Solve the triangle. \( a=8 \quad b=9, G=40^{\circ} \) CN \( \square \) Donst round until the final answer. Then round to the nearest tenth as needed.)

To start, we can utilize the Law of Cosines to find \( c \). According to the Law of Cosines: \[ c^2 = a^2 + b^2 - 2ab \cos(G) \] Plugging in our values: \[ c^2 = 8^2 + 9^2 - 2 \cdot 8 \cdot 9 \cdot \cos(40^{\circ}) \] Calculating each component: \[ c^2 = 64 + 81 - 144 \cdot \cos(40^{\circ}) \] Now, calculate \( \cos(40^{\circ}) \) which is approximately 0.7660. So, \[ c^2 = 145 - 144 \cdot 0.7660 \] \[ c^2 = 145 - 110.784 = 34.216 \] Taking the square root: \[ c \approx \sqrt{34.216} \approx 5.85 \] Next, we can use the Law of Sines to find angles \( A \) and \( B \): \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] First, we find \( \sin A \): \[ \frac{8}{\sin A} = \frac{5.85}{\sin(40^{\circ})} \] With \( \sin(40^{\circ}) \approx 0.6428 \): \[ \frac{8}{\sin A} = \frac{5.85}{0.6428} \] Cross-multiplying yields: \[ 8 \cdot 0.6428 = 5.85 \cdot \sin A \] \[ \sin A \approx \frac{5.1424}{5.85} \approx 0.8787 \] Finding angle \( A \): \[ A \approx \arcsin(0.8787) \approx 61.1^{\circ} \] Now we find angle \( B \): \[ B = 180^{\circ} - A - G \] \[ B \approx 180^{\circ} - 61.1^{\circ} - 40^{\circ} \approx 78.9^{\circ} \] In summary, the results are: \[ A \approx 61.1^{\circ}, \quad B \approx 78.9^{\circ}, \quad c \approx 5.9 \] Final rounded values: \[ A \approx 61.1^{\circ}, \quad B \approx 78.9^{\circ}, \quad c \approx 5.9. \]