💡First, I need to determine the sampling interval used in the systematic sampling method. Given that the smallest two selected numbers are 03 and 09, the interval can be calculated by subtracting these two numbers: 9 - 3 = 6.Next, I calculate the total number of students to be sampled. Since there are 60 students in total and the sampling interval is 6, the number of sampled students is 60 ÷ 6 = 10.Finally, to find the largest number among the selected students, I start from the smallest number (03) and add the product of the sampling interval and the number of intervals (which is one less than the number of sampled students). Therefore, the largest number is 3 + 6 × 9 = 57.

💡Alright, so I have this inequality: ((a^{2} - 4)x^{2} + (a + 2)x - 1 geq 0), and I need to find the range of real numbers (a) such that the solution set for (x) is not empty. Hmm, okay. Let me think about how to approach this.First, I notice that this is a quadratic inequality in terms of (x). The general form of a quadratic inequality is (Ax^{2} + Bx + C geq 0), where (A), (B), and (C) are coefficients. In this case, (A = a^{2} - 4), (B = a + 2), and (C = -1).Now, for a quadratic inequality, the solution set depends on the leading coefficient (A) and the discriminant (Delta = B^{2} - 4AC). If (A > 0), the parabola opens upwards, and the inequality (Ax^{2} + Bx + C geq 0) will have solutions either between the roots or outside the roots, depending on the discriminant. If (A < 0), the parabola opens downwards, and the solutions will be between the roots or outside, again depending on the discriminant. If (A = 0), the inequality becomes linear.So, let me consider different cases based on the value of (A = a^{2} - 4).**Case 1: (A = 0)**If (a^{2} - 4 = 0), then (a = pm 2).- If (a = 2), the inequality becomes ((0)x^{2} + (4)x - 1 geq 0), which simplifies to (4x - 1 geq 0). Solving for (x), we get (x geq frac{1}{4}). So, the solution set is not empty. - If (a = -2), the inequality becomes ((0)x^{2} + (0)x - 1 geq 0), which simplifies to (-1 geq 0). This is never true, so the solution set is empty.Therefore, when (a = 2), the solution set is non-empty, but when (a = -2), it's empty. So, (a = 2) is acceptable, but (a = -2) is not.**Case 2: (A neq 0)**Now, if (a^{2} - 4 neq 0), then (a neq pm 2), and we have a proper quadratic inequality.For the quadratic inequality (Ax^{2} + Bx + C geq 0) to have a non-empty solution set, the quadratic must touch or cross the x-axis. This means the discriminant must be non-negative, i.e., (Delta geq 0).Let's compute the discriminant:[Delta = B^{2} - 4AC = (a + 2)^{2} - 4(a^{2} - 4)(-1)]Simplify this:[Delta = (a^{2} + 4a + 4) - 4(a^{2} - 4)(-1)][Delta = a^{2} + 4a + 4 + 4(a^{2} - 4)][Delta = a^{2} + 4a + 4 + 4a^{2} - 16][Delta = 5a^{2} + 4a - 12]So, for the quadratic to have real roots, we require:[5a^{2} + 4a - 12 geq 0]Let's solve this inequality for (a).First, find the roots of the quadratic equation (5a^{2} + 4a - 12 = 0):Using the quadratic formula:[a = frac{-b pm sqrt{b^{2} - 4ac}}{2a}]Here, (a = 5), (b = 4), and (c = -12):[a = frac{-4 pm sqrt{16 + 240}}{10}][a = frac{-4 pm sqrt{256}}{10}][a = frac{-4 pm 16}{10}]So, the roots are:[a = frac{-4 + 16}{10} = frac{12}{10} = frac{6}{5} = 1.2][a = frac{-4 - 16}{10} = frac{-20}{10} = -2]So, the quadratic (5a^{2} + 4a - 12) factors as (5(a + 2)(a - frac{6}{5})).Since the coefficient of (a^{2}) is positive (5), the parabola opens upwards. Therefore, the inequality (5a^{2} + 4a - 12 geq 0) holds when (a leq -2) or (a geq frac{6}{5}).But wait, in Case 2, we have (a neq pm 2). So, we need to consider (a < -2) or (a geq frac{6}{5}).However, we also need to consider the leading coefficient (A = a^{2} - 4). If (A > 0), the parabola opens upwards, and the inequality (Ax^{2} + Bx + C geq 0) will have solutions either between the roots or outside the roots. If (A < 0), the parabola opens downwards, and the inequality will have solutions between the roots.But in our case, since we only require the solution set to be non-empty, regardless of whether it's between or outside the roots, as long as the quadratic has real roots, the solution set is non-empty.Therefore, as long as the discriminant is non-negative, which we found occurs when (a leq -2) or (a geq frac{6}{5}), the quadratic inequality will have a non-empty solution set.But remember, in Case 1, when (a = 2), the solution set is non-empty, but when (a = -2), it's empty. So, we need to exclude (a = -2) from our solution.Therefore, combining Case 1 and Case 2, the range of (a) is:- (a = 2) (from Case 1)- (a < -2) or (a geq frac{6}{5}) (from Case 2)But wait, in Case 2, we have (a neq pm 2), so (a = 2) is already included in Case 1. So, the overall range is (a < -2) or (a geq frac{6}{5}).But let me double-check if (a = frac{6}{5}) is included. When (a = frac{6}{5}), the discriminant is zero, so the quadratic touches the x-axis at one point, meaning the inequality (Ax^{2} + Bx + C geq 0) will have exactly one solution, so the solution set is not empty. Therefore, (a = frac{6}{5}) is included.Similarly, when (a = 2), as we saw earlier, the solution set is non-empty.But when (a = -2), the solution set is empty, so (a = -2) is excluded.Therefore, the range of (a) is all real numbers less than -2 or greater than or equal to (frac{6}{5}).So, in interval notation, this is:[(-infty, -2) cup left[frac{6}{5}, inftyright)]Looking back at the options:A: ((-2, frac{6}{5}])B: ([-2, frac{6}{5}])C: ((-infty, -2) cup [frac{6}{5}, +infty))D: ((-infty, -2] cup [frac{6}{5}, +infty))So, my conclusion is that the correct answer is C.Wait, but hold on. Let me make sure I didn't make a mistake in interpreting the discriminant.We found that the discriminant is non-negative when (a leq -2) or (a geq frac{6}{5}). However, in Case 2, we have (a neq pm 2). So, (a = -2) is excluded because when (a = -2), the quadratic becomes linear, but in that case, the solution set is empty.Therefore, in the discriminant condition, (a leq -2) includes (a = -2), but in reality, (a = -2) leads to an empty solution set, so we must exclude it.Hence, the correct interval should be (a < -2) or (a geq frac{6}{5}), which is option C.Yes, that makes sense.**Final Answer**boxed{C}

💡Alright, so I've got this problem here where I need to prove that α + 2β = π/2, given that both α and β are acute angles. The problem provides two equations:1. 3 sin²α + 2 sin²β = 12. 3 sin2α - 2 sin2β = 0Okay, let's start by understanding what these equations mean. Both α and β are acute, which means they're between 0 and π/2 radians. That might be useful later on because it tells me that all the trigonometric functions for these angles will be positive.Looking at the second equation first: 3 sin2α - 2 sin2β = 0. Maybe I can solve this for one of the variables. Let's see:3 sin2α = 2 sin2βSo, sin2β = (3/2) sin2α.Hmm, that seems interesting. So, sin2β is proportional to sin2α. Maybe I can express sin2β in terms of sin2α and then substitute that into the first equation.But before I do that, let me recall that sin2θ = 2 sinθ cosθ. So, perhaps I can rewrite sin2α and sin2β using this identity.So, sin2α = 2 sinα cosα, and sin2β = 2 sinβ cosβ.Substituting back into the equation sin2β = (3/2) sin2α, we get:2 sinβ cosβ = (3/2) * 2 sinα cosαSimplifying that:2 sinβ cosβ = 3 sinα cosαDivide both sides by 2:sinβ cosβ = (3/2) sinα cosαHmm, okay. So, sinβ cosβ is equal to (3/2) sinα cosα.I wonder if I can express sinα and sinβ in terms of each other or something. Maybe I can use the first equation for that.The first equation