Solve the triangle \[ \mathrm{a}=8, \mathrm{~b}=11, \mathrm{c}=17 \] \( \mathrm{~A} \approx \square^{\circ} \) (Round to one decimal place as needed) \( \mathrm{B} \approx \mathrm{D}^{\circ} \) (Round to one decimal place as needed) \( \mathrm{C} \approx \square^{\circ} \) (Round to one decimal place as needed.)

To determine the angles \( A \), \( B \), and \( C \) of the triangle with sides \( a = 8 \), \( b = 11 \), and \( c = 17 \), we can use the Law of Cosines. First, we calculate angle \( A \): \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{11^2 + 17^2 - 8^2}{2 \cdot 11 \cdot 17} \] Calculating the squares: \[ \cos A = \frac{121 + 289 - 64}{2 \cdot 11 \cdot 17} = \frac{346}{374} \approx 0.924 \] Now, finding the angle: \[ A \approx \cos^{-1}(0.924) \approx 22.6^{\circ} \] Next, we calculate angle \( B \): \[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{8^2 + 17^2 - 11^2}{2 \cdot 8 \cdot 17} \] Calculating the squares: \[ \cos B = \frac{64 + 289 - 121}{2 \cdot 8 \cdot 17} = \frac{232}{272} \approx 0.853 \] Now, finding the angle: \[ B \approx \cos^{-1}(0.853) \approx 31.9^{\circ} \] Finally, we can find angle \( C \): Since the angles in a triangle sum up to \( 180^{\circ} \): \[ C = 180^{\circ} - A - B = 180^{\circ} - 22.6^{\circ} - 31.9^{\circ} \approx 125.5^{\circ} \] Thus, the measured angles of the triangle are: \[ A \approx 22.6^{\circ}, \quad B \approx 31.9^{\circ}, \quad C \approx 125.5^{\circ} \]