Mathematics - Properties of Triangle Question with Solution | TestHub

Using the sine rule, $\frac{a}{\sin A}=\frac{b}{\sin B}$. We have $\frac{2\sqrt{3}}{\sin 30^{\circ }}=\frac{6}{\sin B}$. This gives $\frac{2\sqrt{3}}{1/2}=\frac{6}{\sin B}$, so $4\sqrt{3}=\frac{6}{\sin B}$. Thus, $\sin B=\frac{6}{4\sqrt{3}}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}$. This implies $B=60^{\circ }$ or $B=120^{\circ }$. If $B=60^{\circ }$, then $A+B=30^{\circ }+60^{\circ }=90^{\circ }<180^{\circ }$, so $C=90^{\circ }$ is valid. If $B=120^{\circ }$, then $A+B=30^{\circ }+120^{\circ }=150^{\circ }<180^{\circ }$, so $C=30^{\circ }$ is valid. Since both values of B lead to valid triangles, there are 2 possible triangles.