is 3 sin²α + 2 sin²β = 1.Let me denote sinα as 'a' and sinβ as 'b' for simplicity. So, the first equation becomes:3a² + 2b² = 1And from the second equation, we have:sinβ cosβ = (3/2) sinα cosαBut since sinβ = b and sinα = a, then cosβ = sqrt(1 - b²) and cosα = sqrt(1 - a²), because α and β are acute, so cosines are positive.So, substituting back into the second equation:b * sqrt(1 - b²) = (3/2) * a * sqrt(1 - a²)Hmm, this seems a bit complicated, but maybe I can square both sides to eliminate the square roots.So, squaring both sides:b² (1 - b²) = (9/4) a² (1 - a²)Expanding both sides:b² - b⁴ = (9/4) a² - (9/4) a⁴Now, let's recall from the first equation that 3a² + 2b² = 1. Maybe I can express one variable in terms of the other.Let's solve for a² from the first equation:3a² = 1 - 2b²So, a² = (1 - 2b²)/3Similarly, I can express a⁴ as (a²)² = [(1 - 2b²)/3]^2Let's compute that:a⁴ = (1 - 4b² + 4b⁴)/9Now, substitute a² and a⁴ into the equation we got from squaring the second equation:b² - b⁴ = (9/4) * [(1 - 2b²)/3] - (9/4) * [(1 - 4b² + 4b⁴)/9]Simplify term by term:First term on the right: (9/4) * [(1 - 2b²)/3] = (9/4)*(1/3 - 2b²/3) = (3/4 - 3b²/2)Second term on the right: (9/4) * [(1 - 4b² + 4b⁴)/9] = (1 - 4b² + 4b⁴)/4So, putting it all together:b² - b⁴ = (3/4 - 3b²/2) - (1/4 - b² + b⁴)Simplify the right side:First, distribute the negative sign:= 3/4 - 3b²/2 - 1/4 + b² - b⁴Combine like terms:3/4 - 1/4 = 2/4 = 1/2-3b²/2 + b² = (-3b²/2 + 2b²/2) = (-b²)/2So, the right side becomes:1/2 - (b²)/2 - b⁴Now, the equation is:b² - b⁴ = 1/2 - (b²)/2 - b⁴Let's subtract 1/2 - (b²)/2 - b⁴ from both sides to bring everything to the left:b² - b⁴ - 1/2 + (b²)/2 + b⁴ = 0Simplify:b² + (b²)/2 - 1/2 = 0Combine like terms:(3b²)/2 - 1/2 = 0Multiply both sides by 2 to eliminate denominators:3b² - 1 = 0So, 3b² = 1 => b² = 1/3 => b = 1/√3 (since β is acute, b is positive)So, sinβ = 1/√3. Therefore, β = arcsin(1/√3). Let's compute that:arcsin(1/√3) is approximately 35.264 degrees.Now, let's find a² from the first equation:a² = (1 - 2b²)/3 = (1 - 2*(1/3))/3 = (1 - 2/3)/3 = (1/3)/3 = 1/9So, a = 1/3 (since α is acute, a is positive)Therefore, sinα = 1/3, so α = arcsin(1/3) ≈ 19.47 degrees.Now, let's check if α + 2β = π/2.Compute α + 2β:α ≈ 19.47 degrees2β ≈ 2*35.264 ≈ 70.528 degreesSo, α + 2β ≈ 19.47 + 70.528 ≈ 90 degrees, which is π/2 radians.So, it seems to hold numerically.But I need to prove it without relying on numerical approximations. Let's see.We have sinα = 1/3 and sinβ = 1/√3.We need to show that α + 2β = π/2.Let me consider the angle addition formula for sine:sin(α + 2β) = sinα cos2β + cosα sin2βBut if α + 2β = π/2, then sin(α + 2β) = sin(π/2) = 1.So, let's compute sin(α + 2β):= sinα cos2β + cosα sin2βWe know sinα = 1/3, sinβ = 1/√3.First, compute cosα and cosβ:cosα = sqrt(1 - sin²α) = sqrt(1 - 1/9) = sqrt(8/9) = 2√2/3cosβ = sqrt(1 - sin²β) = sqrt(1 - 1/3) = sqrt(2/3) = √6/3Now, compute cos2β:cos2β = 1 - 2 sin²β = 1 - 2*(1/3) = 1 - 2/3 = 1/3Similarly, sin2β = 2 sinβ cosβ = 2*(1/√3)*(√6/3) = 2*(√6)/(3√3) = 2*(√2)/3 = 2√2/3Wait, but earlier we had sin2β = (3/2) sin2α.Let's compute sin2α:sin2α = 2 sinα cosα = 2*(1/3)*(2√2/3) = 4√2/9Then, (3/2) sin2α = (3/2)*(4√2/9) = (12√2)/18 = (2√2)/3, which matches sin2β = 2√2/3.Good, consistency here.Now, back to sin(α + 2β):= sinα cos2β + cosα sin2β= (1/3)*(1/3) + (2√2/3)*(2√2/3)= (1/9) + (8/9)= 9/9 = 1So, sin(α + 2β) = 1, which implies α + 2β = π/2 + 2πk, where k is an integer.But since α and β are acute, α + 2β must be less than π/2 + 2*(π/2) = 3π/2, but more precisely, since α < π/2 and β < π/2, α + 2β < π/2 + π = 3π/2. However, given the values we found, α + 2β is exactly π/2.Therefore, α + 2β = π/2.I think that's a solid proof. I used the given equations to find expressions for sinα and sinβ, then computed the necessary cosines and used the angle addition formula to show that sin(α + 2β) = 1, which leads to α + 2β = π/2.

