Mathematics - Properties of Triangle Question with Solution | TestHub

Using the sine rule, $\frac{a}{\sin A}=\frac{b}{\sin B}$. Substituting the given values: $\frac{6}{\sin 30^{\circ }}=\frac{8}{\sin B}$.$\frac{6}{1/2}=\frac{8}{\sin B}⟹12=\frac{8}{\sin B}⟹\sin B=\frac{8}{12}=\frac{2}{3}$. Since $0<\sin B<1$, there are two possible values for $\angle B$: $B_1=\arcsin (2/3)$ (acute) and $B_2=180^{\circ }-\arcsin (2/3)$ (obtuse). For $B_1$, $A+B_1=30^{\circ }+\arcsin (2/3)$. Since $\arcsin (2/3)\approx 41.8^{\circ }$, $A+B_1\approx 71.8^{\circ }<180^{\circ }$. So, a valid triangle exists. For $B_2$, $A+B_2=30^{\circ }+(180^{\circ }-\arcsin (2/3))\approx 30^{\circ }+138.2^{\circ }=168.2^{\circ }<180^{\circ }$. So, a second valid triangle exists. Therefore, two distinct triangles can be formed.