Question

giúp vs ạ

Answer 1

Ta có: \(BC^2=\left(\sqrt3+1\right)\cdot AC^2+\left(\sqrt3-1\right)\cdot AB\cdot AC\)

=>\(AC^2\cdot\sqrt3+AC^2+\sqrt3\cdot AB\cdot AC-AB\cdot AC-AB^2-AC^2=0\)

=>\(AC\cdot\sqrt3\left(AC+AB\right)-AB\left(AB+AC\right)=0\)

=>\(\left(AB+AC\right)\left(AC\sqrt3-AB\right)=0\)

=>\(AC\sqrt3-AB=0\)

=>\(AC\cdot\sqrt3=AB\)

=>\(\frac{AB}{AC}=\sqrt3\)

Xét ΔABC vuông tại A có tan C=\(\frac{AB}{AC}=\sqrt3\)

=>\(\hat{C}=60^0\)

ΔABC vuông tại A

=>\(\hat{B}+\hat{C}=90^0\)

=>\(\hat{B}=90^0-60^0=30^0\)