💡Okay, so I have this problem: I need to figure out how many ways I can choose two distinct squares from an 8x8 chessboard such that the midpoint of the line segment connecting their centers is also the center of another square on the board. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I should visualize the chessboard. It's an 8x8 grid, so there are 64 squares in total. Each square can be identified by its position, like (1,1) for the bottom-left corner and (8,8) for the top-right corner. But wait, the problem mentions the centers of the squares. So, maybe it's better to think of each square as having coordinates that are the centers of these squares.If I consider the chessboard as a grid, each square's center can be represented by coordinates (x, y), where x and y range from 0.5 to 8.5 in increments of 1. For example, the center of the first square would be (0.5, 0.5), the next one (1.5, 0.5), and so on up to (8.5, 8.5). But maybe working with integers would be easier. If I shift the coordinates so that the center of the bottom-left square is at (0,0), then each square's center can be represented as (a, b) where a and b are integers from 0 to 7. That might make calculations simpler.So, if I have two squares with centers at (a, b) and (c, d), the midpoint of the segment connecting them would be at ((a + c)/2, (b + d)/2). For this midpoint to also be the center of another square on the board, both (a + c)/2 and (b + d)/2 must be integers. That means (a + c) and (b + d) must both be even numbers. Ah, so for the sums a + c and b + d to be even, a and c must be both even or both odd, and similarly for b and d. That makes sense because even + even = even and odd + odd = even, while even + odd = odd. So, the coordinates of the two squares must have the same parity in both the x and y directions.Now, how can I count the number of such pairs? Maybe I can categorize the squares based on the parity of their coordinates. Each square's center has coordinates (a, b), where a and b can be either even or odd. So, there are four possible categories:1. Both a and b are even.2. a is even, and b is odd.3. a is odd, and b is even.4. Both a and b are odd.Each of these categories will contain a certain number of squares. Let me figure out how many squares are in each category.Since a and b can each be 0 to 7, inclusive, let's count how many even and odd numbers there are in this range. From 0 to 7, there are four even numbers (0, 2, 4, 6) and four odd numbers (1, 3, 5, 7). So, for each coordinate (a or b), there are four even and four odd possibilities.Therefore, for each category:1. Both a and b even: 4 choices for a, 4 choices for b, so 4*4 = 16 squares.2. a even, b odd: 4*4 = 16 squares.3. a odd, b even: 4*4 = 16 squares.4. Both a and b odd: 4*4 = 16 squares.So, each category has 16 squares. That's good because it means the distribution is even across all categories.Now, if I want to choose two squares such that their midpoint is also a center of a square, I need to choose two squares from the same category. Because only then will both a + c and b + d be even, ensuring the midpoint is an integer coordinate.Therefore, the number of valid pairs is the number of ways to choose two squares from each category, multiplied by the number of categories.The number of ways to choose two squares from a category with 16 squares is given by the combination formula C(n, 2) = n(n - 1)/2. So, for each category, it's C(16, 2) = 16*15/2 = 120.Since there are four categories, each contributing 120 pairs, the total number of valid pairs is 4*120 = 480.Wait, let me double-check that. So, each category has 16 squares, and within each category, the number of pairs is 120. Since there are four such categories, multiplying 120 by 4 gives 480. That seems right.Is there another way to think about this? Maybe considering the entire chessboard and the total number of pairs, then figuring out the fraction that satisfies the midpoint condition.The total number of ways to choose two distinct squares from the chessboard is C(64, 2) = 64*63/2 = 2016. So, there are 2016 total pairs.Out of these, how many satisfy the condition that the midpoint is also a center of a square? We just calculated it as 480. So, 480/2016 simplifies to 5/21. That seems reasonable, but I need to make sure my initial reasoning was correct.Another approach: For the midpoint to be a center, the two squares must be symmetric with respect to some center square. So, for each center square, how many pairs of squares are symmetric about it?Wait, that might complicate things because each pair is counted multiple times depending on how many centers they are symmetric about. Maybe not the best approach.Alternatively, thinking in terms of vectors: The vector from one square to another must be such that it's twice another vector, meaning the difference in coordinates must be even in both x and y directions. So, the displacement between the two squares must be even in both coordinates. That would mean that the two squares are in the same parity category, which aligns with my earlier reasoning.So, yeah, I think my initial calculation is correct. Each category has 16 squares, each contributing 120 pairs, four categories give 480 pairs.Just to ensure I haven't missed anything, let me think about specific examples.Take two squares in the same category, say both with even coordinates. Let's pick (0,0) and (2,2). The midpoint is (1,1), which is the center of another square. Similarly, (0,2) and (2,6) would have a midpoint at (1,4), which is also a center. So, that works.What if I pick two squares from different categories? For example, (0,0) from category 1 and (1,1) from category 4. The midpoint would be (0.5, 0.5), which is not a center of a square on the board. So, that pair doesn't satisfy the condition. Hence, only pairs within the same category work.Therefore, I can be confident that the number of valid pairs is 480.**Final Answer**The number of ways is boxed{480}.

