answer：Alright, so I've got this geometry problem here, and I need to figure out the sum of all possible values for the area of rectangle S. Let me try to visualize the problem first.There's a larger square, and inside it, there are three figures: a rectangle with side lengths 2 and 4, a smaller square with side length 2, and another rectangle S. The smaller square and the 2x4 rectangle are placed in such a way that they don't overlap, leaving a continuous area for rectangle S. I need to find the sum of all possible areas of S, expressed as a fraction m/n where m and n are coprime, and then choose the correct option from A to E.First, let's break down the information:1. The larger square contains three figures: a 2x4 rectangle, a 2x2 square, and rectangle S.2. The smaller square (2x2) and the 2x4 rectangle are placed without overlapping, leaving a continuous area for S.3. We need to find the sum of all possible areas of S.I think the key here is to figure out the possible configurations of the 2x4 rectangle and the 2x2 square inside the larger square, which will determine the area left for S.Let me start by determining the side length of the larger square. Since it contains a 2x4 rectangle, the side length must be at least 4. But it's also containing a 2x2 square, so does that affect the side length? Well, 4 is already larger than 2, so the side length of the larger square is at least 4.But wait, is it exactly 4? Or could it be larger? The problem doesn't specify, so I think it's safe to assume the larger square is just big enough to fit all three figures without overlapping. So, the side length is 4.Let me confirm that. If the larger square has a side length of 4, its area is 16. The area of the 2x4 rectangle is 8, the area of the 2x2 square is 4, so together they take up 12 units of area. That leaves 4 units of area for rectangle S. So, is the area of S always 4? That seems too straightforward, but maybe that's the case.But wait, the problem says "the sum of all possible values for the area of S." So, does that mean there could be more than one possible area for S? Or is it just 4?Let me think about the possible configurations. If the larger square is 4x4, how can the 2x4 rectangle and the 2x2 square be placed without overlapping? There are different ways to arrange them.One way is to place the 2x4 rectangle along one side of the larger square, say the bottom side, and then place the 2x2 square somewhere else. But where? If the 2x4 is along the bottom, then the remaining space on the top would be 4 units wide and 2 units tall. But the 2x2 square can fit into that space, leaving a 2x2 area for S. Wait, but that would make S also 2x2, which is 4. Hmm.Alternatively, could the 2x4 rectangle be placed vertically? If the larger square is 4x4, placing the 2x4 rectangle vertically would take up the entire height of the square, leaving a 2x4 area on the side. But then where would the 2x2 square go? It could be placed in the remaining 2x4 area, but that would overlap with S. Wait, no, the 2x2 square and S have to be placed without overlapping.Wait, maybe I need to draw this out mentally. If the 2x4 is placed vertically on the left side, taking up the entire height of 4, then the remaining width is 2. So, the remaining area is 2x4. But the 2x2 square can be placed in the top right corner, leaving a 2x2 area for S. So, again, S would be 2x2, area 4.Alternatively, could the 2x4 rectangle be placed horizontally, and the 2x2 square be placed in a different spot, leaving a different area for S?Wait, if I place the 2x4 rectangle horizontally at the bottom, taking up 4 units in length and 2 units in height, then the remaining area is 4 units in length and 2 units in height on top. Then, placing the 2x2 square somewhere in that top area, say on the left, would leave a 2x2 area on the right for S. Again, area 4.Is there a way to arrange the 2x4 rectangle and the 2x2 square such that S has a different area? Maybe if they are placed in a different configuration.Wait, what if the 2x4 rectangle is placed diagonally? But no, in a square, placing a rectangle diagonally would complicate things, and the problem doesn't mention anything about diagonals, so I think we can assume all figures are axis-aligned.Alternatively, could the 2x4 rectangle and the 2x2 square overlap in some way that still leaves a continuous area for S? But the problem states they do not overlap, so they must be placed separately.Wait, maybe the larger square isn't necessarily 4x4? Maybe it's larger? Let me reconsider.If the larger square is bigger than 4x4, say 5x5, then the area would be 25. The areas of the 2x4 and 2x2 would still be 8 and 4, so S would be 13. But the problem doesn't specify the size of the larger square, only that it's larger and contains these three figures without overlapping. So, is the side length of the larger square fixed?Wait, the problem says "a larger square contains three figures," so it's not necessarily the minimal square. But without knowing the exact size, how can we determine the area of S? Maybe the side length is fixed because of the way the figures are placed.Wait, if the 2x4 rectangle is placed in the larger square, the side length must be at least 4. If the 2x2 square is placed somewhere else, the side length could still be 4, as the 2x2 can fit in the remaining space.Alternatively, maybe the larger square is 6x6? But that seems too big, and the problem doesn't specify.Wait, perhaps the larger square is determined by the arrangement of the 2x4 and 2x2. Let me think.If the 2x4 is placed horizontally, taking up 4 units in length and 2 units in height, and the 2x2 is placed vertically next to it, taking up 2 units in width and 2 units in height, then the total height would be 2, and the total width would be 4 + 2 = 6. But that would make the larger square 6x6, which seems too big.Alternatively, if the 2x4 is placed vertically, taking up 4 units in height and 2 units in width, and the 2x2 is placed next to it, taking up 2 units in width and 2 units in height, then the total width would be 2 + 2 = 4, and the total height would be 4 + 2 = 6. Again, a 6x6 square.But the problem says it's a larger square, so maybe it's 6x6. But then the area of S would be 36 - 8 - 4 = 24. But that's not one of the options. Hmm.Wait, maybe I'm overcomplicating this. Let's go back to the initial assumption that the larger square is 4x4. Then, the area of S is 4, which is option B. But the problem says "the sum of all possible values for the area of S." So, is there more than one possible area?Wait, maybe depending on how the 2x4 and 2x2 are arranged, S could