1. **State the problem:** We are given a triangle with sides $a=4$, $b=6$, and angle $\alpha = 17^\circ$ opposite side $a$. We need to find the two possible sets of solutions for side $c$ and angles $\beta$, $\gamma$ where $\beta$ is opposite side $b$ and $\gamma$ opposite side $c$. Round angles to the nearest degree and sides to one decimal place. 2. **Given:** - $a=4$ - $b=6$ - $\alpha=17^\circ$ 3. **Use Law of Sines:** $$\frac{a}{\sin \alpha} = \frac{b}{\sin \beta}$$ Solve for $\sin \beta$: $$\sin \beta = \frac{b}{a} \sin \alpha = \frac{6}{4} \times \sin 17^\circ$$ Use $\sin 17^\circ \approx 0.29237170$ (rounded to 8 decimals): $$\sin \beta = 1.5 \times 0.29237170 = 0.43855755$$ 4. **Find angles $\beta_1$ and $\beta_2$:** - $\beta_1 = \sin^{-1}(0.43855755) \approx 26.1^\circ \to 26^\circ$ - $\beta_2 = 180^\circ - 26.1^\circ = 153.9^\circ \to 154^\circ$ 5. **Find corresponding angles $\gamma_1$ and $\gamma_2$:** Since $\alpha + \beta + \gamma = 180^\circ$: - $\gamma_1 = 180^\circ - 17^\circ - 26^\circ = 137^\circ$ - $\gamma_2 = 180^\circ - 17^\circ - 154^\circ = 9^\circ$ 6. **Find sides $c_1$ and $c_2$ using Law of Sines:** $$c = \frac{a \sin \gamma}{\sin \alpha}$$ Use $\sin 137^\circ = \sin 43^\circ \approx 0.6820$ (more accurate: $\sin 137^\circ = \sin (180-137) = \sin 43 = 0.6820$), and $\sin 9^\circ \approx 0.15643447$. Calculate: - $c_1 = \frac{4 \times 0.6820}{0.29237170} = \frac{2.728}{0.29237170} \approx 9.3$ - $c_2 = \frac{4 \times 0.15643447}{0.29237170} = \frac{0.6257}{0.29237170} \approx 2.1$ 7. **Final two solutions sorted ascending by $c$:** - Solution 1: $c_2=2.1$, $\beta_2=154^\circ$, $\gamma_2=9^\circ$ - Solution 2: $c_1=9.3$, $\beta_1=26^\circ$, $\gamma_1=137^\circ$ **Answer:** $$c_1 = 9.3, \quad \beta_1 = 26^\circ, \quad \gamma_1 = 137^\circ$$ $$c_2 = 2.1, \quad \beta_2 = 154^\circ, \quad \gamma_2 = 9^\circ$$