💡Okay, so I need to figure out how many trays Roger can carry at one time. Let's see, the problem says he had to pick up 10 trays from one table and 2 trays from another. So, in total, he needs to pick up 10 + 2 = 12 trays. He made 3 trips to carry all these trays. So, if he made 3 trips, and he carried some number of trays each trip, then the total number of trays he carried is 3 times the number of trays he can carry in one trip. Let me call the number of trays he can carry at a time "T". So, the total number of trays he carried is 3T. But we know that the total number of trays he needed to carry was 12. So, 3T should equal 12. To find T, I can divide both sides of the equation by 3. So, T = 12 / 3 = 4. Therefore, Roger can carry 4 trays at a time.

💡Okay, so I have this problem where I need to find the symmetric point of the point (2, 3, 4) with respect to the xOy plane. Hmm, I'm not entirely sure what that means, but I think it has something to do with reflecting the point across a plane in three-dimensional space. Let me try to break it down.First, I remember that in three dimensions, there are three coordinate planes: xOy, xOz, and yOz. Each of these planes is defined by two axes. For example, the xOy plane is defined by the x-axis and y-axis, and it's essentially the horizontal plane where z is zero. So, reflecting a point across the xOy plane would mean changing its position relative to this plane.I think reflection across a plane involves changing the sign of the coordinate that is perpendicular to that plane. Since the xOy plane is defined by x and y, the z-coordinate is perpendicular to it. That means if I reflect the point (2, 3, 4) across the xOy plane, I should change the sign of the z-coordinate.Let me test this idea. If I have a point (x, y, z), reflecting it across the xOy plane should give me (x, y, -z). Applying that to my point (2, 3, 4), it would become (2, 3, -4). Does that make sense?Wait, let me visualize it. Imagine the xOy plane as the floor. If I have a point above the floor at (2, 3, 4), its reflection should be directly below the floor at the same distance. So, if the original z-coordinate is positive 4, the reflection should have a z-coordinate of negative 4. That seems right.But I should make sure I'm not confusing this with reflections across other planes. For example, reflecting across the xOz plane would involve changing the y-coordinate, and reflecting across the yOz plane would involve changing the x-coordinate. So, in this case, since we're dealing with the xOy plane, only the z-coordinate changes.Let me think of another example to confirm. If I have a point (1, 2, 3), reflecting it across the xOy plane should give me (1, 2, -3). Yes, that follows the same logic. The x and y coordinates remain the same, and the z-coordinate is inverted.Is there a formula or a method to find the reflection of a point across a plane in general? I recall that for reflection across a plane, you can use the formula involving the normal vector of the plane. But in this case, since the xOy plane is one of the coordinate planes, it's simpler because the normal vector is along the z-axis.So, for the xOy plane, the normal vector is (0, 0, 1). The