Solve this question---152. In ∆ABC, A semi-circle with DE as diameter is
drawn such that BC = 26m, the radius R
(in meter) =
A
75°
D
E
R
45°
B
C
(1) 3+√3
(2) 9-√3
(3) 9+√3
(4) 3-√3
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We have triangle ABC where angle B is 45 degrees, angle C is 75 degrees, and side BC is 26 meters. A semicircle with diameter DE is inscribed such that it lies on BC and is tangent to both sides AB and AC. We need to find the radius R of this semicircle.
First, we need to find angle A. Since the sum of angles in any triangle equals 180 degrees, we have angle A plus angle B plus angle C equals 180 degrees. Substituting the known values: angle A plus 45 degrees plus 75 degrees equals 180 degrees. Therefore, angle A equals 180 minus 45 minus 75, which gives us angle A equals 60 degrees.
Now we set up the radius equation. Let O be the center of the semicircle on BC. If BO equals x, then OC equals 26 minus x. Since the semicircle is tangent to side AB, the radius R equals BO times sine of 45 degrees, which is x divided by square root of 2. Similarly, since it's tangent to AC, the radius also equals OC times sine of 75 degrees.
Now we solve the equation. We equate the two expressions for R: x over square root of 2 equals 26 minus x times sine of 75 degrees. First, we calculate sine of 75 degrees using the angle addition formula, which gives us square root of 3 plus 1, all over 2 square root of 2. Substituting this and multiplying both sides by 2 square root of 2, we get 2x equals 26 minus x times square root of 3 plus 1. Expanding and solving for x, we find x equals 13 square root of 3. Therefore, the radius R equals x times sine of 45 degrees, which simplifies to 9 plus square root of 3.
To summarize our solution: We calculated angle A as 60 degrees using the angle sum property. We set up radius equations based on the tangent conditions of the inscribed semicircle. Solving the resulting equation gave us x equals 13 square root of 3. Finally, we calculated the radius R as 9 plus square root of 3 meters, which corresponds to option 3.