have different areas. Let me think.If the 2x4 is placed horizontally at the bottom, and the 2x2 is placed vertically on the right, then S would be the remaining area on the top left, which is 2x2, area 4.Alternatively, if the 2x4 is placed vertically on the left, and the 2x2 is placed horizontally on the top, then S would be the remaining area on the bottom right, which is 2x2, area 4.Wait, is there a way to arrange them such that S is a different rectangle? For example, if the 2x4 is placed in the center, but that would leave two separate areas for S, which contradicts the problem statement that S is a single continuous area.Alternatively, maybe the 2x4 and 2x2 are placed in a way that S is a different rectangle, like 1x4 or something else.Wait, let's consider the larger square as 4x4. The total area is 16. The 2x4 rectangle is 8, the 2x2 is 4, so S is 4. But 4 can be represented as 1x4, 2x2, 4x1, etc. So, the area is fixed, but the dimensions could vary. However, the problem asks for the sum of all possible values for the area of S, not the dimensions. Since the area is fixed at 4, regardless of the configuration, the sum is just 4.But wait, the problem says "the sum of all possible values for the area of S." So, maybe there are multiple areas possible, depending on how the 2x4 and 2x2 are arranged.Wait, let me think again. If the larger square is 4x4, and the 2x4 is placed horizontally, then the remaining area is 4x2. If the 2x2 is placed in the top left corner, then S is 2x2. But if the 2x2 is placed somewhere else, could S be a different area?Wait, no, because the 2x4 and 2x2 have fixed areas, so the remaining area for S must always be 4. So, regardless of where they are placed, as long as they don't overlap, S must occupy 4 units of area.Therefore, the area of S is always 4, so the sum of all possible values is 4.But wait, the options are A)3, B)4, C)5, D)6, E)7. So, the answer should be B)4.But let me double-check. Is there a way to arrange the 2x4 and 2x2 such that S has a different area?Suppose the larger square is 5x5. Then, the area would be 25. The 2x4 is 8, the 2x2 is 4, so S would be 13. But 13 is not an option, and the problem doesn't specify the size of the larger square, so I think the larger square must be 4x4.Alternatively, if the larger square is 6x6, S would be 24, which is also not an option. So, it must be 4x4.Therefore, the area of S is 4, and the sum of all possible values is 4.Wait, but the problem says "the sum of all possible values for the area of S." So, if there's only one possible value, which is 4, then the sum is 4.Yes, that makes sense. So, the answer is B)4.

answer：First, I need to determine the speed of the boat in still water, which I'll denote as ( v ) km/hr.When the boat is moving downstream, its effective speed is the sum of its own speed and the speed of the current. Therefore, the downstream speed is ( v + 3 ) km/hr.The boat travels 33 km downstream in 44 minutes. To use consistent units, I'll convert 44 minutes to hours by dividing by 60, which gives ( frac{44}{60} ) hours.Using the formula for distance, ( text{Distance} = text{Speed} times text{Time} ), I can set up the equation:[ 33 = (v + 3) times frac{44}{60} ]Simplifying ( frac{44}{60} ) to ( frac{11}{15} ), the equation becomes:[ 33 = (v + 3) times frac{11}{15} ]To solve for ( v ), I'll divide both sides by ( frac{11}{15} ):[ frac{33}{frac{11}{15}} = v + 3 ][ 33 times frac{15}{11} = v + 3 ][ 45 = v + 3 ]Finally, subtracting 3 from both sides gives:[ v = 42 text{ km/hr} ]

answer：Alright, so I have this problem about tetrahedrons, and I need to prove that no more than one vertex can have the property that the sum of any two plane angles at that vertex is greater than 180 degrees. Hmm, okay. Let me try to wrap my head around this.First, let me recall what a tetrahedron is. It's a polyhedron with four triangular faces, six edges, and four vertices. Each vertex is connected to three edges, and at each vertex, there are three plane angles formed by the edges meeting at that vertex.The problem is saying that if I look at any vertex of the tetrahedron, and I check the sum of any two of the three plane angles at that vertex, then if this sum is greater than 180 degrees, it can only happen for at most one vertex in the entire tetrahedron. So, I need to show that it's impossible for two or more vertices to have this property simultaneously.Let me try to visualize this. Imagine a tetrahedron with vertices A, B, C, and D. Suppose vertex A has the property that the sum of any two of its plane angles is greater than 180 degrees. That means, for example, angle BAC + angle BAD > 180°, angle BAC + angle CAD > 180°, and angle BAD + angle CAD > 180°. Similarly, if vertex B also had this property, then angle ABC + angle ABD > 180°, angle ABC + angle CBD > 180°, and angle ABD + angle CBD > 180°.Wait, but if both A and B have this property, doesn't that mean that the angles around A and B are each contributing to the overall angles of the tetrahedron? Maybe this would cause some kind of conflict because the sum of angles around a point in 3D space is 360 degrees. So, if at vertex A, the sum of any two angles is greater than 180°, that might imply that each pair of angles at A is more than 180°, which could lead to the total sum around A being more than 360°, which isn't possible.But hold on, the sum of all three angles at a vertex in a tetrahedron is actually less than 360°, right? Because in a flat plane, three angles meeting at a point would sum to 360°, but in a tetrahedron, the faces are triangles, so the angles are "curved" inward, making the sum less than 360°. So, if the sum of any two angles at a vertex is greater than 180°, that might mean that each individual angle is more than 90°, but not necessarily.Let me think about this more carefully. Suppose at vertex A, the three plane angles are α, β, and γ. The condition is that α + β > 180°, α + γ > 180°, and β + γ > 180°. If I add all three inequalities together, I get 2(α + β + γ) > 540°, which simplifies to α + β + γ > 270°. But since the sum of the angles at a vertex in a tetrahedron is less than 360°, this tells me that if a vertex has this property, the sum of its three angles is between 270° and 360°.Similarly, if another vertex, say B, also has this property, then the sum of its three angles would also be between 270° and 360°. But the total sum of all angles at all four vertices of the tetrahedron is fixed. Wait, what is the total sum of all angles in a tetrahedron?Each face is a triangle, and the sum of angles in a triangle is 180°, so for four faces, that's 4 * 180° = 