1. **State the problem:** We have triangle ABC with sides AB = 12 m, AC = 10 m, and BC = 15 m. We need to find the size of angle BAC (angle at vertex A). 2. **Identify sides relative to angle BAC:** Angle BAC is at vertex A, so the side opposite this angle is BC = 15 m. The other two sides are AB = 12 m and AC = 10 m. 3. **Use the Law of Cosines:** The Law of Cosines states: $$a^2 = b^2 + c^2 - 2bc \cos(A)$$ where $a$ is the side opposite angle $A$, and $b$, $c$ are the other two sides. Here, $a = BC = 15$, $b = AB = 12$, and $c = AC = 10$. 4. **Plug in the values:** $$15^2 = 12^2 + 10^2 - 2 \times 12 \times 10 \times \cos(\angle BAC)$$ 5. **Calculate squares:** $$225 = 144 + 100 - 240 \cos(\angle BAC)$$ 6. **Simplify the right side:** $$225 = 244 - 240 \cos(\angle BAC)$$ 7. **Isolate the cosine term:** $$240 \cos(\angle BAC) = 244 - 225$$ $$240 \cos(\angle BAC) = 19$$ 8. **Solve for cosine:** $$\cos(\angle BAC) = \frac{19}{240} \approx 0.07917$$ 9. **Find the angle:** $$\angle BAC = \cos^{-1}(0.07917)$$ 10. **Calculate the angle in degrees:** $$\angle BAC \approx 85.5^\circ$$ **Final answer:** The size of angle BAC is approximately **85.5 degrees**.