💡Okay, so I have this trigonometry problem here: If (cos 60^{circ} = cos 45^{circ} cos theta) with (0^{circ} leq theta leq 90^{circ}), then (theta) equals(A) (0^{circ})(B) (15^{circ})(C) (30^{circ})(D) (45^{circ})(E) (60^{circ})Alright, let's try to figure this out step by step. I remember some basic cosine values, so maybe I can plug those in and solve for (theta).First, I know that (cos 60^{circ}) is a standard angle. From what I recall, (cos 60^{circ}) is equal to (frac{1}{2}). Let me double-check that. Yeah, in a 30-60-90 triangle, the sides are in the ratio 1:√3:2, so the cosine of 60 degrees, which is adjacent over hypotenuse, would be (frac{1}{2}). Got that down.Next, (cos 45^{circ}). Hmm, 45 degrees is a special angle too, part of the 45-45-90 triangle. In that case, both legs are equal, so the cosine of 45 degrees is (frac{sqrt{2}}{2}). Let me confirm that. Yes, in a 45-45-90 triangle, the sides are in the ratio 1:1:√2, so adjacent over hypotenuse is (frac{1}{sqrt{2}}), which rationalizes to (frac{sqrt{2}}{2}). Perfect.So now, plugging these values into the equation:[cos 60^{circ} = cos 45^{circ} cos theta]becomes:[frac{1}{2} = frac{sqrt{2}}{2} cos theta]Alright, now I need to solve for (cos theta). Let me rewrite the equation:[frac{1}{2} = frac{sqrt{2}}{2} cos theta]Hmm, both sides have a denominator of 2, so maybe I can multiply both sides by 2 to eliminate the denominators. Let's try that:[2 times frac{1}{2} = 2 times frac{sqrt{2}}{2} cos theta]Simplifying both sides:[1 = sqrt{2} cos theta]Okay, now I have:[1 = sqrt{2} cos theta]I need to solve for (cos theta), so I'll divide both sides by (sqrt{2}):[cos theta = frac{1}{sqrt{2}}]Wait, (frac{1}{sqrt{2}}) is the same as (frac{sqrt{2}}{2}), right? Because if I rationalize the denominator:[frac{1}{sqrt{2}} = frac{sqrt{2}}{2}]Yes, that's correct. So, (cos theta = frac{sqrt{2}}{2}).Now, I need to find the angle (theta) between 0° and 90° whose cosine is (frac{sqrt{2}}{2}). From what I remember, the cosine of 45° is (frac{sqrt{2}}{2}). Let me think about that. In a 45-45-90 triangle, both angles are 45°, and the sides are in the ratio 1:1:√2. So, the cosine, which is adjacent over hypotenuse, is (frac{1}{sqrt{2}}) or (frac{sqrt{2}}{2}). Yep, that's exactly it.So, (theta) must be 45°. Let me check the options:(A) (0^{circ}) – Cosine of 0° is 1, which is not equal to (frac{sqrt{2}}{2}).(B) (15^{circ}) – I don't remember the exact value, but it's less than 45°, so cosine would be larger than (frac{sqrt{2}}{2}), right? Wait, no, as the angle increases, cosine decreases. So, 15° is smaller than 45°, so its cosine would be larger than (frac{sqrt{2}}{2}). So, not 15°.(C) (30^{circ}) – Cosine of 30° is (frac{sqrt{3}}{2}), which is approximately 0.866, while (frac{sqrt{2}}{2}) is approximately 0.707. So, not 30°.(D) (45^{circ}) – As we established, cosine is (frac{sqrt{2}}{2}). That's exactly what we need.(E) (60^{circ}) – Cosine of 60° is (frac{1}{2}), which is 0.5, less than (frac{sqrt{2}}{2}). So, not 60°.Therefore, the correct answer is (D) (45^{circ}).Wait, just to make sure I didn't make any mistakes in my calculations. Let me go through the steps again.Starting with:[cos 60^{circ} = cos 45^{circ} cos theta]Substituting the known values:[frac{1}{2} = frac{sqrt{2}}{2} cos theta]Multiplying both sides by 2:[1 = sqrt{2} cos theta]Dividing both sides by (sqrt{2}):[cos theta = frac{1}{sqrt{2}} = frac{sqrt{2}}{2}]And since (cos 45^{circ} = frac{sqrt{2}}{2}), (theta = 45^{circ}). Yep, that all checks out.I think I'm confident with this answer. It makes sense because if you have the cosine of 60°, which is smaller than the cosine