BM.cos\widehat{ABC}}=\dfrac{a\sqrt{5}}{3}\)

b.

\(A+B+C=180^0\Rightarrow cosA=-cos\left(B+C\right)=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow BC=\sqrt{AB^2+AC^2-2AB.AC.cosA}=2\sqrt{2}\)