Solve the triangle. \( \mathrm{a}=24.8 \quad \mathrm{~b}=15.2 \quad \mathrm{c}=36.9 \) What is the degree measure of angle A? \( \square^{\circ} \) (Simplify your answer. Type an integer or a decimal. Round to the nearest tenth if needed.) What is the degree measure of angle B ? \( \square^{\circ} \) (Simplify your answer. Type an integer or a decimal. Round to the nearest tenth if needed.) What is the degree measure of angle C ? \( \square^{\circ} \) (Simplify your answer. Type an integer or a decimal. Round to the nearest tenth if needed.)
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To solve the triangle with sides \( a = 24.8 \), \( b = 15.2 \), and \( c = 36.9 \), we can use the Law of Cosines to find the angles. 1. **Finding angle A**: \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{15.2^2 + 36.9^2 - 24.8^2}{2 \cdot 15.2 \cdot 36.9} \] \[ \cos A = \frac{231.04 + 1368.61 - 615.04}{2 \cdot 15.2 \cdot 36.9} = \frac{984.61}{1122.96} \approx 0.8765 \] Thus, \[ A \approx \cos^{-1}(0.8765) \approx 28.9^{\circ} \] 2. **Finding angle B**: \[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{24.8^2 + 36.9^2 - 15.2^2}{2 \cdot 24.8 \cdot 36.9} \] \[ \cos B = \frac{615.04 + 1368.61 - 231.04}{2 \cdot 24.8 \cdot 36.9} = \frac{1752.61}{1835.76} \approx 0.9546 \] Thus, \[ B \approx \cos^{-1}(0.9546) \approx 17.1^{\circ} \] 3. **Finding angle C**: Since the sum of angles in a triangle is \( 180^{\circ} \): \[ C = 180^{\circ} - A - B \approx 180^{\circ} - 28.9^{\circ} - 17.1^{\circ} \approx 134^{\circ} \] Final answers: \( A \approx 28.9^{\circ} \) \( B \approx 17.1^{\circ} \) \( C \approx 134.0^{\circ} \)