1. **Problem statement:** Given triangle ABC with AC = AB and angles DCA and ACB supplementary, find (a) $\angle ACB$, (b) $\angle ABC$, and (c) $\angle CAB$. 2. **Key facts and formulas:** - Since AC = AB, triangle ABC is isosceles with $\angle ABC = \angle CAB$. - Supplementary angles sum to 180°, so $\angle DCA + \angle ACB = 180^\circ$. - Given $\angle DCA = 115^\circ$, then $\angle ACB = 180^\circ - 115^\circ = 65^\circ$. 3. **Find $\angle ACB$:** $$\angle ACB = 180^\circ - 115^\circ = 65^\circ$$ 4. **Find $\angle ABC$ and $\angle CAB$:** - Let $x = \angle ABC = \angle CAB$ (since AC = AB). - Sum of angles in triangle ABC is 180°: $$x + x + 65^\circ = 180^\circ$$ $$2x + 65^\circ = 180^\circ$$ $$2x = 180^\circ - 65^\circ = 115^\circ$$ $$x = \frac{115^\circ}{2} = 57.5^\circ$$ 5. **Final answers:** - (a) $\angle ACB = 65^\circ$ - (b) $\angle ABC = 57.5^\circ$ - (c) $\angle CAB = 57.5^\circ$