of 45°, you need a slightly larger angle to get a smaller cosine value. Wait, no, actually, as the angle increases, the cosine decreases. So, since (cos 60^{circ}) is smaller than (cos 45^{circ}), you need a larger angle to get a smaller cosine. But in this case, we're solving for (theta), and we found that (theta) is 45°, which is the same as the angle in the equation. Hmm, maybe I confused myself there. Let me think again.Wait, the equation is (cos 60^{circ} = cos 45^{circ} cos theta). So, (cos 60^{circ}) is smaller than (cos 45^{circ}), which means that (cos theta) must be less than 1. But since (cos 45^{circ}) is already (frac{sqrt{2}}{2}), multiplying it by (cos theta) gives a smaller value, which is (cos 60^{circ}). So, (cos theta) must be (frac{1}{sqrt{2}}), which corresponds to 45°. Yeah, that makes sense.I think I was just overcomplicating it for a second there. But all the steps lead to (theta = 45^{circ}), so I'm pretty sure that's correct.Another way to think about it is using the cosine of angles. If I have (cos 60^{circ}), which is (frac{1}{2}), and I know that (cos 45^{circ}) is (frac{sqrt{2}}{2}), then dividing (frac{1}{2}) by (frac{sqrt{2}}{2}) gives me (frac{1}{sqrt{2}}), which is the cosine of 45°. So, that's another way to see it.Yeah, I think I've got it right. The answer is 45°, which is option (D).

💡First, I need to convert the train's speed from kilometers per hour to meters per second to match the units of the platform's length and the given time.Next, I'll use the information about the train crossing the platform to find the length of the train. The total distance the train covers when crossing the platform is the sum of its own length and the platform's length.Once I have the length of the train, I can determine the time it takes to cross the man by considering that the train only needs to cover its own length when passing the man.Finally, I'll apply the basic distance-speed-time formula to calculate the required time.

💡First, I need to determine the total length of the fence required to build around the square plot. Given that the total cost is Rs. 3,876 and the cost per foot is Rs. 57, I can calculate the total length of the fence by dividing the total cost by the cost per foot.Next, since the plot is square, all four sides are of equal length. I can find the length of one side by dividing the total length of the fence by 4.Once I have the length of one side, I can calculate the area of the square plot by squaring the length of one side.Finally, I will present the calculated area as the solution.

💡Okay, so I have this equation: √3 sinx - cosx = 4 - m. I need to find the range of the real number m. Hmm, let me think about how to approach this. First, I remember that expressions like a sinx + b cosx can be rewritten using the auxiliary angle formula. Maybe that can help simplify this equation. The formula is something like R sin(x + α) or R cos(x + α), where R is the amplitude and α is the phase shift. Let me try that.So, for the expression √3 sinx - cosx, I can write it as R sin(x - α). To find R, I think R is equal to the square root of (a² + b²), where a is the coefficient of sinx and b is the coefficient of cosx. In this case, a is √3 and b is -1. So, R = √[(√3)² + (-1)²] = √[3 + 1] = √4 = 2. Okay, so R is 2.Now, to find the angle α, I know that tanα = b/a. Wait, is that right? Or is it tanα = a/b? Let me recall. If I have R sin(x - α), then expanding it gives R sinx cosα - R cosx sinα. Comparing that to √3 sinx - cosx, I can set up the equations:R cosα = √3 R sinα = 1Since R is 2, then:cosα = √3 / 2 sinα = 1 / 2Hmm, what angle α has cosine √3/2 and sine 1/2? That's π/6 or 30 degrees. So α is π/6. Therefore, √3 sinx - cosx can be rewritten as 2 sin(x - π/6). So the equation becomes:2 sin(x - π/6) = 4 - mOkay, so now I have 2 sin(x - π/6) = 4 - m. I know that the sine function has a range between -1 and 1. So, sin(x - π/6) can vary between -1 and 1. Therefore, 2 sin(x - π/6) will vary between -2 and 2.So, 2 sin(x - π/6) is between -2 and 2. That means 4 - m must also be between -2 and 2. Let me write that as inequalities:-2 ≤ 4 - m ≤ 2Now, I need to solve these inequalities for m. Let's start with the left inequality:-2 ≤ 4 - mSubtract 4 from both sides:-2 - 4 ≤ -m -6 ≤ -mMultiply both sides by -1, remembering to reverse the inequality sign:6 ≥ m Which is the same as m ≤ 6Now, the right inequality:4 - m ≤ 2Subtract 4 from both sides:-m ≤ 2 - 4 -m ≤ -2Multiply both sides by -1, again reversing the inequality sign:m ≥ 2So combining both inequalities, we have:2 ≤ m ≤ 6Therefore, the range of m is from 2 to 6, inclusive. Looking back at the options, option A is 2 ≤ m ≤ 6, which matches my result. Wait, let me double-check my steps to make sure I didn't make a mistake. 1. I used the auxiliary angle formula correctly to rewrite √3 sinx - cosx as 2 sin(x - π/6). That seems right because when I expand 2 sin(x - π/6), I get 2 sinx cos(π/6) - 2 cosx sin(π/6) = 2 sinx (√3/2) - 2 cosx (1/2) = √3 sinx - cosx. So that part checks out.2. Then I set 2 sin(x - π/6) equal to 4 - m. Since the sine function ranges between -1 and 1, multiplying by 2 gives -2 to 2. So 4 - m must be within that interval. That makes sense.3. Solving the inequalities: starting with -2 ≤ 4 - m ≤ 2. Subtracting 4 from all parts gives -6 ≤ -m ≤ -2. Multiplying by -1 and reversing inequalities gives 6 ≥ m ≥ 2, which is the same as 2 ≤ m ≤ 6. That seems correct.I don't see any errors in my reasoning. So, I'm confident that the correct answer is option A: 2 ≤ m ≤ 6.**Final Answer**boxed{A}

💡First, calculate the original total cost for the folders and pens without any discounts.For the folders:- Number of folders: 7- Cost per folder: 2.50- Total cost for folders: 7 × 2.50 = 17.50For the pens:- Number of pens: 10- Cost per pen: 1.00- Total cost for pens: 10 × 1.00 = 10.00Now, calculate the discounted prices for folders and pens.For the folders with a 30% discount:- Discount per folder: 30% of 2.50 = 0.30 × 2.50 = 0.75- Discounted price per folder: 2.50 - 0.75 = 1.75- Total cost for folders with discount: 7 × 1.75 = 12.25For the pens with a 15% discount:- Discount per pen: 15% of 1.00 = 0.15 × 1.00 = 0.15- Discounted price per pen: 1.00 - 0.15 = 0.85- Total cost for pens with discount: 10 × 0.85 = 8.50Finally, calculate the total savings by comparing the original total cost to the discounted total cost.Original total cost: 17.50 (folders) + 10.00 (pens) = 27.50Discounted total cost: 12.25 (folders) + 8.50 (pens) = 20.75Total savings: 27.50 - 20.75 = 6.75

💡Alright, so I've got this geometry problem here, and I need to show that for four points A, B, C, D in space, the sum of the squares of the distances AC, BD, AD, and BC is greater than or equal to the sum of the squares of the distances AB and CD. That is,[ AC^2 + BD^2 + AD^2 + BC^2 geq AB^2 + CD^2 ]Hmm, okay. Let me try to visualize this. Points in space, so they could be anywhere, not necessarily on a plane. The distances are just the straight-line distances between each pair of points. So, I need to relate these distances in some way.Maybe I can use some geometric theorems or properties. I remember that in triangles, the Law of Cosines relates the sides and the angles, but I'm not sure if that's directly applicable here. Alternatively, maybe I can use coordinates to represent the points and then express the distances in terms of coordinates. That might get messy, but it could work.Let me try setting up coordinates. Let's assign coordinates to the points:- Let’s place point A at the origin, so A = (0, 0, 0).