💡Okay, so I have this problem here: there are 400 students in a school, and each student sent 200 greetings to other students for New Year. I need to find the minimum number of pairs of students who have mutually greeted each other. Hmm, okay, let me try to wrap my head around this.First off, let's break down what's given. There are 400 students, and each one sends out 200 greetings. So, each student is sending greetings to 200 others. But the question is about mutual greetings, meaning both student A sends a greeting to student B and student B sends a greeting back to student A. We need to find the minimum number of such pairs.Alright, so maybe I should think about this in terms of graph theory. If I consider each student as a node, and each greeting as a directed edge from one node to another, then a mutual greeting would be a pair of directed edges going both ways between two nodes. So, essentially, we're looking for the minimum number of such bidirectional edges in this graph.Given that there are 400 nodes and each node has an out-degree of 200, the total number of directed edges in the graph is 400 * 200 = 80,000. Now, in an undirected graph, the maximum number of edges possible is C(400, 2) = (400 * 399)/2 = 79,800. Wait a minute, that's interesting. The total number of directed edges is 80,000, which is just slightly more than the maximum number of undirected edges.So, if we think about it, if we could somehow arrange all the greetings to be one-way, we'd only need 79,800 directed edges. But since we have 80,000, which is 200 more, that suggests that there must be some mutual greetings. Each mutual greeting adds an extra directed edge beyond the one-way connections.Let me formalize this a bit. Let’s denote the number of mutual pairs as M. Each mutual pair contributes 2 directed edges (A to B and B to A). The remaining directed edges are one-way, contributing 1 directed edge each. So, the total number of directed edges can be expressed as:Total directed edges = 2*M + (Total one-way edges)But the total one-way edges can't exceed the maximum number of undirected edges, which is 79,800. So, we have:80,000 = 2*M + (79,800 - M)Wait, is that right? Let me check. If M is the number of mutual pairs, then the number of one-way edges would be the total possible undirected edges minus M, because each mutual pair uses up one undirected edge. So, the one-way edges would be 79,800 - M, and each contributes 1 directed edge. The mutual pairs contribute 2 directed edges each, so 2*M.So, total directed edges = 2*M + (79,800 - M) = M + 79,800But we know the total directed edges are 80,000, so:M + 79,800 = 80,000Therefore, M = 80,000 - 79,800 = 200So, the minimum number of mutual pairs is 200.Wait, does that make sense? Let me think about it again. If we have 400 students, each sending 200 greetings, that's 80,000 directed edges. The maximum number of one-way edges without any mutual greetings would be 79,800. So, we have 200 extra directed edges, which must come from mutual greetings. Each mutual greeting adds one extra directed edge beyond the one-way connection. So, 200 extra directed edges mean 200 mutual pairs.Yeah, that seems to check out. So, the minimum number of mutual pairs is 200.I guess another way to think about it is that if you try to minimize mutual greetings, you want to maximize the number of one-way greetings. But since the total number of greetings exceeds the maximum possible one-way connections, you have to have some mutuals. The difference between the total greetings and the maximum one-way connections gives you the number of mutual pairs.So, yeah, I think 200 is the right answer.

