1. **State the problem:** We have a triangle with angles 95° and 45°, and side length 6 opposite the 95° angle (side A). We need to find the third angle $a$, and the lengths of sides $a$ and $b$. 2. **Find the third angle:** The sum of angles in a triangle is 180°. $$a = 180 - 95 - 45 = 40$$ 3. **Use the Law of Sines:** $$\frac{a}{\sin(95^\circ)} = \frac{b}{\sin(45^\circ)} = \frac{6}{\sin(40^\circ)}$$ 4. **Calculate side $a$:** $$a = \frac{6 \times \sin(95^\circ)}{\sin(40^\circ)}$$ Calculate intermediate step: $$a = \frac{6 \times \sin(95^\circ)}{\sin(40^\circ)} \approx \frac{6 \times 0.9962}{0.6428}$$ Simplify with cancellation: $$a = \frac{6 \times 0.9962}{0.6428} \approx 9.30$$ 5. **Calculate side $b$:** $$b = \frac{6 \times \sin(45^\circ)}{\sin(40^\circ)}$$ Calculate intermediate step: $$b = \frac{6 \times 0.7071}{0.6428}$$ Simplify with cancellation: $$b = \frac{6 \times 0.7071}{0.6428} \approx 6.60$$ **Final answers:** - Angle $a = 40$ degrees - Side $a \approx 9.30$ - Side $b \approx 6.60$