Question: Find the length of side $c$. a = $8$ B c = [?] $73^\circ$ C b = $14$ A Law of Cosines: $$c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$$ Round your answer to the nearest hundredth. Triangle graph, position_hint=top-center: side a=$8$ on segment BC, side b=$14$ on segment CA, angle C=$73^\circ$ at vertex C, and unknown side c on segment BA.

1. **State the problem:** We need to find the length of side $c$ in a triangle where sides $a=8$, $b=14$, and angle $C=73^\circ$ are given. 2. **Formula used:** The Law of Cosines states: $$c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$$ This formula relates the lengths of the sides of a triangle to the cosine of one of its angles. 3. **Substitute the known values:** $$c^2 = 8^2 + 14^2 - 2 \times 8 \times 14 \times \cos(73^\circ)$$ 4. **Calculate each term:** $$8^2 = 64$$ $$14^2 = 196$$ Calculate $\cos(73^\circ)$ (using a calculator): $$\cos(73^\circ) \approx 0.29237$$ 5. **Plug in the cosine value:** $$c^2 = 64 + 196 - 2 \times 8 \times 14 \times 0.29237$$ Calculate the product: $$2 \times 8 \times 14 = 224$$ So: $$c^2 = 64 + 196 - 224 \times 0.29237$$ 6. **Multiply:** $$224 \times 0.29237 \approx 65.011$$ 7. **Simplify:** $$c^2 = 64 + 196 - 65.011 = 260.989$$ 8. **Find $c$ by taking the square root:** $$c = \sqrt{260.989}$$ 9. **Calculate the square root:** $$c \approx 16.15$$ 10. **Final answer:** The length of side $c$ is approximately **16.15** (rounded to the nearest hundredth).