25. In $\triangle A B C$, let $A B-c, B C=a$ and $A C=b$. Suppose that $\frac{b}{c-a}-\frac{a}{b+c}=1$, find the value of the greatest angle of $\triangle A B C$ in degrees.
25. Answer: $120^{\circ}$.
It is given that $\frac{b}{c-a}-\frac{a}{b+c}=1$. This can be rearranged into $b^{2}+a^{2}-c^{2}=-a b$. By using cosine rule for triangle, $\cos C=\frac{b^{2}+a^{2}-c^{2}}{2 a b}=-\frac{1}{2}$. Hence $C=120^{\circ}$ and must be the greatest angle.