Question: Solve for $c$. $b = 15$ $20^\circ$ $125^\circ$ $C = [?]$ Law of Sines: $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$ Round your answer to the nearest hundredth. Graph/shape: triangle CBA with C at bottom-left, B at bottom-center, and A at upper-right; side CA is labeled $b = 15$, side BA is labeled $c$, angle at C is $20^\circ$, and angle at B is $125^\circ$; position_hint = center

1. **State the problem:** We need to find the length $c$ in triangle CBA where $b = 15$, angle $C = 20^\circ$, and angle $B = 125^\circ$. 2. **Find angle $A$:** The sum of angles in a triangle is $180^\circ$. $$A = 180^\circ - B - C = 180^\circ - 125^\circ - 20^\circ = 35^\circ$$ 3. **Recall the Law of Sines:** $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$ Here, side $b$ is opposite angle $B$, side $c$ is opposite angle $C$, and side $a$ is opposite angle $A$. 4. **Set up the ratio to find $c$:** $$\frac{\sin B}{b} = \frac{\sin C}{c} \implies c = b \times \frac{\sin C}{\sin B}$$ 5. **Substitute known values:** $$c = 15 \times \frac{\sin 20^\circ}{\sin 125^\circ}$$ 6. **Calculate sine values:** $$\sin 20^\circ \approx 0.3420$$ $$\sin 125^\circ \approx 0.8192$$ 7. **Calculate $c$:** $$c = 15 \times \frac{0.3420}{0.8192} = 15 \times 0.4175 = 6.2625$$ 8. **Round to nearest hundredth:** $$c \approx 6.26$$ **Final answer:** $c = 6.26$