152. In ∆ABC, A semi-circle with DE as diameter is
drawn such that BC = 26m, the radius R
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In this problem, we have triangle ABC with angles A equals 75 degrees and B equals 45 degrees. A semi-circle with diameter DE is drawn inside the triangle, where D is on side AB and E is on side AC. We're told that BC equals 26 meters, and we need to find the radius of the semi-circle.
Let's start by finding angle C. Since the sum of angles in a triangle is 180 degrees, angle C equals 180 minus 75 minus 45, which is 60 degrees. Next, we need to calculate the height h from vertex A to side BC. Using trigonometry in this triangle with angles 75, 45, and 60 degrees, we can determine that the height h equals 13 times the quantity 3 minus square root of 3. This height will be crucial for finding the radius of our semi-circle.
Now, let's establish the relationship between the radius R and the height h. Since DE is parallel to BC, triangles ADE and ABC are similar. This means the ratio of corresponding parts is equal. The distance from A to the center of the semi-circle is h minus R, where h is the height from A to BC. Using the property of similar triangles, we can write that h minus R divided by h equals DE divided by BC. Since DE equals 2R and BC equals 26, we get h minus R divided by h equals 2R divided by 26, which simplifies to R divided by 13. Solving this equation, we find that R equals 13h divided by h plus 13.
In the final step, we substitute the value of h, which is 13 times the quantity 3 minus square root of 3, into our formula for R. After substituting, we get R equals 13 squared times 3 minus square root of 3, divided by 13 times 4 minus square root of 3. To simplify this expression, we multiply both numerator and denominator by 4 plus square root of 3, which is the conjugate of the denominator. After performing the algebraic operations and simplifying, we get R equals 9 minus square root of 3. This is our final answer for the radius of the semi-circle.
To summarize what we've learned: We solved a geometric problem involving a triangle with angles 75, 45, and 60 degrees, containing a semi-circle with diameter DE. We first calculated the height from vertex A to side BC, which equals 13 times the quantity 3 minus square root of 3. Then, using the property of similar triangles, we established that the radius R equals 13h divided by h plus 13. After substituting the value of h and simplifying, we found that the radius equals 9 minus square root of 3, which is approximately 7.27 meters. This problem demonstrates how geometric relationships and algebraic techniques can be combined to solve complex mathematical challenges.