1. **Problem statement:** Given a triangle with side lengths $a=45$ mm, $b=60$ mm, and $c=102$ mm, Claudia calculated the angles $\alpha$, $\beta$, and $\gamma$ using the law of cosines and sines but obtained $\alpha \approx 11.85^\circ$, $\gamma \approx 27.75^\circ$, and $\beta \approx 140.40^\circ$. We need to show why this solution is incorrect, explain Claudia's mistake, and solve the problem correctly. 2. **Claudia's formula for $\alpha$:** $$\alpha = \arccos\left(\frac{a^2 - c^2 - b^2}{-2bc}\right)$$ This is a rearrangement of the law of cosines: $$a^2 = b^2 + c^2 - 2bc \cos(\alpha)$$ 3. **Check Claudia's substitution:** Calculate the numerator: $$a^2 - c^2 - b^2 = 45^2 - 102^2 - 60^2 = 2025 - 10404 - 3600 = 2025 - 14004 = -11979$$ Calculate the denominator: $$-2bc = -2 \times 60 \times 102 = -12240$$ So, $$\frac{a^2 - c^2 - b^2}{-2bc} = \frac{-11979}{-12240} = 0.9789$$ 4. **Calculate $\alpha$ correctly:** $$\alpha = \arccos(0.9789) \approx 11.85^\circ$$ This matches Claudia's value. 5. **Claudia's mistake:** The law of cosines formula is: $$a^2 = b^2 + c^2 - 2bc \cos(\alpha)$$ Rearranged correctly for $\cos(\alpha)$: $$\cos(\alpha) = \frac{b^2 + c^2 - a^2}{2bc}$$ But Claudia used: $$\cos(\alpha) = \frac{a^2 - c^2 - b^2}{-2bc}$$ which is algebraically equivalent but can cause confusion. 6. **Check if the triangle inequality holds:** For any triangle, the sum of any two sides must be greater than the third: $$a + b > c \Rightarrow 45 + 60 = 105 > 102 \quad \checkmark$$ $$a + c > b \Rightarrow 45 + 102 = 147 > 60 \quad \checkmark$$ $$b + c > a \Rightarrow 60 + 102 = 162 > 45 \quad \checkmark$$ So the sides can form a triangle. 7. **Calculate $\gamma$ using the law of sines:** Claudia used: $$\frac{\sin(\gamma)}{c} = \frac{\sin(\alpha)}{a} \Rightarrow \sin(\gamma) = \frac{c}{a} \sin(\alpha)$$ Calculate: $$\sin(\alpha) = \sin(11.85^\circ) \approx 0.205$$ $$\sin(\gamma) = \frac{102}{45} \times 0.205 = 2.333 \times 0.205 = 0.478$$ But $\sin(\gamma)$ cannot be greater than 1, so this is valid. Calculate $\gamma$: $$\gamma = \arcsin(0.478) \approx 28.6^\circ$$ Claudia's value was $27.75^\circ$, close but slightly off. 8. **Calculate $\beta$:** $$\beta = 180^\circ - \alpha - \gamma = 180^\circ - 11.85^\circ - 28.6^\circ = 139.55^\circ$$ 9. **Why Claudia's solution is wrong:** The angle $\beta$ is very large ($> 90^\circ$), but the side opposite $\beta$ is $b=60$ mm, which is smaller than side $c=102$ mm opposite $\gamma$. In a triangle, larger angles are opposite longer sides. Here, $\beta$ is largest but opposite a smaller side than $c$. This contradicts the triangle inequality and angle-side relationship. 10. **Correct approach: Use law of cosines to find $\alpha$ properly:** $$\cos(\alpha) = \frac{b^2 + c^2 - a^2}{2bc} = \frac{60^2 + 102^2 - 45^2}{2 \times 60 \times 102} = \frac{3600 + 10404 - 2025}{12240} = \frac{11979}{12240} = 0.9789$$ $$\alpha = \arccos(0.9789) = 11.85^\circ$$ (This matches Claudia's $\alpha$.) 11. **Calculate $\beta$ using law of cosines:** $$\cos(\beta) = \frac{a^2 + c^2 - b^2}{2ac} = \frac{45^2 + 102^2 - 60^2}{2 \times 45 \times 102} = \frac{2025 + 10404 - 3600}{9180} = \frac{8809}{9180} = 0.9599$$ $$\beta = \arccos(0.9599) = 16.0^\circ$$ 12. **Calculate $\gamma$:** $$\gamma = 180^\circ - \alpha - \beta = 180^\circ - 11.85^\circ - 16.0^\circ = 152.15^\circ$$ 13. **Check angle-side relationship:** Largest angle $\gamma = 152.15^\circ$ opposite longest side $c=102$ mm, which is consistent. 14. **Summary:** - Claudia incorrectly applied the law of sines after calculating $\alpha$, leading to inconsistent angles. - Correct method is to use law of cosines for each angle. **Final correct angles:** $$\alpha \approx 11.85^\circ, \quad \beta \approx 16.0^\circ, \quad \gamma \approx 152.15^\circ$$