Mathematics - Properties of Triangle Question with Solution | TestHub

Using the sine rule, we have $\frac{a}{\sin A}=\frac{b}{\sin B}$. Substituting the given values, $\frac{2\sqrt{2}}{\sin 45^{\circ }}=\frac{b}{\sin 60^{\circ }}$. This becomes $\frac{2\sqrt{2}}{1/\sqrt{2}}=\frac{b}{\sqrt{3}/2}$. Simplifying, $2\sqrt{2}\cdot \sqrt{2}=b\cdot \frac{2}{\sqrt{3}}$, which means $4=b\cdot \frac{2}{\sqrt{3}}$. Solving for $b$, we get $b=\frac{4\sqrt{3}}{2}=2\sqrt{3}$.