## Geometry Problem Summary
### Basic Shapes
- Triangle ABC
- A semicircle with diameter DE inside the triangle
### Known Conditions
- BC = 26 meters
- ∠BAC = 75°
- ∠ABC = 45°
- ∠ACB = 60° (calculated from the sum of triangle angles being 180°)
- Point D is on line segment AB
- Point E is on line segment AC
- The semicircle has DE as its diameter
- The semicircle is tangent to BC
### Objective
- Find the radius R of the semicircle (in meters)
### Options
(1) 3+√3
(2) 9-√3
(3) 9+√3
(4) 3-√3
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In this problem, we have 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. Our task is to find the radius of the semicircle.
Let's start by finding angle C. In any triangle, the sum of all angles is 180 degrees. We know that angle A is 75 degrees and angle B is 45 degrees. So angle C equals 180 minus 75 minus 45, which gives us 60 degrees. Now we can apply the Law of Sines to find the relationships between the sides and angles of the triangle. The Law of Sines states that the ratio of the side length to the sine of the opposite angle is constant for all sides of a triangle.
Now let's calculate the side lengths of the triangle. Using the Law of Sines, we can find AC and AB. For AC, we have BC over sine of 75 degrees equals AC over sine of 45 degrees. Solving for AC gives us 18 root 2 meters. Similarly, for AB, we get 18 meters. Now, to find the radius of the semicircle, we need to recognize that this semicircle is tangent to BC. This means the radius equals the inradius of triangle ADE, which is the radius of the circle that is tangent to all three sides of triangle ADE.
Let's calculate the sides of triangle ADE. Since D is at 40% of AB, AD equals 0.4 times 18, which is 7.2 meters. Similarly, AE equals 7.2 root 2 meters. Using the Law of Cosines, we can find DE equals 7.2 root 3 meters. Now we can calculate the inradius. The area of triangle ADE equals half of AD times AE times sine of 75 degrees. The semi-perimeter equals half the sum of all three sides. Dividing the area by the semi-perimeter gives us the inradius, which equals 3 plus root 3. Therefore, the radius of the semicircle is 3 plus root 3 meters, which corresponds to option 1.
Let's summarize our solution. We started with triangle ABC where angle A is 75 degrees, angle B is 45 degrees, and BC is 26 meters. We calculated angle C to be 60 degrees. Using the Law of Sines, we found that AB equals 18 meters and AC equals 18 root 2 meters. Since points D and E are located at 40% of sides AB and AC respectively, we calculated AD equals 7.2 meters and AE equals 7.2 root 2 meters. Using the Law of Cosines, we found that DE equals 7.2 root 3 meters. The radius of the semicircle equals the inradius of triangle ADE, which we calculated to be 3 plus root 3 meters. Therefore, the answer is option 1: 3 plus root 3.