What is the nature of Triangle if AB/AC=1/2 angle (BAC)=60° What is the nature of  Triangle if $\frac{AB}{AC}=\frac12$ and $\angle BAC=60^{\circ}$?. Can we use ratio between side lengths?

Hint: apply the law of cosines, and you get: $$BC^2=AB^2+AC^2-2AB\cdot AC \cos\angle BAC\\=AB^2+4AB^2-2AB\cdot2AB\frac12=AB^2+4AB^2-2AB^2=3AB^2\Rightarrow \\BC=AB\sqrt{3} \qquad AC=2AB$$
we must conclude that $\Delta ABC$ is a $30°-60°$ rectangular triangle with the hypothenuse coincident with $AC$.