720°. However, each angle at a vertex is shared by three faces, so we have to be careful not to double-count. Actually, each angle at a vertex is part of one face, so the total sum of all angles at all vertices is equal to the sum of all face angles, which is 4 * 180° = 720°.So, if two vertices each have their angles summing to more than 270°, that would mean that the total sum contributed by these two vertices is more than 540°, leaving only 720° - 540° = 180° for the remaining two vertices. But each of the remaining two vertices must have their angles summing to at least something reasonable. Wait, in a tetrahedron, each vertex has three angles, each of which is part of a triangle, so each angle is less than 180°, but how much less?Actually, in a tetrahedron, the angles at each vertex are all less than 180°, but their sum is less than 360°, as I thought earlier. So, if two vertices have their angles summing to more than 270°, that leaves less than 180° for the other two vertices. But each of those other two vertices has three angles, so the sum of their angles would have to be less than 180°, meaning that each of their angles would have to be less than 60°, perhaps? That seems possible, but maybe not.Wait, no, because each angle is part of a triangle, so each angle must be greater than 0°, but there's no strict upper limit except less than 180°. However, if the sum of the angles at a vertex is less than 180°, that would mean that each angle is less than 60°, which is very sharp. But is that possible?I think it's possible, but maybe the problem is that if two vertices have their angles summing to more than 270°, then the other two vertices have their angles summing to less than 180°, which might cause some other constraints to be violated.Alternatively, maybe I should think about the dihedral angles or something else. Wait, no, the problem is about plane angles at the vertices, not dihedral angles.Let me try another approach. Suppose that two vertices, say A and B, both have the property that the sum of any two plane angles at those vertices is greater than 180°. Then, as I derived earlier, the sum of the three angles at A is greater than 270°, and similarly for B. So, the total sum of angles at A and B is greater than 540°, leaving less than 180° for the sum of angles at C and D.But each of C and D has three angles, so the sum of angles at C and D is less than 180°, meaning that each of their angles is less than 60°, as I thought before. But then, looking at the faces of the tetrahedron, each face is a triangle, and the sum of angles in a triangle is 180°. So, for example, consider the face ABC. The angles at A, B, and C in this face must sum to 180°. But if the angles at A and B are each greater than, say, 90°, then the angle at C would have to be less than 0°, which is impossible.Wait, that doesn't make sense. Let me think again. If at vertex A, the angles are α, β, γ, and at vertex B, the angles are δ, ε, ζ, then in the face ABC, the angles at A and B are α and δ, respectively, and the angle at C is something else. But if α + β > 180°, that doesn't necessarily mean that α is greater than 90°, because β could be very large as well.Hmm, maybe I need to consider specific angles. Let's suppose that at vertex A, the three angles are 100°, 100°, and 80°, so that the sum of any two is greater than 180°: 100 + 100 = 200 > 180, 100 + 80 = 180, which is not greater than 180. Oh, wait, that doesn't satisfy the condition because 100 + 80 is exactly 180, not greater. So, maybe 101°, 101°, and 78°, so that 101 + 101 = 202 > 180, 101 + 78 = 179, which is less than 180. Hmm, that's still not satisfying the condition for all pairs.Wait, maybe I need all pairs to be greater than 180°, so all three sums: α + β > 180°, α + γ > 180°, and β + γ > 180°. So, for that, each angle must be greater than 90°, because if any angle were 90° or less, then the sum of that angle with another angle would be at most 90° + something, which might not exceed 180°.Wait, no, that's not necessarily true. For example, if α = 91°, β = 91°, then α + β = 182° > 180°, but γ could be 88°, so α + γ = 179°, which is less than 180°, which doesn't satisfy the condition. So, to have all three sums greater than 180°, each pair must sum to more than 180°, which would require that each angle is greater than 90°, because if any angle were 90° or less, then adding it to another angle that's also 90° or less would give at most 180°, which doesn't satisfy the condition.Wait, actually, no. Suppose α = 100°, β = 100°, γ = 80°. Then α + β = 200° > 180°, α + γ = 180°, which is not greater than 180°, so that doesn't work. So, to have all three sums greater than 180°, each angle must be greater than 90°, because if any angle were 90° or less, then adding it to another angle that's also 90° or less would give at most 180°, which doesn't satisfy the condition.Wait, but if all three angles are greater than 90°, then their sum would be greater than 270°, as I thought earlier. So, if two vertices each have their angles summing to more than 270°, then the total sum of angles at those two vertices is more than 540°, leaving less than 180° for the other two vertices. But each of those other two vertices has three angles, so their sum would have to be less than 180°, meaning that each of their angles is less than 60°, as I thought before.But then, considering the faces of the tetrahedron, each face is a triangle, and the sum of angles in a triangle is 180°. So, for example, consider the face ABC. The angles at A, B, and C in this face must sum to 180°. But if the angles at A and B are each greater than 90°, then the angle at C would have to be less than 0°, which is impossible. Wait, that can't be right.Wait, no, because the angles at A and B in the face ABC are not the same as the plane angles at A and B in the tetrahedron. The plane angles at A and B are the angles between the edges meeting at those vertices, but in the face ABC, the angles at A and B are the angles of the triangle ABC, which are different from the plane angles of the tetrahedron.Hmm, I think I'm getting confused here. Let me clarify. In the tetrahedron, at each vertex, there are three plane angles, which are the angles between the edges meeting at that vertex. These are different from the face angles, which are the angles of the triangular faces.So, for example, at vertex A, the plane angles are between edges AB and AC, AB and AD, and AC and AD. These are different from the angles of the face ABC, which is a triangle with vertices A, B, and C.Therefore, the angles at A in the face ABC are not the same as the plane angles