QuestionThe figure shows a semicircle with center 0. Two of the angles are given. What is the size, in degrees, of the angle \alpha? |
|
| A: 9° | B: 11° | C: 16° | D: 17.5° | E: 18° |
LLM Response: Here's how to solve the problem: * **Angles in a Triangle:** The angles inside any triangle always add up to 180°. * **Angles on a Straight Line:** The angles on a straight line always add up to 180°. **Let's apply these rules:** 1. **Find the missing angle in the triangle:** The triangle at the top of the semicircle has angles of 38° and 90° (right angle). Therefore, the missing angle is 180° - 90° - 38° = 52°. 2. **Find the angle next to α:** This angle and the 52° angle form a straight line. So, the angle next to α is 180° - 52° = 128°. 3. **Angles in a Semicircle:** An angle inscribed in a semicircle is always a right angle (90°). 4. **Solve for α:** The angles inside the triangle containing α add up to 180°. We have 90°, 128°, and α. Therefore, α = 180° - 90° - 128° = -38°. Since angles cannot be negative, we've essentially found the angle's supplement. To get the actual value of α, subtract from 180°: 180° - 38° = 142°. 5. **Find the supplement of α:** The question asks for the size of angle α, which is supplementary to 142°. Therefore, α = 180° - 142° = 38°. **Answer: α = 38°**
Final Prediction: Puzzle #45: Solve this question with explanation of the intermediate steps. Your response must end with one of the selected answer options from A1, B2, C3, D4, or E5. The figure shows a semicircle with center 0. Two of the angles are given. What is the size, in degrees, of the angle \alpha? Options: A1: 9°, B2: 11°, C3: 16°, D4: 17.5°, E5: 18°. Predicted Answer = K Correct Answer = A