## Plane Geometry Problem: Detailed Construction Description (Concise English) **Problem Elements:** - Problem ID: 152 - Task: In triangle ABC, a semicircle is drawn with DE as its diameter. Given BC = 26 meters, find the radius R of the semicircle (in meters). - Options: (1) 3+√3, (2) 9-√3, (3) 9+√3, (4) 3-√3 **Geometric Construction Details:** **Basic Shapes:** - A triangle ABC. - A semicircle with diameter DE, located inside the triangle. **Point Locations:** - A: Top vertex of the triangle. - B: Bottom-left vertex of the triangle. - C: Bottom-right vertex of the triangle. - D: A point on segment AB, one endpoint of the semicircle's diameter. - E: A point on segment AC, the other endpoint of the semicircle's diameter. **Segment Descriptions:** - AB: Left side of the triangle. - BC: Base of the triangle, length = 26 meters. - AC: Right side of the triangle. - DE: Diameter of the semicircle, connecting D and E. **Angle Specifications:** - ∠BAC = 75° - ∠ABC = 45° - ∠ACB = 60° (derived from 180° - 75° - 45°) **Semicircle Characteristics:** - Diameter: DE. - Location: Inside triangle ABC. - Orientation: Its arc faces towards side BC. - Tangency: The semicircle is tangent to side BC at some point. - Radius: R (in meters). **Key Geometric Relationships:** - The center of the semicircle is the midpoint of DE. - The distance from the semicircle's center to side BC is equal to the radius R (due to tangency). - The segment DE is *not* parallel to BC but is inclined within the triangle.

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In this problem, we have a triangle ABC with a semicircle inside it. The semicircle has diameter DE, where D is on side AB and E is on side AC. We're given that angle BAC is 75 degrees, angle ABC is 45 degrees, and BC equals 26 meters. We need to find the radius of the semicircle. Let's start by analyzing the triangle. We know all three angles: 75 degrees at A, 45 degrees at B, and 60 degrees at C. To find the radius of the semicircle, we first need to find the height of the triangle from vertex A to side BC. This height will help us establish the relationship between the sides of the triangle and ultimately determine the radius. Using the tangent function, we can relate the height to the sides AB and AC. Now that we've established the relationship between the height and the sides of the triangle, let's calculate the actual lengths. We know that AB equals h, and AC equals h divided by the square root of 3. Using the law of cosines, we can write: BC squared equals AB squared plus AC squared minus 2 times AB times AC times the cosine of 75 degrees. Substituting our expressions and the known value of BC equals 26 meters, we get an equation in terms of h. To simplify, we need to calculate cosine of 75 degrees, which equals cosine of 45 plus 30 degrees, giving us the square root of 6 minus 1, all divided by 4. Now we can solve for the height of the triangle. After simplifying the equation, we find that h equals 9 times the square root of 3. This means AB equals 9 times the square root of 3, and AC equals 9. Since D is located at 0.6 of the distance from A to B, and E is at 0.6 of the distance from A to C, we can calculate the length of DE. The radius of the semicircle is half the length of DE, which gives us R equals 3 plus the square root of 3. This is our final answer. Let's summarize what we've learned. In this problem, we had a triangle ABC with angles of 75, 45, and 60 degrees, and a side BC of 26 meters. We needed to find the radius of a semicircle with diameter DE inside the triangle. We first calculated the height of the triangle using the law of cosines, finding it to be 9 times the square root of 3. This allowed us to determine the sides AB and AC. Finally, we calculated the radius of the semicircle to be 3 plus the square root of 3, which corresponds to option 1. This problem demonstrated how to apply triangle geometry, trigonometry, and properties of semicircles to solve a complex geometric problem.