reflection formula across a plane with normal vector n and a point P can be given by:P' = P - 2 * ((P · n) / ||n||²) * nBut since n is (0, 0, 1) and ||n||² is 1, this simplifies to:P' = (x, y, z) - 2 * z * (0, 0, 1) = (x, y, z - 2z) = (x, y, -z)Which confirms what I thought earlier. So, applying this to (2, 3, 4), we get (2, 3, -4).Just to be thorough, let me consider if there's any other aspect I might be missing. For instance, what if the plane wasn't one of the coordinate planes? Then, the reflection would be more complicated, involving more steps and possibly solving equations. But since it's the xOy plane, it's straightforward.Another way to think about it is geometrically. The xOy plane divides the space into two halves: the upper half where z is positive and the lower half where z is negative. Reflecting a point across this plane means moving it to the opposite half at the same distance from the plane. So, if the original point is 4 units above the xOy plane, the reflected point should be 4 units below it, which again gives us z = -4.I also remember that reflecting a point across a plane results in the plane being the perpendicular bisector of the segment joining the original point and its reflection. So, the midpoint between (2, 3, 4) and its reflection should lie on the xOy plane. Let's check that.The midpoint M between (2, 3, 4) and (2, 3, -4) is:M = ((2 + 2)/2, (3 + 3)/2, (4 + (-4))/2) = (2, 3, 0)Which indeed lies on the xOy plane since z = 0. That confirms that the reflection is correct.Is there any chance I could have mixed up the coordinates? For example, changing the x or y instead of z? I don't think so because the xOy plane is defined by x and y, so the reflection should only affect the coordinate perpendicular to it, which is z.Let me also recall that in two dimensions, reflecting a point across the x-axis changes the y-coordinate's sign, and reflecting across the y-axis changes the x-coordinate's sign. Extending this to three dimensions, reflecting across the xOy plane (which is like the "floor" in 3D space) would change the z-coordinate's sign, similar to how reflecting across the x-axis changes y in 2D.So, putting it all together, the symmetric point of (2, 3, 4) with respect to the xOy plane is (2, 3, -4). I think that's the correct answer.

💡Okay, so I have this problem where I need to find the value of ( a + b + c ) given that ( a ), ( b ), and ( c ) are positive integers satisfying the following equations:[ab + c + 10 = 51][bc + a + 10 = 51][ac + b + 10 = 51]Hmm, let me start by simplifying each equation. If I subtract 10 from both sides of each equation, I get:1. ( ab + c = 41 )2. ( bc + a = 41 )3. ( ac + b = 41 )So now I have three equations:1. ( ab + c = 41 )2. ( bc + a = 41 )3. ( ac + b = 41 )I need to find positive integers ( a ), ( b ), and ( c ) that satisfy all three equations. Let me see how I can approach this.First, maybe I can subtract some equations to eliminate variables. Let's subtract the first equation from the second:[(bc + a) - (ab + c) = 41 - 41][bc + a - ab - c = 0][bc - ab + a - c = 0]Factor terms:[b(c - a) + (a - c) = 0][(b - 1)(c - a) = 0]So, either ( b - 1 = 0 ) or ( c - a = 0 ). That means either ( b = 1 ) or ( c = a ).Similarly, let's subtract the second equation from the third:[(ac + b) - (bc + a) = 41 - 41][ac + b - bc - a = 0][ac - bc + b - a = 0]Factor terms:[c(a - b) + (b - a) = 0][(c - 1)(a - b) = 0]So, either ( c - 1 = 0 ) or ( a - b = 0 ). That means either ( c = 1 ) or ( a = b ).And subtracting the first equation from the third:[(ac + b) - (ab + c) = 41 - 41][ac + b - ab - c = 0][ac - ab + b - c = 0]Factor terms:[a(c - b) + (b - c) = 0][(a - 1)(c - b) = 0]So, either ( a - 1 = 0 ) or ( c - b = 0 ). That means either ( a = 1 ) or ( c = b ).Putting all these together, we have several possibilities:1. From the first subtraction: ( b = 1 ) or ( c = a ).2. From the second subtraction: ( c = 1 ) or ( a = b ).3. From the third subtraction: ( a = 1 ) or ( c = b ).So, let's consider the cases.**Case 1: ( a = 1 )**If ( a = 1 ), let's substitute into the first equation:[1 cdot b + c = 41 implies b + c = 41]Now, substitute ( a = 1 ) into the second equation:[b cdot c + 1 = 41 implies bc = 40]So, we have ( b + c = 41 ) and ( bc = 40 ). Let's solve for ( b ) and ( c ).From ( b + c = 41 ), we can express ( c = 41 - b ). Substitute into ( bc = 40 ):[b(41 - b) = 40][41b - b^2 = 40][b^2 - 41b + 40 = 0]Let's solve this quadratic equation:[b = frac{41 pm sqrt{1681 - 160}}{2} = frac{41 pm sqrt{1521}}{2} = frac{41 pm 39}{2}]So, ( b = frac{41 + 39}{2} = 40 ) or ( b = frac{41 - 39}{2} = 1 ).If ( b = 40 ), then ( c = 41 - 40 = 1 ).If ( b = 1 ), then ( c = 41 - 1 = 40 ).So, we have two possibilities:1. ( a = 1 ), ( b = 40 ), ( c = 1 )2. ( a = 1 ), ( b = 1 ), ( c = 40 )Let's check these in the third equation ( ac + b = 41 ):For the first possibility: ( 1 cdot 1 + 40 = 1 + 40 = 41 ). That works.For the second possibility: ( 1 cdot 40 + 1 = 40 + 1 = 41 ). That also works.So, both are valid solutions. Therefore, in this case, ( a + b + c ) can be either ( 1 + 40 + 1 = 42 ) or ( 1 + 1 + 40 = 42 ). Wait, both give the same sum. Hmm, interesting.But let me check if these satisfy all original equations.First possibility: ( a = 1 ), ( b = 40 ), ( c = 1 )Check ( ab + c + 10 = 1 cdot 40 + 1 + 10 = 40 + 1 + 10 = 51 ). Good.Check ( bc + a + 10 = 40 cdot 1 + 1 + 10 = 40 + 1 + 10 = 51 ). Good.Check ( ac + b + 10 = 1 cdot 1 + 40 + 10 = 1 + 40 + 10 = 51 ). Good.Second possibility: ( a = 1 ), ( b = 1 ), ( c = 40 )Check ( ab + c + 10 = 1 cdot 1 + 40 + 10 = 1 + 40 + 10 = 51 ). Good.Check ( bc + a + 10 = 1 cdot 40 + 1 + 10 = 40 + 1 + 10 = 51 ). Good.Check ( ac + b + 10 = 1 cdot 40 + 1 + 10 = 40 + 1 + 10 = 51 ). Good.So both are valid. So, in this case, ( a + b + c = 42 ).But wait, the problem says "positive integers", so both are acceptable. Hmm.But let me check other cases as well.**Case 2: ( b = 1 )**If ( b = 1 ), substitute into the first equation:[a cdot 1 + c = 41 implies a + c = 41]Substitute ( b = 1 ) into the second equation:[1 cdot c + a = 41 implies c + a = 41]Same as the first equation, so no new information.Substitute ( b = 1 ) into the third equation:[a cdot c + 1 = 41 implies ac = 40]So, similar to Case 1, we have ( a + c = 41 ) and ( ac = 40 ). Solving this, same as before:( c = 41 - a ), so:[a(41 - a) = 40][41a - a^2 = 40][a^2 - 41a + 40 = 0]Which leads to ( a = 40 ) or ( a = 1 ), hence ( c = 1 ) or ( c = 40 ).So, the solutions are:1. ( a = 40 ), ( b = 1 ), ( c = 1 )2. ( a = 1 ), ( b = 1 ), ( c = 40 )Same as in Case 1, leading to ( a + b + c = 42 ).