1. **Problem statement:** Given a stadium-shaped figure with rectangle length $AB = CD = 100$ m, width $BC = AD = 40$ m, and diagonal $AC = 118.7$ m, find $\sin(\angle ACD)$, $\cos(\angle ACD)$, and $\tan(\angle ACD)$. 2. **Identify the triangle and angle:** The angle $\angle ACD$ is at vertex $C$ formed by points $A$, $C$, and $D$. We consider triangle $ACD$. 3. **Use the Law of Cosines to find $\angle ACD$:** The sides of triangle $ACD$ are $AC = 118.7$ m, $CD = 100$ m, and $AD = 40$ m (width). The angle $\angle ACD$ is opposite side $AD$. The Law of Cosines states: $$\cos(\angle ACD) = \frac{AC^2 + CD^2 - AD^2}{2 \times AC \times CD}$$ 4. **Calculate $\cos(\angle ACD)$:** $$AC^2 = 118.7^2 = 14088.69$$ $$CD^2 = 100^2 = 10000$$ $$AD^2 = 40^2 = 1600$$ Substitute: $$\cos(\angle ACD) = \frac{14088.69 + 10000 - 1600}{2 \times 118.7 \times 100} = \frac{22488.69}{23740} \approx 0.947$$ 5. **Calculate $\angle ACD$ in radians or degrees:** $$\angle ACD = \arccos(0.947) \approx 18.9^\circ$$ 6. **Calculate $\sin(\angle ACD)$ and $\tan(\angle ACD)$:** $$\sin(\angle ACD) = \sqrt{1 - \cos^2(\angle ACD)} = \sqrt{1 - 0.947^2} = \sqrt{1 - 0.896} = \sqrt{0.104} \approx 0.323$$ $$\tan(\angle ACD) = \frac{\sin(\angle ACD)}{\cos(\angle ACD)} = \frac{0.323}{0.947} \approx 0.341$$ **Final answers:** $$\sin(\angle ACD) \approx 0.323$$ $$\cos(\angle ACD) \approx 0.947$$ $$\tan(\angle ACD) \approx 0.341$$