at A in the tetrahedron. So, my earlier reasoning about the face angles might not directly apply.Wait, but maybe I can relate the plane angles to the face angles somehow. Let me think. Each face angle is part of a triangle, and the plane angles at a vertex are related to the dihedral angles, which are the angles between the faces.Hmm, this is getting complicated. Maybe I need to use some geometric properties or inequalities.I recall that in a tetrahedron, the sum of the face angles at a vertex is less than 360°, as I thought earlier. So, if at vertex A, the sum of the plane angles is greater than 270°, that leaves less than 90° for the remaining angles at A, but wait, no, the sum of the plane angles is already greater than 270°, so the remaining angles would have to be negative, which is impossible.Wait, no, the sum of the plane angles at a vertex is less than 360°, so if the sum is greater than 270°, that's still possible, but leaves less than 90° for the remaining angles? Wait, no, the sum is already greater than 270°, so the remaining angles would have to be negative, which is impossible. So, that can't happen.Wait, I'm getting confused again. Let me try to formalize this.Let’s denote the three plane angles at vertex A as α, β, γ. The condition is that α + β > 180°, α + γ > 180°, and β + γ > 180°. Adding these three inequalities, we get 2(α + β + γ) > 540°, so α + β + γ > 270°. But since the sum of the plane angles at a vertex is less than 360°, we have 270° < α + β + γ < 360°.Similarly, if vertex B also has this property, then the sum of its three plane angles, say δ + ε + ζ, also satisfies 270° < δ + ε + ζ < 360°.Now, the total sum of all plane angles at all four vertices is equal to the sum of all face angles, which is 4 * 180° = 720°, as each face is a triangle with angles summing to 180°, and there are four faces.So, if vertices A and B each have their plane angles summing to more than 270°, then the total sum contributed by A and B is more than 540°, leaving less than 720° - 540° = 180° for the sum of the plane angles at vertices C and D.But each of C and D has three plane angles, so the sum of their angles is less than 180°, meaning that each of their angles is less than 60°, as I thought earlier.But now, consider the faces of the tetrahedron. Each face is a triangle, and the sum of its angles is 180°. For example, consider face ABC. The angles at A, B, and C in this face must sum to 180°. But the plane angles at A and B are part of this face, right? Wait, no, the plane angles at A and B are the angles between the edges meeting at those vertices, not the angles of the face itself.Wait, so the angle at A in face ABC is different from the plane angle at A in the tetrahedron. The plane angle at A is the angle between edges AB and AC, which is the same as the angle at A in face ABC. Wait, no, actually, in the face ABC, the angle at A is the same as the plane angle between edges AB and AC in the tetrahedron. So, yes, the face angles are the same as the plane angles at the vertices.Wait, that makes sense. So, in face ABC, the angle at A is the plane angle between AB and AC, which is one of the plane angles at vertex A. Similarly, the angle at B in face ABC is the plane angle between BA and BC, which is one of the plane angles at vertex B.Therefore, the sum of the angles in face ABC is equal to the sum of the plane angles at A, B, and C in that face. But since each face is a triangle, the sum is 180°, so the sum of the plane angles at A, B, and C in face ABC is 180°.But wait, if the plane angles at A and B are each part of multiple faces, then the sum of all plane angles at A across all faces would be more than just the sum of the angles in one face.Wait, no, each plane angle at a vertex is part of only one face. For example, the plane angle at A between AB and AC is part of face ABC, the plane angle at A between AB and AD is part of face ABD, and the plane angle at A between AC and AD is part of face ACD.So, each plane angle at a vertex is part of exactly one face. Therefore, the sum of all plane angles at all vertices is equal to the sum of all face angles, which is 4 * 180° = 720°, as I thought earlier.So, going back, if vertices A and B each have their plane angles summing to more than 270°, then the total sum contributed by A and B is more than 540°, leaving less than 180° for the sum of the plane angles at C and D.But each of C and D has three plane angles, so the sum of their angles is less than 180°, meaning that each of their angles is less than 60°, as I thought before.But now, consider face ABC. The sum of the plane angles at A, B, and C in this face is 180°. But if the plane angles at A and B are each part of this face, and if the plane angles at A and B are each greater than 90°, as required for their sums to be greater than 180°, then the plane angle at C in this face would have to be less than 0°, which is impossible.Wait, that can't be right. If the plane angles at A and B in face ABC are each greater than 90°, then their sum would be greater than 180°, leaving less than 0° for the angle at C, which is impossible because angles can't be negative.Therefore, this leads to a contradiction, meaning that our initial assumption that two vertices can have this property is false. Hence, no more than one vertex of a tetrahedron can have the property that the sum of any two plane angles at that vertex is greater than 180°.I think that makes sense. So, the key idea is that if two vertices had this property, the sum of their plane angles would exceed 540°, leaving less than 180° for the other two vertices, which would force the angles in some faces to be negative, which is impossible. Therefore, only one vertex can have this property.

answer：First, I need to determine the duration of each of the first two classes that David attends each week. The course lasts for 24 weeks, and David spends a total of 336 hours on the course.Each week, David has:- 2 classes of unknown duration (let's denote this duration as ( x ) hours per class)- 1 four-hour class- 4 hours of homeworkSo, the total time spent per week is:- ( 2x ) hours for the two classes- 4 hours for the additional class- 4 hours for homeworkThis adds up to ( 2x + 4 + 4 = 2x + 8 ) hours per week.Over 24 weeks, the total time spent is:[ (2x + 8) times 24 text{ hours} ]We know that the total time spent on the course is 336 hours, so we can set up the equation:[ (2x + 8) times 24 = 336 ]Expanding the equation:[ 48x + 192 = 336 ]Subtracting 192 from both sides:[ 48x = 144 ]Dividing both sides by 48:[ x = 3 ]Therefore, each of the first two classes is 3 hours long.