**Case 3: ( c = a )**From the first subtraction, if ( c = a ), let's substitute into the equations.First equation: ( ab + c = 41 ) becomes ( ab + a = 41 implies a(b + 1) = 41 ).Since 41 is a prime number, the factors are 1 and 41. So, possible cases:1. ( a = 1 ), ( b + 1 = 41 implies b = 40 )2. ( a = 41 ), ( b + 1 = 1 implies b = 0 ). But ( b ) must be positive, so discard this.So, ( a = 1 ), ( b = 40 ), ( c = a = 1 ). Which is the same as in Case 1.**Case 4: ( a = b )**From the second subtraction, if ( a = b ), substitute into the equations.First equation: ( ab + c = 41 ) becomes ( a^2 + c = 41 ).Second equation: ( bc + a = 41 ) becomes ( ac + a = 41 implies a(c + 1) = 41 ).Third equation: ( ac + b = 41 ) becomes ( ac + a = 41 implies a(c + 1) = 41 ).So, from the second and third equations, we have ( a(c + 1) = 41 ). Since 41 is prime, ( a ) must be 1 or 41.If ( a = 1 ), then ( c + 1 = 41 implies c = 40 ). Then, from the first equation, ( 1^2 + 40 = 41 ), which is correct.If ( a = 41 ), then ( c + 1 = 1 implies c = 0 ), which is invalid since ( c ) must be positive.So, only solution is ( a = b = 1 ), ( c = 40 ). Which is again the same as in Case 1.**Case 5: ( c = b )**From the third subtraction, if ( c = b ), substitute into the equations.First equation: ( ab + c = 41 ) becomes ( ab + b = 41 implies b(a + 1) = 41 ).Again, since 41 is prime, possible cases:1. ( b = 1 ), ( a + 1 = 41 implies a = 40 )2. ( b = 41 ), ( a + 1 = 1 implies a = 0 ). Invalid.So, ( b = 1 ), ( a = 40 ), ( c = b = 1 ). Which is same as in Case 1.So, in all cases, the solutions are either ( (1, 40, 1) ) or ( (1, 1, 40) ), both leading to ( a + b + c = 42 ).Wait, but in the initial problem statement, the user had a different solution, where ( a = b = 1 ), ( c = 39 ), leading to ( a + b + c = 41 ). But in my solution, I get 42. Hmm, that's conflicting.Let me double-check my calculations.Wait, in the problem statement, the user wrote:"Given the modified equations: - ( ab + c + 10 = 51 ) - ( bc + a + 10 = 51 ) - ( ac + b + 10 = 51 )Start by simplifying these to: - ( ab + c = 41 ) - ( bc + a = 41 ) - ( ac + b = 41 )Next, the first equality implies: - ( ab + c - bc - a = b(a - c) - (a - c) = 0 Rightarrow (b - 1)(a - c) = 0 )By symmetry, from the updated equations: - ( (b - 1)(a - c) = 0 ) - ( (c - 1)(b - a) = 0 ) - ( (a - 1)(c - b) = 0 )Inferring symmetric results, assume ( a = b ). Substitute into the first equation: - ( a^2 + c = 41 Rightarrow a^2 + c = ac + a ) - This simplifies to ( a^2 + c = a(c + 1) = 41 )Given ( 41 ) is prime and considering ( a, c ) are positive integers, possibilities are either ( a = 1 ) or ( a = 41 ). With ( a = 41 ), then ( c + 1 = 1 Rightarrow c = 0 )—impossible under positive integer constraints. It leaves ( a = b = 1 ), and substituting, we have ( 1 + 1 + c = 41 Rightarrow c = 39 ).Calculating ( a + b + c ):[ a + b + c = 1 + 1 + 39 = 41 ]Therefore, after modifications, ( a + b + c = boxed{41} );"Wait, in their solution, they assumed ( a = b ), leading to ( c = 39 ). But in my solution, I