1. **State the problem:** We are given a triangle ABC with sides |AB| = 12 cm, |AC| = 10 cm, and angle $\angle ABC = 37^\circ$. We need to find the size of angle $\theta = \angle BCA$ to the nearest degree. 2. **Formula used:** We will use the Law of Cosines to find side |BC| first, then use the Law of Sines or Law of Cosines again to find angle $\theta$. The Law of Cosines states: $$ c^2 = a^2 + b^2 - 2ab \cos(C) $$ where $c$ is the side opposite angle $C$. 3. **Apply Law of Cosines to find |BC|:** Let $|BC| = a$, $|AC| = b = 10$, $|AB| = c = 12$, and angle $B = 37^\circ$. $$ a^2 = b^2 + c^2 - 2bc \cos(B) = 10^2 + 12^2 - 2 \times 10 \times 12 \times \cos(37^\circ) $$ Calculate: $$ a^2 = 100 + 144 - 240 \times \cos(37^\circ) $$ Using $\cos(37^\circ) \approx 0.7986$: $$ a^2 = 244 - 240 \times 0.7986 = 244 - 191.664 = 52.336 $$ So, $$ a = \sqrt{52.336} \approx 7.23 \text{ cm} $$ 4. **Use Law of Cosines again to find angle $\theta = \angle BCA$:** Now, $\theta$ is opposite side |AB| = 12 cm. Using Law of Cosines: $$ 12^2 = 7.23^2 + 10^2 - 2 \times 7.23 \times 10 \times \cos(\theta) $$ Calculate: $$ 144 = 52.336 + 100 - 144.6 \cos(\theta) $$ Simplify: $$ 144 = 152.336 - 144.6 \cos(\theta) $$ Rearranged: $$ 144.6 \cos(\theta) = 152.336 - 144 = 8.336 $$ Intermediate step with cancellation: $$ \cancel{144.6} \cos(\theta) = 8.336 $$ Solve for $\cos(\theta)$: $$ \cos(\theta) = \frac{8.336}{144.6} \approx 0.0577 $$ 5. **Find $\theta$:** $$ \theta = \cos^{-1}(0.0577) \approx 86.7^\circ $$ 6. **Final answer:** $\theta \approx 87^\circ$ (to the nearest degree).