answer：Okay, so I have this function f(x) = 2 cos(ω/2 x) sin(ω/2 x - π/3) + √3/2, and ω is greater than 0. The problem says that the distance between two adjacent symmetric axes of the graph of f(x) is π/2. I need to find the interval where f(x) is monotonically decreasing, and then solve another part where f(θ/2) = -3/5 and θ is in the interval [-π/6, 5π/6], and find sin(θ - 5π/6).Alright, let's start with part (1). First, I think I need to simplify the function f(x). It looks like a product of cosine and sine functions, so maybe I can use a trigonometric identity to simplify it. The identity I remember is sin(A - B) = sin A cos B - cos A sin B, but here I have a product of cos and sin. Wait, there's another identity: 2 cos A sin B = sin(A + B) + sin(B - A). Let me try that.So, f(x) = 2 cos(ω/2 x) sin(ω/2 x - π/3) + √3/2. Let me set A = ω/2 x and B = ω/2 x - π/3. Then, 2 cos A sin B = sin(A + B) + sin(B - A). Let's compute A + B and B - A.A + B = ω/2 x + ω/2 x - π/3 = ω x - π/3.B - A = (ω/2 x - π/3) - ω/2 x = -π/3.So, 2 cos A sin B = sin(ω x - π/3) + sin(-π/3). Since sin(-π/3) is -√3/2, so 2 cos A sin B = sin(ω x - π/3) - √3/2.Therefore, f(x) = [sin(ω x - π/3) - √3/2] + √3/2 = sin(ω x - π/3). Okay, so f(x) simplifies to sin(ω x - π/3). That's much simpler.Now, the problem mentions the distance between two adjacent symmetric axes of the graph of f(x) is π/2. Hmm, symmetric axes for a sine function. For a standard sine function, sin(kx), the graph has symmetry about its midlines, which occur at the maxima and minima. Wait, actually, for sin(kx + φ), the function is symmetric about its peaks and troughs. So, the distance between two adjacent symmetric axes would be half the period, right?Because the period of sin(kx + φ) is 2π/k, so the distance between two adjacent peaks or troughs is π/k. But here, the distance between two adjacent symmetric axes is given as π/2. So, π/k = π/2, which implies k = 2. But in our function, f(x) = sin(ω x - π/3), so k is ω. Therefore, ω = 2.So, ω = 2. Therefore, f(x) = sin(2x - π/3).Now, part (1) asks for the interval where f(x) is monotonically decreasing. For a sine function, it's increasing where its derivative is positive and decreasing where its derivative is negative.Let's compute the derivative of f(x). f'(x) = 2 cos(2x - π/3). So, f'(x) = 2 cos(2x - π/3).We need to find where f'(x) < 0, which is where cos(2x - π/3) < 0.So, cos(θ) < 0 when θ is in (π/2 + 2πk, 3π/2 + 2πk) for integers k.Therefore, 2x - π/3 ∈ (π/2 + 2πk, 3π/2 + 2πk).Let's solve for x:π/2 + 2πk < 2x - π/3 < 3π/2 + 2πkAdd π/3 to all parts:π/2 + π/3 + 2πk < 2x < 3π/2 + π/3 + 2πkCompute π/2 + π/3: π/2 is 3π/6, π/3 is 2π/6, so together 5π/6.Similarly, 3π/2 + π/3: 3π/2 is 9π/6, π/3 is 2π/6, so together 11π/6.So:5π/6 + 2πk < 2x < 11π/6 + 2πkDivide all parts by 2:5π/12 + πk < x < 11π/12 + πkTherefore, the function f(x) is monotonically decreasing on intervals (5π/12 + πk, 11π/12 + πk) for integers k.But the question says "the interval where f(x) is monotonically decreasing." Since the function is periodic, it's decreasing on each of these intervals. So, I think the answer is all intervals of the form [5π/12 + πk, 11π/12 + πk], where k is an integer.Wait, but in the derivative, we had strict inequality, so f'(x) < 0, so f(x) is strictly decreasing on open intervals. But when writing intervals where it's decreasing, sometimes people include the endpoints where the derivative is zero. But in this case, since at x = 5π/12 + πk and x = 11π/12 + πk, the derivative is zero, so the function is neither increasing nor decreasing at those exact points. So, maybe it's safer to write open intervals.But the problem says "monotonically decreasing," which usually allows for intervals where the function is non-increasing, including points where the derivative is zero. Hmm, but in reality, at those points, the function is at a maximum or minimum. So, I think it's better to write the intervals as closed intervals, including the endpoints where the derivative is zero, because at those points, the function is not increasing.Wait, but actually, at the endpoints, the function is at a peak or trough, so it's not increasing or decreasing. So, if we include those points in the interval where it's decreasing, it's a bit ambiguous. Maybe the question expects the open intervals where the function is strictly decreasing.But since the problem says "monotonically decreasing," which can include intervals where the function is non-increasing, but in reality, at the endpoints, it's stationary. So, perhaps the answer is written as closed intervals.But to be precise, since the derivative is zero at those points, and the function is decreasing on the open intervals, I think it's better to write open intervals. But I'm not sure. Maybe the answer expects closed intervals.Wait, let me check. If I take x = 5π/12, then f'(x) = 2 cos(2*(5π/12) - π/3) = 2 cos(5π/6 - π/3) = 2 cos(π/2) = 0. Similarly, at x = 11π/12, f'(x) = 2 cos(2*(11π/12) - π/3) = 2 cos(11π/6 - π/3) = 2 cos(3π/2) = 0. So, at these points, the derivative is zero, so the function is not increasing or decreasing there. So, the function is decreasing on the open intervals (5π/12 + πk, 11π/12 + πk). So, I think the answer should be open intervals.But the problem says "the interval where f(x) is monotonically decreasing." Since the function is periodic, it's decreasing on infinitely many intervals. So, the answer is all intervals of the form (5π/12 + πk, 11π/12 + πk), where k is any integer.But in the problem