1. **State the problem:** We are given a triangle ABC inscribed in a circle with center O. Given angles: - Angle at B ($\angle ABC$) = 165° - Angle at C ($\angle BCA$) = 25° We need to find the angles $\angle ABC$ and $\angle BAC$. 2. **Recall the triangle angle sum rule:** The sum of interior angles in any triangle is 180°. $$\angle ABC + \angle BCA + \angle BAC = 180^\circ$$ 3. **Substitute the known values:** $$165^\circ + 25^\circ + \angle BAC = 180^\circ$$ 4. **Calculate $\angle BAC$:** $$\angle BAC = 180^\circ - 165^\circ - 25^\circ = 180^\circ - 190^\circ = -10^\circ$$ 5. **Interpretation:** A negative angle is impossible in this context, so the given angles cannot all be interior angles of triangle ABC. 6. **Re-examine the problem:** The problem states angle at B is 165°, which is likely an exterior angle or an angle outside the triangle. 7. **Use the exterior angle theorem:** The exterior angle at B equals the sum of the two opposite interior angles. If $\angle ABC$ is the interior angle at B, and the given 165° is the exterior angle at B, then: $$165^\circ = \angle BAC + \angle BCA$$ Given $\angle BCA = 25^\circ$, then: $$\angle BAC = 165^\circ - 25^\circ = 140^\circ$$ 8. **Find $\angle ABC$ using triangle sum:** $$\angle ABC = 180^\circ - \angle BAC - \angle BCA = 180^\circ - 140^\circ - 25^\circ = 15^\circ$$ **Final answers:** - $\angle ABC = 15^\circ$ - $\angle BAC = 140^\circ$