1. **State the problem:** We have a triangle with angles 95° and 45°, and one side of length 8 opposite the 95° angle (side A). We need to find the remaining angle $a$ and the sides $a$ and $b$. 2. **Find the missing angle:** The sum of angles in a triangle is 180°. $$a = 180 - 95 - 45 = 40$$ 3. **Use the Law of Sines:** $$\frac{a}{\sin(95^\circ)} = \frac{b}{\sin(45^\circ)} = \frac{8}{\sin(40^\circ)}$$ 4. **Calculate side $a$ opposite 40°:** $$a = \frac{8 \times \sin(40^\circ)}{\sin(95^\circ)}$$ 5. **Calculate side $b$ opposite 45°:** $$b = \frac{8 \times \sin(45^\circ)}{\sin(95^\circ)}$$ 6. **Evaluate values:** $$\sin(95^\circ) \approx 0.9962, \sin(40^\circ) \approx 0.6428, \sin(45^\circ) \approx 0.7071$$ 7. **Calculate $a$:** $$a = \frac{8 \times 0.6428}{0.9962} \approx \frac{5.1424}{0.9962} \approx 5.16$$ 8. **Calculate $b$:** $$b = \frac{8 \times 0.7071}{0.9962} \approx \frac{5.6568}{0.9962} \approx 5.68$$ **Final answers:** - Missing angle $a = 40^\circ$ - Side $a \approx 5.16$ - Side $b \approx 5.68$