answer:To prove the given formulas for the area of a triangle, we will use the Law of Sines theorem. 1. **Step 1**: Recall the Law of Sines theorem, which is stated as follows: frac{a}{sin alpha} = frac{b}{sin beta} = frac{c}{sin gamma} = 2R Here, ( a, b, c ) are the sides of the triangle opposite to the angles ( alpha, beta, gamma ) respectively, and ( R ) is the radius of the circumcircle. 2. **Step 2**: Let's isolate ( b ). According to the Law of Sines: frac{b}{sin beta} = frac{a}{sin alpha} Solving for ( b ): b = frac{a sin beta}{sin alpha} 3. **Step 3**: The area ( S ) of the triangle can be calculated using the formula: S = frac{1}{2} ab sin gamma 4. **Step 4**: Substitute the expression for ( b ) from Step 2 into the area formula: S = frac{1}{2} a left(frac{a sin beta}{sin alpha}right) sin gamma Simplifying the expression: S = frac{a^2 sin beta sin gamma}{2 sin alpha} 5. **Conclusion**: We have thus proven the first formula: S = frac{a^2 sin beta sin gamma}{2 sin alpha} 6. **Step 5**: To prove the second formula, we start by substituting the expressions for ( a ) and ( b ) using the relation from the Law of Sines. We have: a = 2R sin alpha quad text{and} quad b = 2R sin beta 7. **Step 6**: Substitute these expressions into the area formula ( S = frac{1}{2} ab sin gamma ): S = frac{1}{2} cdot (2R sin alpha) cdot (2R sin beta) cdot sin gamma Simplify the expression: S = frac{1}{2} cdot 4R^2 sin alpha sin beta sin gamma S = 2R^2 sin alpha sin beta sin gamma 8. **Conclusion**: We have thus proven the second formula: S = 2R^2 sin alpha sin beta sin gamma Hence, both formulas for the area of a triangle have been derived and proved: boxed{S = frac{a^2 sin beta sin gamma}{2 sin alpha} quad text{and} quad S = 2R^2 sin alpha sin beta sin gamma}

answer:1. **Setting up the Coordinate System:** We will use a coordinate system with the origin at the center of the square (ABCD) and the (x)-axis parallel to the line (l). Let the vertices of the square be (A(x, y)), (B(y, -x)), (C(-x, -y)), and (D(-y, x)). 2. **Coordinates of Points (Q) and (P):** - Point (Q) is the midpoint of side (AB): [ Q = left( frac{x + y}{2}, frac{y - x}{2} right) ] - Point (P) is the foot of the perpendicular dropped from (D) onto the fixed line (l), defined by (y = a). Therefore, its coordinates are: [ P = (-y, a) ] 3. **Midpoint of Segment (PQ):** The midpoint of segment (PQ) can be found by taking the average of the coordinates of (P) and (Q): [ M = left( frac{-y + frac{x + y}{2}}{2}, frac{a + frac{y - x}{2}}{2} right) ] Simplifying the coordinates, we get: [ M = left( frac{-2y + x + y}{4}, frac{2a + y - x}{4} right) = left( frac{x - y}{4}, frac{2a + y - x}{4} right) ] 4. **General Equation for the Geometric Locus:** Denoting ( t = frac{x - y}{4} ), the coordinates of ( M ) become: [ M = left( t, frac{2a + y - x}{4} right) = left( t, -t + frac{a}{2} right) ] Hence, the geometrical locus (GML) of the midpoints of segments (PQ) is described by: [ (t, -t + frac{a}{2}) ] 5. **Constraints on ( t ):** It's important to note that due to the rotation and the constraints of the coordinates: [ x, y in [- sqrt{2(a^2 + b^2)}, sqrt{2(a^2 + b^2)}] ] But for the practical applications within the limits of the problem, ( t ) can vary accordingly within possible transformations of the square. # Conclusion [ boxed{(t, -t + frac{a}{2})} ]

answer:1. Let ( S_n ) be the sum of the weights of cheese bought after the ( n )-th customer. We need to determine if the salesgirl can declare, after each of the first 10 customers, that there is just enough cheese for the next 10 customers if each customer buys a portion of cheese equal to the average weight of the previous purchases. 2. The total weight of cheese sold after ( n ) customers is ( S_n ). The average weight of the portions of cheese sold to the first ( n ) customers is ( frac{S_n}{n} ). 3. The salesgirl declares that there is enough cheese for the next 10 customers if each of them buys a portion of cheese equal to the average weight of the previous purchases. This means that the remaining cheese, ( 20 - S_n ), should be equal to ( 10 times frac{S_n}{n} ). 4. Setting up the equation: [ 20 - S_n = 10 times frac{S_n}{n} ] Simplifying, we get: [ 20 - S_n = frac{10S_n}{n} ] Multiplying both sides by ( n ) to clear the fraction: [ 20n - nS_n = 10S_n ] Combining like terms: [ 20n = 10S_n + nS_n ] [ 20n = S_n (n + 10) ] Solving for ( S_n ): [ S_n = frac{20n}{n + 10} ] 5. We need to check if the amount of cheese bought by any customer is positive. To do this, we calculate the difference between the cheese bought by the ( n )-th customer and the ( (n-1) )-th customer: [ S_n - S_{n-1} = frac{20n}{n + 10} - frac{20(n-1)}{(n-1) + 10} ] Simplifying the expression: [ S_n - S_{n-1} = 20 left( frac{n}{n + 10} - frac{n-1}{n + 9} right) ] Finding a common denominator: [ S_n - S_{n-1} = 20 left( frac{n(n + 9) - (n-1)(n + 10)}{(n + 10)(n + 9)} right) ] Simplifying the numerator: [ n(n + 9) - (n-1)(n + 10) = n^2 + 9n - (n^2 + 10n - n - 10) = n^2 + 9n - n^2 - 9n + 10 = 10 ] Thus: [ S_n - S_{n-1} = 20 left( frac{10}{(n + 10)(n + 9)} right) ] Since the denominator is always positive for ( n geq 1 ), ( S_n - S_{n-1} ) is positive, meaning the amount of cheese bought by any customer is positive. 6. To find the amount of cheese left in the store after the first 10 customers have made their purchases, we plug ( n = 10 ) into the expression for ( S_n ): [ S_{10} = frac{20 times 10}{10 + 10} = frac{200}{20} = 10 text{ kg} ] Therefore, the amount of cheese left in the store is: [ 20 - S_{10} = 20 - 10 = 10 text{ kg} ] The final answer is ( boxed{10} ) kg.

answer:Since frac{sin C}{sin B} = frac{c}{b}, we have sin C = frac{sqrt{3}}{2}. As c > b, we have C = 60^circ or C = 120^circ. (I) If C = 60^circ, then A = 180^circ - 30^circ - 60^circ = 90^circ. Thus, S_{triangle ABC} = frac{1}{2}bc = frac{1}{2} times 1 times sqrt{3} = boxed{frac{sqrt{3}}{2}}. (II) If C = 120^circ, then A = 180^circ - 30^circ - 120^circ = 30^circ. Thus, S_{triangle ABC} = frac{1}{2}bc sin A = frac{1}{2} times 1 times sqrt{3} times frac{1}{2} = boxed{frac{sqrt{3}}{4}}. Therefore, the possible solutions are C = 60^circ, S_{triangle ABC} = frac{sqrt{3}}{2} or C = 120^circ, S_{triangle ABC} = frac{sqrt{3}}{4}.