statement, it's written as "the interval," singular. Maybe it's expecting one interval, but since the function is periodic, it's decreasing on each of these intervals. So, perhaps the answer is written as [5π/12 + kπ, 11π/12 + kπ], k ∈ ℤ.But I'm not sure if it's open or closed. Maybe I should write both possibilities, but I think the standard is to write open intervals when the derivative is strictly negative.Wait, actually, when we talk about intervals where a function is increasing or decreasing, we usually include the endpoints where the derivative is zero because at those points, the function is not increasing or decreasing, but the interval is still considered as a whole. So, maybe it's acceptable to write closed intervals.But in this case, since the derivative is zero at the endpoints, the function is not decreasing at those exact points, so strictly speaking, the function is decreasing on the open intervals. So, I think the answer is (5π/12 + kπ, 11π/12 + kπ), k ∈ ℤ.But let me check the problem statement again. It says "the interval where f(x) is monotonically decreasing." Since it's a periodic function, it's decreasing on infinitely many intervals, so the answer is a set of intervals, each of length π/2, starting at 5π/12 + kπ and ending at 11π/12 + kπ.So, I think the answer is x ∈ (5π/12 + kπ, 11π/12 + kπ) for any integer k.But in the problem, it's written as "the interval," so maybe they expect one specific interval, but since the function is periodic, it's decreasing on each of these intervals. So, I think the answer is all intervals of the form (5π/12 + kπ, 11π/12 + kπ), k ∈ ℤ.Alright, moving on to part (2). Given that f(θ/2) = -3/5, and θ ∈ [-π/6, 5π/6], find sin(θ - 5π/6).First, let's recall that f(x) = sin(2x - π/3). So, f(θ/2) = sin(2*(θ/2) - π/3) = sin(θ - π/3).So, sin(θ - π/3) = -3/5.We need to find sin(θ - 5π/6). Let's express θ - 5π/6 in terms of θ - π/3.θ - 5π/6 = (θ - π/3) - π/2.So, sin(θ - 5π/6) = sin[(θ - π/3) - π/2].Using the sine subtraction formula: sin(A - B) = sin A cos B - cos A sin B.So, sin[(θ - π/3) - π/2] = sin(θ - π/3) cos(π/2) - cos(θ - π/3) sin(π/2).We know that cos(π/2) = 0 and sin(π/2) = 1, so this simplifies to:sin(θ - π/3)*0 - cos(θ - π/3)*1 = -cos(θ - π/3).Therefore, sin(θ - 5π/6) = -cos(θ - π/3).So, we need to find cos(θ - π/3). We know that sin(θ - π/3) = -3/5, and θ ∈ [-π/6, 5π/6]. Let's find the range of θ - π/3.θ ∈ [-π/6, 5π/6], so θ - π/3 ∈ [-π/6 - π/3, 5π/6 - π/3] = [-π/2, π/2].So, θ - π/3 is in the interval [-π/2, π/2], which is the range where sine and cosine are defined, and cosine is non-negative in this interval because from -π/2 to π/2, cosine is positive.Therefore, cos(θ - π/3) is positive, and since sin(θ - π/3) = -3/5, we can find cos(θ - π/3) using the Pythagorean identity:sin²x + cos²x = 1.So, cos²(θ - π/3) = 1 - sin²(θ - π/3) = 1 - (9/25) = 16/25.Therefore, cos(θ - π/3) = 4/5 (since it's positive in this interval).Therefore, sin(θ - 5π/6) = -cos(θ - π/3) = -4/5.So, the value is -4/5.Wait, let me double-check. We had sin(θ - π/3) = -3/5, and θ - π/3 is in [-π/2, π/2], so θ - π/3 is in the fourth quadrant (since sine is negative and cosine is positive). Therefore, cos(θ - π/3) is indeed 4/5, so sin(θ - 5π/6) = -4/5.Yes, that seems correct.So, summarizing:(1) The function f(x) is monotonically decreasing on intervals (5π/12 + kπ, 11π/12 + kπ) for any integer k.(2) sin(θ - 5π/6) = -4/5.I think that's the solution.

answer：Okay, so I have this problem about two similar right triangles. The smaller one has an area of 10 square inches and a hypotenuse of 10 inches. The larger one has an area of 250 square inches. I need to find the sum of the lengths of the legs of the larger triangle.First, I remember that similar triangles have their sides in proportion. So, if I can find the scale factor between the two triangles, I can use that to find the sides of the larger triangle.But before that, I need to figure out the sides of the smaller triangle. It's a right triangle with area 10 and hypotenuse 10. Let me denote the legs as 'a' and 'b'. Since it's a right triangle, the area is (1/2)*a*b = 10. So, a*b = 20.Also, the hypotenuse is 10, so by Pythagoras, a^2 + b^2 = 10^2 = 100.Now, I have two equations:1. a*b = 202. a^2 + b^2 = 100I need to solve these to find 'a' and 'b'. Maybe I can express one variable in terms of the other from the first equation and substitute into the second.From the first equation, b = 20/a. Substitute into the second equation:a^2 + (20/a)^2 = 100Simplify:a^2 + 400/a^2 = 100Multiply both sides by a^2 to eliminate the denominator:a^4 + 400 = 100a^2Bring all terms to one side:a^4 - 100a^2 + 400 = 0This looks like a quadratic in terms of a^2. Let me set x = a^2:x^2 - 100x + 400 = 0Use the quadratic formula:x = [100 ± sqrt(10000 - 1600)] / 2x = [100 ± sqrt(8400)] / 2sqrt(8400) = sqrt(100*84) = 10*sqrt(84) = 10*sqrt(4*21) = 20*sqrt(21)So,x = [100 ± 20*sqrt(21)] / 2x = 50 ± 10*sqrt(21)Since x = a^2, it must be positive. So both solutions are positive, but since a < b, I think we take the smaller x for a^2.So, a^2 = 50 - 10*sqrt(21)and b^2 = 50 + 10*sqrt(21)Therefore, a = sqrt(50 - 10*sqrt(21)) and b = sqrt(50 + 10*sqrt(21))Hmm, that seems complicated. Maybe there's a simpler way or perhaps the triangle is a special one.Wait, maybe I made a mistake earlier. Let me check.I had a^4 - 100a^2 + 400 = 0Let me try factoring this equation. Maybe it factors into (a^2 - something)(a^2 - something else) = 0.Looking for two numbers that multiply to 400 and add up to -100. Hmm, 80 