💡Okay, so I have this problem where there are seven equilateral triangles, each with side length 2, arranged in a row. The way they're arranged is such that the midpoint of the base of one triangle is a vertex of the next triangle. I need to find the area of the region covered by the union of these seven triangles. The options are given from A to E, with various multiples of sqrt(3)/2. First, I need to visualize how these triangles are arranged. Since each triangle is equilateral with side length 2, all sides are 2 units, and all angles are 60 degrees. When they say the midpoint of the base of one triangle is a vertex of the next, I imagine that each subsequent triangle is attached at the midpoint of the previous one's base. So, it's like a chain of triangles connected at their midpoints.Let me sketch this out mentally. The first triangle is standing on its base. The next triangle has one of its vertices at the midpoint of the first triangle's base. So, the second triangle is kind of leaning on the first one. Similarly, the third triangle is attached at the midpoint of the second triangle's base, and so on until the seventh triangle.Now, since each triangle is attached at the midpoint, the overlapping area between two adjacent triangles must be calculated. Because if I just multiply the area of one triangle by seven, I'll be overcounting the overlapping regions. So, I need to find the total area covered by subtracting the overlapping areas.First, let's calculate the area of a single equilateral triangle. The formula for the area of an equilateral triangle with side length 'a' is (sqrt(3)/4) * a^2. So, plugging in a = 2, the area is (sqrt(3)/4) * 4 = sqrt(3). So, each triangle has an area of sqrt(3).Therefore, seven such triangles would have a total area of 7 * sqrt(3) if there were no overlaps. But since they are overlapping, the actual area covered will be less. So, I need to figure out how much area is overlapping and subtract that from the total.Now, how much do these triangles overlap? Each triangle is attached at the midpoint of the base of the previous one. So, the overlapping region between two adjacent triangles is a smaller triangle. Let me figure out the dimensions of this overlapping region.Since the midpoint divides the base into two equal parts, each of length 1 (since the total base is 2). So, the overlapping region is a triangle with side length 1. Because when you attach a triangle at the midpoint, the overlapping area is a smaller equilateral triangle where each side is half the length of the original triangle.So, the area of each overlapping triangle is (sqrt(3)/4) * (1)^2 = sqrt(3)/4. Since each overlap is between two triangles, and there are seven triangles arranged in a row, how many overlaps are there?Well, between seven triangles arranged in a row, there are six overlaps. For example, between triangle 1 and 2, triangle 2 and 3, ..., triangle 6 and 7. So, six overlaps in total.Therefore, the total overlapping area is 6 * (sqrt(3)/4) = (6/4) * sqrt(3) = (3/2) * sqrt(3).So, the total area covered by the union of the seven triangles is the total area without considering overlaps minus the overlapping area. That is:Total area = 7 * sqrt(3) - (3/2) * sqrt(3)Let me compute that:7 * sqrt(3) is equal to (14/2) * sqrt(3). So, subtracting (3/2) * sqrt(3) from that gives:(14/2 - 3/2) * sqrt(3) = (11/2) * sqrt(3)So, the area is (11 * sqrt(3)) / 2.Looking back at the options, that's option B.Wait, let me double-check my reasoning. Each overlap is indeed a small triangle of side length 1, so area sqrt(3)/4, and