3. In triangle $A B C$, we are given that $A C=5 \sqrt{2}, B C=5$ and $\angle B A C=30^{\circ}$. What is the largest possible size in degrees of $\angle A B C$ ?
Solution
135

Let $\angle A B C=\theta$. By the sine rule, $\frac{\sin \theta}{5 \sqrt{2}}=\frac{\sin 30}{5}$, which simplifies to $\frac{0.5}{5}$. And so $\sin \theta=$ $\frac{0.5 \times 5 \sqrt{2}}{5}=\frac{\sqrt{2}}{2}$. Therefore $\theta=45,135$ and so the largest possible value of $\theta$ is 135 .