and 20. Because 80 + 20 = 100, but with negative signs, it would be -80 and -20.So, (a^2 - 80)(a^2 - 20) = 0Yes, that works because (a^2 - 80)(a^2 - 20) = a^4 - 20a^2 -80a^2 + 1600 = a^4 -100a^2 +1600, but wait, that's not matching the original equation which is a^4 -100a^2 +400=0.Oh, so my factoring was incorrect. Maybe I need to adjust.Alternatively, perhaps I can factor it as (a^2 - something)^2 = 0, but that doesn't seem to fit.Wait, maybe I made a mistake in the substitution earlier.Let me go back.We had a*b = 20 and a^2 + b^2 = 100.I can also recall that (a + b)^2 = a^2 + 2ab + b^2. So, (a + b)^2 = 100 + 40 = 140. Therefore, a + b = sqrt(140) = 2*sqrt(35). But I don't know if that helps directly.Alternatively, maybe I can use the fact that for a right triangle, the legs can be expressed in terms of the hypotenuse and the angles. But without knowing the angles, that might not help.Wait, maybe I can assume that the triangle is similar to a 6-8-10 triangle, which is a common right triangle. Let me check.In a 6-8-10 triangle, the area is (1/2)*6*8 = 24, but our smaller triangle has an area of 10, which is less. So, it's not a 6-8-10 triangle.Alternatively, maybe it's a scaled-down version. Let me see.If the hypotenuse is 10, and the area is 10, perhaps the legs are shorter than 6 and 8.Wait, let me think differently. Maybe I can use the ratio of areas to find the scale factor.The areas are 10 and 250, so the ratio is 250/10 = 25. Since the ratio of areas is the square of the scale factor, the scale factor is sqrt(25) = 5.So, the larger triangle is 5 times bigger in each side than the smaller one.Therefore, if I can find the legs of the smaller triangle, I can multiply each by 5 to get the legs of the larger triangle, and then sum them up.But to do that, I need to find the legs of the smaller triangle.Wait, earlier I tried solving and got a^2 = 50 ± 10*sqrt(21). Let me compute that numerically to see what the legs are.Compute sqrt(21) ≈ 4.5837So,a^2 = 50 - 10*4.5837 ≈ 50 - 45.837 ≈ 4.163So, a ≈ sqrt(4.163) ≈ 2.04Similarly,b^2 = 50 + 10*4.5837 ≈ 50 + 45.837 ≈ 95.837So, b ≈ sqrt(95.837) ≈ 9.79Wait, but if a ≈ 2.04 and b ≈ 9.79, then a^2 + b^2 ≈ 4.16 + 95.84 ≈ 100, which checks out.And a*b ≈ 2.04*9.79 ≈ 20, which also checks out.So, the legs of the smaller triangle are approximately 2.04 and 9.79 inches.But I need exact values, not approximate.Wait, earlier I had a^2 = 50 - 10*sqrt(21) and b^2 = 50 + 10*sqrt(21). So,a = sqrt(50 - 10*sqrt(21)) and b = sqrt(50 + 10*sqrt(21))But these expressions can be simplified.Let me see if sqrt(50 ± 10*sqrt(21)) can be expressed as sqrt(x) ± sqrt(y).Assume sqrt(50 ± 10*sqrt(21)) = sqrt(x) ± sqrt(y)Then, squaring both sides:50 ± 10*sqrt(21) = x + y ± 2*sqrt(x*y)So, we have:x + y = 50and2*sqrt(x*y) = 10*sqrt(21)Divide both sides by 2:sqrt(x*y) = 5*sqrt(21)Square both sides:x*y = 25*21 = 525So, we have:x + y = 50x*y = 525We can solve for x and y.Let me set up the quadratic equation:t^2 - 50t + 525 = 0Using quadratic formula:t = [50 ± sqrt(2500 - 2100)] / 2t = [50 ± sqrt(400)] / 2t = [50 ± 20] / 2So,t = (50 + 20)/2 = 70/2 = 35ort = (50 - 20)/2 = 30/2 = 15So, x = 35 and y = 15, or vice versa.Therefore,sqrt(50 ± 10*sqrt(21)) = sqrt(35) ± sqrt(15)But we need to check the sign.Since a < b, and a = sqrt(50 - 10*sqrt(21)) ≈ 2.04, and b = sqrt(50 + 10*sqrt(21)) ≈ 9.79, we can write:a = sqrt(35) - sqrt(15)andb = sqrt(35) + sqrt(15)Because:(sqrt(35) - sqrt(15))^2 = 35 + 15 - 2*sqrt(35*15) = 50 - 2*sqrt(525) = 50 - 2*sqrt(25*21) = 50 - 10*sqrt(21)Similarly,(sqrt(35) + sqrt(15))^2 = 35 + 15 + 2*sqrt(525) = 50 + 10*sqrt(21)Perfect! So, the legs of the smaller triangle are sqrt(35) - sqrt(15) and sqrt(35) + sqrt(15).Now, since the scale factor is 5, the legs of the larger triangle are 5*(sqrt(35) - sqrt(15)) and 5*(sqrt(35) + sqrt(15)).Therefore, the sum of the legs of the larger triangle is:5*(sqrt(35) - sqrt(15)) + 5*(sqrt(35) + sqrt(15)) = 5*sqrt(35) - 5*sqrt(15) + 5*sqrt(35) + 5*sqrt(15) = 10*sqrt(35)Wait, that simplifies nicely. The -5*sqrt(15) and +5*sqrt(15) cancel out, leaving 10*sqrt(35).But wait, earlier I thought the sum was 30*sqrt(5). Did I make a mistake?Wait, let me check my steps.I had a = sqrt(35) - sqrt(15) and b = sqrt(35) + sqrt(15). So, a + b = 2*sqrt(35). Then, scaling by 5, the sum becomes 10*sqrt(35).But earlier, I thought the sum was 30*sqrt(5). Hmm, which one is correct?Wait, let's compute sqrt(35) and sqrt(5):sqrt(35) ≈ 5.916sqrt(5) ≈ 2.236So, 10*sqrt(35) ≈ 59.16And 30*sqrt(5) ≈ 67.08But from the approximate values earlier, the legs of the smaller triangle were approximately 2.04 and 9.79, summing to about 11.83. Scaling by 5 gives approximately 59.15, which matches 10*sqrt(35).But wait, earlier I thought the legs were 2*sqrt(5) and 4*sqrt(5), summing to 6*sqrt(5), and scaling by 5 gives 30*sqrt(5). But that was based on incorrect assumption that the triangle was 6-8-10 scaled down, which it's not.So, the correct approach is the one where I solved for a and b exactly, leading to the sum of the legs of the larger triangle being 10*sqrt(35).But wait, let me double-check the area.If the legs of the smaller triangle are sqrt(35) - sqrt(15) and sqrt(35) + sqrt(15), then their product is:(sqrt(35) - sqrt(15))(sqrt(35) + sqrt(15)) = 35 - 15 = 20, which matches a*b = 20.And their squares sum to 50 -10*sqrt(21) + 50 +10*sqrt(21) = 100, which matches a^2 + b^2 = 100.So, that's correct.Therefore, the sum of the legs of the larger triangle is 10*sqrt(35).But wait, the problem asks for the sum