there are six overlaps. So, 6 * sqrt(3)/4 is 3*sqrt(3)/2. Then, subtracting that from 7*sqrt(3) gives 7*sqrt(3) - 1.5*sqrt(3) = 5.5*sqrt(3), which is 11/2 * sqrt(3). That seems correct.Alternatively, maybe I can think of the figure as a larger polygon. Since each triangle is attached at the midpoint, the overall shape might be a larger equilateral triangle or some other polygon. But with seven triangles, it's more likely a sort of zig-zag shape.But regardless, the overlapping areas are six small triangles, each of area sqrt(3)/4, so total overlapping area is 3*sqrt(3)/2. Subtracting that from the total area gives the correct union area.So, I think my answer is correct. The area is (11*sqrt(3))/2, which is option B.**Final Answer**boxed{B}

💡Okay, so I have this problem where there's a circle with two chords starting from the same point. The chords are 20 cm and 26 cm long, and the angle between them is 36 degrees and 38 minutes. I need to find the radius of the circle. Hmm, let me think about how to approach this.First, I remember that in a circle, the length of a chord is related to the radius and the central angle subtended by the chord. The formula for the length of a chord is ( c = 2r sinleft(frac{theta}{2}right) ), where ( c ) is the chord length, ( r ) is the radius, and ( theta ) is the central angle in radians. But in this problem, the angle given is between the two chords, not the central angles subtended by each chord. So, I need to figure out how to relate this angle to the central angles.Wait, maybe I can use the Law of Cosines here. Since the two chords form a triangle with the angle between them, I can find the length of the third side of this triangle, which would be the distance between the endpoints of the two chords. Then, I can relate this distance to the radius using the chord length formula.Let me write down what I know:- Chord 1: 20 cm- Chord 2: 26 cm- Angle between them: 36 degrees 38 minutesFirst, I need to convert the angle from degrees and minutes to just degrees to make calculations easier. 38 minutes is ( frac{38}{60} ) degrees, which is approximately 0.6333 degrees. So, the total angle is 36.6333 degrees.Now, I'll use the Law of Cosines to find the distance between the endpoints of the two chords. The Law of Cosines states that for a triangle with sides a, b, and c, and angle C opposite side c, ( c^2 = a^2 + b^2 - 2abcos(C) ).Applying this to our problem:- Let the two chords be sides a and b: 20 cm and 26 cm.- The angle between them is C: 36.6333 degrees.- The distance between the endpoints is c.So, plugging into the formula:( c^2 = 20^2 + 26^2 - 2 times 20 times 26 times cos(36.6333^circ) )Calculating each part:- ( 20^2 = 400 )- ( 26^2 = 676 )- ( 2 times 20 times 26 = 1040 )- ( cos(36.6333^circ) ) is approximately 0.805 (I can use a calculator for this)So, ( c^2 = 400 + 676 - 1040 times 0.805 )Calculating further:- ( 400 + 676 = 1076 )- ( 1040 times 0.805 = 836.2 )- Therefore, ( c^2 = 1076 - 836.2 = 239.8 )- So, ( c approx sqrt{239.8} approx 15.49 ) cmNow, this distance c is the length of the chord that connects the endpoints of the two given chords. Using the chord length formula again, ( c = 2r sinleft(frac{theta}{2}right) ), where ( theta ) is the central angle subtended by this chord.But wait, I don't know the central angle for this chord. Hmm, maybe I need to relate the given angle between the two chords to the central angles.Let me denote the central angles subtended by the two chords as ( theta_1 ) and ( theta_2 ). The angle between the chords at the circumference is 36.6333 degrees, which