1. **State the problem:** Given a triangle with side $b=8.5$ inches, side $c=18.4$ inches, and angle $A=53.213^\circ$, solve for the remaining side $a$ and angles $B$ and $C$. 2. **Relevant formula and rules:** Use the Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ Since we know $b$, $c$, and $A$, we can find angle $B$ using: $$\sin B = \frac{b \sin A}{c}$$ Important: If $\sin B > 1$, no triangle exists. If $\sin B < 1$, there may be one or two solutions depending on the angle. 3. **Calculate $\sin B$:** $$\sin B = \frac{8.5 \times \sin 53.213^\circ}{18.4}$$ Calculate $\sin 53.213^\circ$: $$\sin 53.213^\circ \approx 0.799$$ So, $$\sin B = \frac{8.5 \times 0.799}{18.4} = \frac{6.7915}{18.4} \approx 0.369$$ 4. **Find angle $B$:** $$B = \sin^{-1}(0.369) \approx 21.645^\circ$$ 5. **Check for second solution:** Since $\sin B = \sin(180^\circ - B)$, the second possible angle is: $$B' = 180^\circ - 21.645^\circ = 158.355^\circ$$ 6. **Check if second solution is valid:** Sum of angles must be $180^\circ$: For $B'$, sum with $A$ is: $$53.213^\circ + 158.355^\circ = 211.568^\circ > 180^\circ$$ Not possible, so only one solution exists. 7. **Find angle $C$:** $$C = 180^\circ - A - B = 180^\circ - 53.213^\circ - 21.645^\circ = 105.142^\circ$$ 8. **Find side $a$ using Law of Sines:** $$a = c \times \frac{\sin A}{\sin C} = 18.4 \times \frac{0.799}{\sin 105.142^\circ}$$ Calculate $\sin 105.142^\circ$: $$\sin 105.142^\circ \approx 0.966$$ So, $$a = 18.4 \times \frac{0.799}{0.966} \approx 18.4 \times 0.827 = 15.22$$ 9. **Round answers:** $$B \approx 21.645^\circ \to 21.645^\circ$$ $$C \approx 105.142^\circ \to 105.142^\circ$$ $$a \approx 15.2$$ inches **Final conclusion:** There is only one possible solution for the triangle with $$B \approx 21.645^\circ, C \approx 105.142^\circ, a \approx 15.2$$ inches.