considered ( a = 1 ) and ( b = 40 ), leading to ( c = 1 ), and vice versa, giving ( a + b + c = 42 ).So, which one is correct?Wait, let me check the equations again.If ( a = 1 ), ( b = 40 ), ( c = 1 ):Check all three original equations:1. ( ab + c + 10 = 1*40 + 1 + 10 = 40 + 1 + 10 = 51 ). Correct.2. ( bc + a + 10 = 40*1 + 1 + 10 = 40 + 1 + 10 = 51 ). Correct.3. ( ac + b + 10 = 1*1 + 40 + 10 = 1 + 40 + 10 = 51 ). Correct.Similarly, ( a = 1 ), ( b = 1 ), ( c = 40 ):1. ( ab + c + 10 = 1*1 + 40 + 10 = 1 + 40 + 10 = 51 ). Correct.2. ( bc + a + 10 = 1*40 + 1 + 10 = 40 + 1 + 10 = 51 ). Correct.3. ( ac + b + 10 = 1*40 + 1 + 10 = 40 + 1 + 10 = 51 ). Correct.So, both solutions are valid, giving ( a + b + c = 42 ).But in the user's solution, they got ( c = 39 ), which would give ( a + b + c = 41 ). Let me see how that happened.In their solution, they assumed ( a = b ), leading to ( a^2 + c = 41 ) and ( a(c + 1) = 41 ). Then, they concluded ( a = 1 ), ( c = 39 ). But wait, if ( a = 1 ), then ( a(c + 1) = 1*(39 + 1) = 40 ), which is not 41. Wait, that's a mistake.Wait, let me re-examine their steps:They wrote:"Substitute into the first equation: - ( a^2 + c = 41 Rightarrow a^2 + c = ac + a ) - This simplifies to ( a^2 + c = a(c + 1) = 41 )"Wait, that step seems off. Let me see:From ( a = b ), first equation: ( a^2 + c = 41 ).Second equation: ( bc + a = 41 ) becomes ( ac + a = 41 implies a(c + 1) = 41 ).So, from the second equation, ( a(c + 1) = 41 ).From the first equation, ( a^2 + c = 41 ).So, equate ( a^2 + c = a(c + 1) ):[a^2 + c = a c + a][a^2 - a c + c - a = 0][a(a - c) + (c - a) = 0][(a - 1)(a - c) = 0]So, either ( a = 1 ) or ( a = c ).If ( a = 1 ), then from ( a(c + 1) = 41 ), ( c + 1 = 41 implies c = 40 ). Then, from the first equation, ( 1^2 + 40 = 41 ). Correct.If ( a = c ), then from ( a(c + 1) = 41 ), ( a(a + 1) = 41 ). Since 41 is prime, ( a(a + 1) = 41 ) implies ( a = 40 ) and ( a + 1 = 1 ), which is impossible, or ( a = 1 ) and ( a + 1 = 41 ), which is ( a = 40 ). Wait, that doesn't make sense.Wait, if ( a = c ), then ( a(a + 1) = 41 ). Since 41 is prime, the only way is ( a = 1 ) and ( a + 1 = 41 ), but ( 1 + 1 = 2 neq 41 ). Alternatively, ( a = 40 ) and ( a + 1 = 1 ), which is impossible. So, only solution is ( a = 1 ), ( c = 40 ).Therefore, in this case, ( a = b = 1 ), ( c = 40 ), leading to ( a + b + c = 42 ).But the user's solution said ( c = 39 ). That must be an error. Because if ( a = 1 ), ( c = 40 ), not 39.So, in conclusion, the correct value of ( a + b + c ) is 42, not 41.But wait, in my initial solving, I got 42, but the user's solution got 41. So, perhaps the user made a mistake in their calculation.Alternatively, maybe I missed something. Let me check again.Wait, in the user's solution, they wrote:"Substitute into the first equation: - ( a^2 + c = 41 Rightarrow a^2 + c = ac + a ) - This simplifies to ( a^2 + c = a(c + 1) = 41 )"Wait, that step is confusing. Let me re-examine.From ( a = b ), first equation is ( a^2 + c = 41 ).Second equation is ( bc + a = 41 ), which becomes ( a c + a = 41 implies a(c + 1) = 41 ).So, they set ( a^2 + c = a(c + 1) ):[a^2 + c = a c + a][a^2 - a c + c - a = 0][a(a - c) + (c - a) = 0][(a - 1)(a - c) = 0]So, either ( a = 1 ) or ( a = c ).If ( a = 1 ), then ( c = 40 ).If ( a = c ), then ( a(a + 1) = 41 ). Since 41 is prime, ( a(a + 1) = 41 ) implies ( a = 40 ) and ( a + 1 = 1 ), which is impossible, or ( a = 1 ) and ( a + 1 = 41 ), which is ( a = 40 ). Wait, that's not correct.Wait, if ( a = c ), then ( a(a + 1) = 41 ). So, ( a^2 + a - 41 = 0 ). Solving this quadratic:[a = frac{-1 pm sqrt{1 + 164}}{2} = frac{-1 pm sqrt{165}}{2}]Which is not an integer. So, no solution here. Therefore, only ( a = 1 ), ( c = 40 ).Thus, ( a = b = 1 ), ( c = 40 ), leading to ( a + b + c = 42 ).Therefore, the correct answer should be 42, not 41. The user's solution had an error in their calculation when they concluded ( c = 39 ). It should be ( c = 40 ).So, in conclusion, after carefully solving, the value of ( a + b + c ) is 42.

💡Okay, so I need to find the value of sec(330 degrees). Hmm, I remember that secant is the reciprocal of cosine, so sec(θ) = 1/cos(θ). That means I can find sec(330°) by first finding cos(330°) and then taking its reciprocal.Alright, let me think about where 330 degrees is on the unit circle. I know that 360 degrees is a full circle, so 330 degrees is just 30 degrees less than 360. That places it in the fourth quadrant of the unit circle. In the fourth quadrant, cosine is positive, and sine is negative. So, cos(330°) should be positive.Now, I remember that for angles in the fourth quadrant, we can express them as 360° - θ, where θ is the reference angle. In this case, θ would be 30°, because 360° - 30° = 330°. So, cos(330°) is the same as cos(30°), but since cosine is positive in the fourth quadrant, it's just cos(30°).I think cos(30°) is one of those special angles. Let me recall, cos(30°) is √3/2. So, cos(330°) should also be √3/2.Now, going back to sec(330°), which is 1 divided by cos(330°). So, that would be 1 divided by (√3/2). Dividing by a fraction is the same as multiplying by its reciprocal, so 1 * (2/√3) equals 2/√3.But wait, usually, we rationalize the denominator when expressing such answers. To rationalize 2/√3, I can multiply both the numerator and the denominator by √3. That gives me (2√3)/(√3 * √3) which simplifies to (2√3)/3.So, putting it all together, sec(330°) is 2√3/3.Let me double-check my steps to make sure I didn't make any mistakes. I started by recognizing that secant is the reciprocal of cosine, located 330° in the fourth quadrant, determined the reference angle was 30°, found cos(30°) is √3/2, took the reciprocal to get sec(330°) as 2/√3, and then rationalized the denominator to 2√3/3. Everything seems to add up.Just to be thorough, I can also think about the coordinates on the unit circle for 330°. The coordinates should be (cos(330°), sin(330°)). For 30°, the coordinates are (√3/2, 1/2), so for 330°, which is in the fourth quadrant, the x-coordinate remains positive, and the y-coordinate becomes negative. So, cos(330°) is indeed √3/2, and sin(330°) is -1/2. Therefore, taking the reciprocal of √3/2 gives me 2/√3, which is the same as 2√3/3 after rationalizing.I think that's solid. I don't see any errors in my reasoning. So, I'm confident that sec(330°) is 2√3/3.**Final Answer**The value of sec(330^circ) is boxed{dfrac{2sqrt{3}}{3}}.

