1. The problem states: Given the semicircles with radii 35 meters and 42.5 meters, and an angle of 100° between two line segments of lengths 20 inches and 12.5 inches, solve question 23. However, the exact question 23 is not provided, so we will interpret it as finding the length of the chord or arc related to the angle and segments. 2. To solve problems involving angles and segments, we use the Law of Cosines formula: $$c^2 = a^2 + b^2 - 2ab \cos(\theta)$$ where $a$ and $b$ are the lengths of the two sides, $\theta$ is the angle between them, and $c$ is the length of the side opposite the angle. 3. Given $a = 20$, $b = 12.5$, and $\theta = 100^\circ$, we calculate $c$: $$c^2 = 20^2 + 12.5^2 - 2 \times 20 \times 12.5 \times \cos(100^\circ)$$ 4. Calculate each term: $$20^2 = 400$$ $$12.5^2 = 156.25$$ $$\cos(100^\circ) \approx -0.1736$$ 5. Substitute values: $$c^2 = 400 + 156.25 - 2 \times 20 \times 12.5 \times (-0.1736)$$ $$c^2 = 556.25 + 86.8$$ $$c^2 = 643.05$$ 6. Take the square root: $$c = \sqrt{643.05} \approx 25.35$$ 7. Therefore, the length of the side opposite the 100° angle is approximately 25.35 inches. This completes the solution for the first question found in the message.