1. **Problem statement:** We are given a triangle ABC with sides AB=10 cm, BC=8 cm, and CA=6 cm. We need to verify if it is a right-angled triangle and solve for its angles. 2. **Check if the triangle is right-angled:** According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. 3. Identify the longest side as the potential hypotenuse: AB = 10 cm. 4. Calculate squares: $$AB^2 = 10^2 = 100$$ $$BC^2 = 8^2 = 64$$ $$CA^2 = 6^2 = 36$$ 5. Check Pythagorean theorem: $$BC^2 + CA^2 = 64 + 36 = 100$$ Since $$AB^2 = BC^2 + CA^2$$, triangle ABC is right-angled at point C. 6. **Find the angles:** Use trigonometric ratios. 7. Angle at A ($\angle A$): $$\cos \angle A = \frac{CA}{AB} = \frac{6}{10} = 0.6$$ $$\angle A = \cos^{-1}(0.6) \approx 53.13^\circ$$ 8. Angle at B ($\angle B$): $$\cos \angle B = \frac{BC}{AB} = \frac{8}{10} = 0.8$$ $$\angle B = \cos^{-1}(0.8) \approx 36.87^\circ$$ 9. Angle at C ($\angle C$) is the right angle: $$\angle C = 90^\circ$$ **Final answer:** Triangle ABC is right-angled at C with angles approximately $53.13^\circ$ at A, $36.87^\circ$ at B, and $90^\circ$ at C.