💡Okay, so I have this geometry problem here, and I need to figure out which statement is incorrect. Let me read it again carefully.In space, α represents a plane, and m, n represent two lines. Among the following statements, the incorrect one is (   )A: If m is parallel to α, and m, n are not parallel, then n is not parallel to α.B: If m is parallel to α, and m, n are not perpendicular, then n is not perpendicular to α.C: If m is perpendicular to α, and m, n are not parallel, then n is not perpendicular to α.D: If m is perpendicular to α, and m, n are not perpendicular, then n is not parallel to α.Alright, so I need to analyze each statement one by one and see which one is incorrect. Let me start with option A.**Option A**: If m is parallel to α, and m, n are not parallel, then n is not parallel to α.Hmm, okay. So, m is parallel to the plane α. That means m doesn't intersect α and lies outside of it. Now, m and n are not parallel. So, n could be in various positions relative to α. Let me visualize this.If m is parallel to α, then n could be parallel to α as well, even if m and n aren't parallel. For example, imagine two different lines on a different plane that's parallel to α. They don't have to be parallel to each other. So, n could still be parallel to α even if m and n aren't parallel. That means the conclusion that n is not parallel to α might not hold. So, maybe A is incorrect.But let me check the other options to be sure.**Option B**: If m is parallel to α, and m, n are not perpendicular, then n is not perpendicular to α.Alright, so m is parallel to α, and m and n aren't perpendicular. Does that mean n can't be perpendicular to α?Well, if m is parallel to α, then m lies in a direction that's parallel to the plane. If n were perpendicular to α, then n would be orthogonal to every line in α, including m. But if m and n aren't perpendicular, that would mean n can't be perpendicular to α because that would require them to be orthogonal. So, this seems correct.**Option C**: If m is perpendicular to α, and m, n are not parallel, then n is not perpendicular to α.Okay, m is perpendicular to α, so m is like a vertical line sticking out of the plane. If m and n are not parallel, does that mean n can't be perpendicular to α?Wait, if m is perpendicular to α, then any line perpendicular to α would have to be parallel to m, right? Because all lines perpendicular to α are parallel to each other. So, if m and n are not parallel, then n can't be perpendicular to α. That makes sense. So, C is correct.**Option D**: If m is perpendicular to α, and m, n are not perpendicular, then n is not parallel to α.Hmm, m is perpendicular to α, so m is like a vertical line. If m and n are not perpendicular, does that mean n can't be parallel to α?Wait, if n is parallel to α, then n lies within a plane parallel to α or is parallel to some line in α. But m is perpendicular to α, so m is perpendicular to every line in α. If n is parallel to α, then m would be perpendicular to n as well. But the statement says m and n are not perpendicular, so n can't be parallel to α. That seems correct.So, going back to option A. I think my initial thought was right. If m is parallel to α, and m and n are not parallel, n could still be parallel to α. For example, imagine two different lines on a different plane that's parallel to α. They don't have to be parallel to each other, but both can be parallel to α. So, the conclusion that n is not parallel to α isn't necessarily true. Therefore, statement A is incorrect.Let me just double-check my reasoning for A. If m is parallel to α, then m doesn't intersect α and lies outside. If n is also parallel to α, it can lie in a different position but still be parallel. Since m and n are not parallel, they can still both be parallel to α without being parallel to each other. So, yes, A is incorrect.**Final Answer**boxed{A}