1. **State the problem:** We are given triangle $\triangle JKL$ with sides $j=12$ cm, $k=9$ cm, and $l=9.75$ cm. We need to find the measure of angle $\angle J$ using the Law of Cosines. 2. **Recall the Law of Cosines formula:** $$c^2 = a^2 + b^2 - 2ab \cos(C)$$ where $c$ is the side opposite angle $C$. 3. **Assign sides and angle:** Here, side $j=12$ is opposite angle $J$, so: $$12^2 = 9^2 + 9.75^2 - 2 \times 9 \times 9.75 \times \cos(\angle J)$$ 4. **Calculate squares:** $$144 = 81 + 95.0625 - 175.5 \cos(\angle J)$$ 5. **Combine constants:** $$144 = 176.0625 - 175.5 \cos(\angle J)$$ 6. **Isolate cosine term:** $$144 - 176.0625 = -175.5 \cos(\angle J)$$ $$-32.0625 = -175.5 \cos(\angle J)$$ 7. **Divide both sides by $-175.5$:** $$\cancel{-32.0625} \div \cancel{-175.5} = \cancel{-175.5} \cos(\angle J) \div \cancel{-175.5}$$ $$\cos(\angle J) = \frac{32.0625}{175.5} \approx 0.1827$$ 8. **Find angle $J$ by taking inverse cosine:** $$\angle J = \cos^{-1}(0.1827) \approx 79.5^\circ$$ **Final answer:** $$\boxed{\angle J \approx 79.5^\circ}$$ **Note:** The original problem incorrectly used sine instead of cosine in the Law of Cosines formula, which caused an impossible value for sine. The Law of Cosines uses cosine, not sine, for the included angle.