is related to the central angles by the formula ( text{Angle at circumference} = frac{1}{2}(theta_1 + theta_2) ). So, ( 36.6333^circ = frac{1}{2}(theta_1 + theta_2) ), which means ( theta_1 + theta_2 = 73.2666^circ ).But I also know the lengths of the chords:- For the 20 cm chord: ( 20 = 2r sinleft(frac{theta_1}{2}right) )- For the 26 cm chord: ( 26 = 2r sinleft(frac{theta_2}{2}right) )Simplifying these:- ( sinleft(frac{theta_1}{2}right) = frac{10}{r} )- ( sinleft(frac{theta_2}{2}right) = frac{13}{r} )Now, I have two equations:1. ( theta_1 + theta_2 = 73.2666^circ )2. ( sinleft(frac{theta_1}{2}right) = frac{10}{r} )3. ( sinleft(frac{theta_2}{2}right) = frac{13}{r} )This seems a bit complicated, but maybe I can express ( theta_1 ) and ( theta_2 ) in terms of r and then solve for r.Alternatively, perhaps I can use the fact that the chord c we found earlier (approximately 15.49 cm) subtends a central angle which we can relate to the radius. Let's denote this central angle as ( theta_c ). Then, using the chord length formula:( 15.49 = 2r sinleft(frac{theta_c}{2}right) )But I don't know ( theta_c ) either. Maybe there's another relationship I can use. Since the three chords form a triangle, perhaps the sum of the central angles equals 360 degrees? Wait, no, because the chords are from the same point, the central angles would add up to the total angle around that point, which is 360 degrees. But in this case, we have two chords, so the sum of their central angles plus the central angle of the third chord (c) should equal 360 degrees.Wait, that might not be correct. Actually, the central angles correspond to the arcs between the points where the chords meet the circle. Since the two original chords are from the same point, the central angles for those chords and the central angle for the chord c should add up to 360 degrees. So, ( theta_1 + theta_2 + theta_c = 360^circ ).But I already have ( theta_1 + theta_2 = 73.2666^circ ), so ( theta_c = 360^circ - 73.2666^circ = 286.7334^circ ). That seems too large for a central angle, which should be less than 180 degrees if we're considering the smaller arc. Maybe I made a mistake here.Wait, perhaps the central angle for chord c is actually the angle between the two original chords, but measured at the center. So, if the angle at the circumference is 36.6333 degrees, then the central angle would be twice that, which is 73.2666 degrees. That makes more sense because the central angle is twice the inscribed angle subtended by the same arc.So, if the central angle for chord c is 73.2666 degrees, then using the chord length formula:( 15.49 = 2r sinleft(frac{73.2666^circ}{2}right) )Calculating ( frac{73.2666^circ}{2} = 36.6333^circ )So, ( sin(36.6333^circ) approx 0.595 )Therefore, ( 15.49 = 2r times 0.595 )Simplifying:( 15.49 = 1.19r )So, ( r = frac{15.49}{1.19} approx 13.02 ) cmWait, that seems reasonable. Let me check if this radius makes sense with the original chords.Using the chord length formula for the 20 cm chord:( 20 = 2r sinleft(frac{theta_1}{2}right) )So, ( sinleft(frac{theta_1}{2}right) = frac{10}{13.02} approx 0.768 )Therefore, ( frac{theta_1}{2} approx arcsin(0.768) approx 50^circ )So, ( theta_1 approx 100^circ )Similarly, for the 26 cm chord:( 26 = 2r sinleft(frac{theta_2}{2}right) )So, ( sinleft(frac{theta_2}{2}right) = frac{13}{13.02} approx 0.998 )Therefore, ( frac{theta_2}{2} approx arcsin(0.998) approx 86^circ )So, ( theta_2 approx 172^circ )But wait, earlier we had ( theta_1 + theta_2 = 73.2666^circ ), but here ( theta_1 + theta_2 approx 100^circ + 172^circ = 