What is the nature of Triangle if AB/AC=1/2 angle (BAC)=60°

Question

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

Answer 1

Using common notation, $BC=a,CA=b,AB=c$

We have $\displaystyle\frac cb=\frac12\iff b=2c\ \ \ \ (1) $

Applying cosine rule $$\cos\angle BAC=\frac{b^2+c^2-a^2}{2bc}$$

Using $(1)$ $$\frac{(2c)^2+c^2-a^2}{2(2c)c}=\cos60^\circ=\frac12\implies a^2=\cdots$$

Answer 2

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 $AC$.