of the legs of the larger triangle, which is 5*a + 5*b = 5*(a + b). Since a + b = 2*sqrt(35), then 5*(a + b) = 10*sqrt(35).Alternatively, since the scale factor is 5, the sum of the legs scales by 5 as well.But earlier, I thought the sum was 30*sqrt(5). Let me see where that came from.I think I made a mistake in assuming the legs were 2*sqrt(5) and 4*sqrt(5), which would give a hypotenuse of sqrt( (2√5)^2 + (4√5)^2 ) = sqrt(20 + 80) = sqrt(100) = 10, which is correct. And the area would be (1/2)*(2√5)*(4√5) = (1/2)*8*5 = 20, but the area is supposed to be 10. So, that's double the area. Therefore, those legs would actually correspond to a triangle with area 20, not 10. So, my initial assumption was wrong.Therefore, the correct legs are sqrt(35) - sqrt(15) and sqrt(35) + sqrt(15), leading to the sum of the legs of the larger triangle being 10*sqrt(35).But wait, 10*sqrt(35) is approximately 59.16, which seems quite large. Let me check the scale factor again.The area ratio is 250/10 = 25, so the scale factor is sqrt(25) = 5. So, each side is multiplied by 5.If the smaller triangle has legs approximately 2.04 and 9.79, then the larger triangle has legs approximately 10.2 and 48.95, summing to approximately 59.15, which matches 10*sqrt(35).But let me see if there's a simpler exact form.Wait, sqrt(35) is already simplified, so 10*sqrt(35) is the exact value.Alternatively, maybe I can express it in terms of sqrt(5).But 35 = 7*5, so sqrt(35) = sqrt(7*5) = sqrt(7)*sqrt(5). So, 10*sqrt(35) = 10*sqrt(7)*sqrt(5) = 10*sqrt(5)*sqrt(7). But that doesn't simplify further.Alternatively, maybe I made a mistake in the earlier steps.Wait, let me go back to the beginning.We have two similar right triangles. The smaller has area 10 and hypotenuse 10. The larger has area 250.Since they are similar, the ratio of their areas is the square of the ratio of their corresponding sides. So, ratio of areas = 250/10 = 25, so ratio of sides = 5.Therefore, each side of the larger triangle is 5 times the corresponding side of the smaller triangle.So, if I can find the legs of the smaller triangle, multiply each by 5, and sum them.So, the key is to find the legs of the smaller triangle.Given that the smaller triangle is a right triangle with area 10 and hypotenuse 10.Let me denote the legs as a and b.So,(1/2)*a*b = 10 => a*b = 20anda^2 + b^2 = 100We can solve these equations.Let me express b = 20/a and substitute into the second equation:a^2 + (20/a)^2 = 100Multiply both sides by a^2:a^4 + 400 = 100a^2Rearrange:a^4 - 100a^2 + 400 = 0Let x = a^2:x^2 - 100x + 400 = 0Using quadratic formula:x = [100 ± sqrt(10000 - 1600)] / 2x = [100 ± sqrt(8400)] / 2sqrt(8400) = sqrt(100*84) = 10*sqrt(84) = 10*sqrt(4*21) = 20*sqrt(21)So,x = [100 ± 20*sqrt(21)] / 2x = 50 ± 10*sqrt(21)So, a^2 = 50 - 10*sqrt(21) and b^2 = 50 + 10*sqrt(21)Therefore, a = sqrt(50 - 10*sqrt(21)) and b = sqrt(50 + 10*sqrt(21))As before, we can express sqrt(50 ± 10*sqrt(21)) as sqrt(35) ± sqrt(15), as shown earlier.So, a = sqrt(35) - sqrt(15) and b = sqrt(35) + sqrt(15)Therefore, the legs of the smaller triangle are sqrt(35) - sqrt(15) and sqrt(35) + sqrt(15)Sum of legs of smaller triangle: (sqrt(35) - sqrt(15)) + (sqrt(35) + sqrt(15)) = 2*sqrt(35)Therefore, sum of legs of larger triangle: 5*(2*sqrt(35)) = 10*sqrt(35)So, the sum is 10*sqrt(35)But wait, earlier I thought it was 30*sqrt(5). Let me see if 10*sqrt(35) can be expressed as 30*sqrt(5).Compute 10*sqrt(35) = 10*sqrt(5*7) = 10*sqrt(5)*sqrt(7) ≈ 10*2.236*2.6458 ≈ 59.16And 30*sqrt(5) ≈ 30*2.236 ≈ 67.08These are different values, so they can't be equal. Therefore, 10*sqrt(35) is the correct answer.Wait, but let me check if I can express 10*sqrt(35) in terms of sqrt(5). Since 35 = 5*7, it's already in simplest form.Therefore, the sum of the legs of the larger triangle is 10*sqrt(35) inches.But wait, let me think again. Maybe there's a simpler way.Alternatively, since the triangles are similar, the ratio of their areas is 25, so the ratio of their corresponding linear measurements is 5.Therefore, if I can find the sum of the legs of the smaller triangle, multiply by 5 to get the sum for the larger.Sum of legs of smaller triangle: a + bWe have a*b = 20 and a^2 + b^2 = 100We can find (a + b)^2 = a^2 + 2ab + b^2 = 100 + 40 = 140Therefore, a + b = sqrt(140) = 2*sqrt(35)Therefore, sum of legs of larger triangle = 5*(a + b) = 5*2*sqrt(35) = 10*sqrt(35)Yes, that's consistent with earlier.So, the answer is 10*sqrt(35)But wait, let me check if 10*sqrt(35) can be simplified further or expressed differently.Since 35 is 5*7, and both are primes, sqrt(35) is already simplified.Therefore, the sum is 10*sqrt(35) inches.But wait, earlier I thought the answer was 30*sqrt(5). Let me see why.If I mistakenly thought the legs were 2*sqrt(5) and 4*sqrt(5), their sum would be 6*sqrt(5), and scaling by 5 gives 30*sqrt(5). But that was based on incorrect legs, as that would give an area of 20, not 10.Therefore, the correct answer is 10*sqrt(35)But let me check numerically:sqrt(35) ≈ 5.91610*sqrt(35) ≈ 59.16Earlier, I had the legs of the smaller triangle as approximately 2.04 and 9.79, summing to approximately 11.83. Scaling by 5 gives approximately 59.15, which matches.Therefore, the sum of the legs of the larger triangle is 10*sqrt(35) inches.But wait, let me see if the problem expects the answer in a different form.Alternatively, maybe I can rationalize or present it differently, but I think 10*sqrt(35) is the simplest exact form.Therefore, the final answer is 10*sqrt(35) inches.