272^circ ), which doesn't match. That means I must have made a mistake somewhere.Let me go back. The central angle for chord c is actually the angle between the two original chords, which is 73.2666 degrees, not the angle at the circumference. So, using that, I found the radius to be approximately 13.02 cm. But when I checked with the original chords, the central angles don't add up correctly.Maybe I need to consider that the central angles for the two original chords and the central angle for chord c should add up to 360 degrees. So, ( theta_1 + theta_2 + theta_c = 360^circ ). We have ( theta_c = 73.2666^circ ), so ( theta_1 + theta_2 = 360^circ - 73.2666^circ = 286.7334^circ ). But earlier, I thought ( theta_1 + theta_2 = 73.2666^circ ), which was incorrect.So, actually, ( theta_1 + theta_2 = 286.7334^circ ). Now, using the chord length formulas:- ( 20 = 2r sinleft(frac{theta_1}{2}right) )- ( 26 = 2r sinleft(frac{theta_2}{2}right) )Let me denote ( alpha = frac{theta_1}{2} ) and ( beta = frac{theta_2}{2} ). Then, ( alpha + beta = frac{theta_1 + theta_2}{2} = frac{286.7334^circ}{2} = 143.3667^circ ).So, we have:1. ( sin(alpha) = frac{10}{r} )2. ( sin(beta) = frac{13}{r} )3. ( alpha + beta = 143.3667^circ )Now, I can use the sine addition formula or some trigonometric identities to solve for r. Let me express ( beta = 143.3667^circ - alpha ), then substitute into the second equation:( sin(143.3667^circ - alpha) = frac{13}{r} )Using the sine of a difference:( sin(143.3667^circ - alpha) = sin(143.3667^circ)cos(alpha) - cos(143.3667^circ)sin(alpha) )We know ( sin(alpha) = frac{10}{r} ), so ( cos(alpha) = sqrt{1 - left(frac{10}{r}right)^2} ).Similarly, ( sin(143.3667^circ) approx sin(180^circ - 36.6333^circ) = sin(36.6333^circ) approx 0.595 )And ( cos(143.3667^circ) = -cos(36.6333^circ) approx -0.805 )So, substituting back:( 0.595 times sqrt{1 - left(frac{10}{r}right)^2} - (-0.805) times frac{10}{r} = frac{13}{r} )Simplifying:( 0.595 times sqrt{1 - left(frac{10}{r}right)^2} + 0.805 times frac{10}{r} = frac{13}{r} )( 0.595 times sqrt{1 - left(frac{10}{r}right)^2} + frac{8.05}{r} = frac{13}{r} )Subtract ( frac{8.05}{r} ) from both sides:( 0.595 times sqrt{1 - left(frac{10}{r}right)^2} = frac{13 - 8.05}{r} )( 0.595 times sqrt{1 - left(frac{10}{r}right)^2} = frac{4.95}{r} )Divide both sides by 0.595:( sqrt{1 - left(frac{10}{r}right)^2} = frac{4.95}{0.595r} approx frac{8.322}{r} )Square both sides:( 1 - left(frac{10}{r}right)^2 = left(frac{8.322}{r}right)^2 )( 1 - frac{100}{r^2} = frac{69.25}{r^2} )Multiply both sides by ( r^2 ):( r^2 - 100 = 69.25 )( r^2 = 169.25 )( r approx sqrt{169.25} approx 13.01 ) cmSo, the radius is approximately 13.01 cm. Let me check if this makes sense with the original chords.For the 20 cm chord:( 20 = 2 times 13.01 times sinleft(frac{theta_1}{2}right) )( sinleft(frac{theta_1}{2}right) = frac{10}{13.01} approx 0.768 )( frac{theta_1}{2} approx 50^circ )( theta_1 approx 100^circ )For the 26 cm chord:( 26 = 2 times 13.01 times sinleft(frac{theta_2}{2}right) )( sinleft(frac{theta_2}{2}right) = frac{13}{13.01} approx 0.999 )( frac{theta_2}{2} approx 86.8^circ )( theta_2 approx 173.6^circ )Now, ( theta_1 + theta_2 approx 100^circ + 173.6^circ = 273.6^circ ), which is close to the earlier calculated 286.7334^circ, but not exact. This slight discrepancy might be due to rounding errors in the calculations. However, considering the approximations made, especially in the sine values and angle conversions, 13.01 cm seems a reasonable estimate for the radius.Therefore, the radius of the circle is approximately 13.02 cm.