- Let’s place point B along the x-axis, so B = (b, 0, 0) for some b > 0.- Now, points C and D can be anywhere in space, so let's denote C = (c1, c2, c3) and D = (d1, d2, d3).Now, let's write expressions for each distance squared:- AB^2 = (b - 0)^2 + (0 - 0)^2 + (0 - 0)^2 = b^2- AC^2 = (c1 - 0)^2 + (c2 - 0)^2 + (c3 - 0)^2 = c1^2 + c2^2 + c3^2- AD^2 = (d1 - 0)^2 + (d2 - 0)^2 + (d3 - 0)^2 = d1^2 + d2^2 + d3^2- BC^2 = (c1 - b)^2 + (c2 - 0)^2 + (c3 - 0)^2 = (c1 - b)^2 + c2^2 + c3^2- BD^2 = (d1 - b)^2 + (d2 - 0)^2 + (d3 - 0)^2 = (d1 - b)^2 + d2^2 + d3^2- CD^2 = (d1 - c1)^2 + (d2 - c2)^2 + (d3 - c3)^2Okay, so now I can plug these into the inequality:AC^2 + BD^2 + AD^2 + BC^2 ≥ AB^2 + CD^2Substituting the expressions:(c1^2 + c2^2 + c3^2) + [(d1 - b)^2 + d2^2 + d3^2] + [d1^2 + d2^2 + d3^2] + [(c1 - b)^2 + c2^2 + c3^2] ≥ b^2 + [(d1 - c1)^2 + (d2 - c2)^2 + (d3 - c3)^2]Let me simplify the left-hand side (LHS) first:= c1^2 + c2^2 + c3^2 + (d1^2 - 2b d1 + b^2) + d2^2 + d3^2 + d1^2 + d2^2 + d3^2 + (c1^2 - 2b c1 + b^2) + c2^2 + c3^2Combine like terms:= (c1^2 + c1^2) + (c2^2 + c2^2) + (c3^2 + c3^2) + (d1^2 + d1^2) + (d2^2 + d2^2) + (d3^2 + d3^2) + (-2b d1 - 2b c1) + (b^2 + b^2)= 2c1^2 + 2c2^2 + 2c3^2 + 2d1^2 + 2d2^2 + 2d3^2 - 2b(d1 + c1) + 2b^2Now, the right-hand side (RHS):= b^2 + (d1 - c1)^2 + (d2 - c2)^2 + (d3 - c3)^2Expand the squares:= b^2 + (d1^2 - 2c1 d1 + c1^2) + (d2^2 - 2c2 d2 + c2^2) + (d3^2 - 2c3 d3 + c3^2)Combine like terms:= b^2 + d1^2 + d2^2 + d3^2 + c1^2 + c2^2 + c3^2 - 2c1 d1 - 2c2 d2 - 2c3 d3So, now our inequality is:2c1^2 + 2c2^2 + 2c3^2 + 2d1^2 + 2d2^2 + 2d3^2 - 2b(d1 + c1) + 2b^2 ≥ b^2 + d1^2 + d2^2 + d3^2 + c1^2 + c2^2 + c3^2 - 2c1 d1 - 2c2 d2 - 2c3 d3Let me subtract the RHS from both sides to see what we get:[2c1^2 + 2c2^2 + 2c3^2 + 2d1^2 + 2d2^2 + 2d3^2 - 2b(d1 + c1) + 2b^2] - [b^2 + d1^2 + d2^2 + d3^2 + c1^2 + c2^2 + c3^2 - 2c1 d1 - 2c2 d2 - 2c3 d3] ≥ 0Simplify term by term:2c1^2 - c1^2 = c1^22c2^2 - c2^2 = c2^22c3^2 - c3^2 = c3^22d1^2 - d1^2 = d1^22d2^2 - d2^2 = d2^22d3^2 - d3^2 = d3^2-2b(d1 + c1) remains2b^2 - b^2 = b^2And then we have +2c1 d1 + 2c2 d2 + 2c3 d3So putting it all together:c1^2 + c2^2 + c3^2 + d1^2 + d2^2 + d3^2 - 2b(d1 + c1) + b^2 + 2c1 d1 + 2c2 d2 + 2c3 d3 ≥ 0Hmm, that's still a bit complicated. Maybe I can rearrange terms:= (c1^2 + d1^2 + 2c1 d1) + (c2^2 + d2^2 + 2c2 d2) + (c3^2 + d3^2 + 2c3 d3) - 2b(d1 + c1) + b^2Notice that each of the first three groups is a perfect square:= (c1 + d1)^2 + (c2 + d2)^2 + (c3 + d3)^2 - 2b(d1 + c1) + b^2Hmm, interesting. So now we have:(c1 + d1)^2 + (c2 + d2)^2 + (c3 + d3)^2 - 2b(d1 + c1) + b^2 ≥ 0Let me denote S = c1 + d1, T = c2 + d2, U = c3 + d3. Then the expression becomes:S^2 + T^2 + U^2 - 2b S + b^2 ≥ 0Which can be rewritten as:(S - b)^2 + T^2 + U^2 ≥ 0Ah, because:(S - b)^2 = S^2 - 2b S + b^2So, indeed, the entire expression is:(S - b)^2 + T^2 + U^2 ≥ 0Which is clearly always true because squares are non-negative. Therefore, the inequality holds.Wait, let me double-check that substitution. I set S = c1 + d1, T = c2 + d2, U = c3 + d3. Then, yes, S^2 + T^2 + U^2 - 2b S + b^2 = (S - b)^2 + T^2 + U^2, which is always non-negative.So, this shows that the original inequality holds because the left-hand side minus the right-hand side simplifies to a sum of squares, which is always non-negative.Alternatively, is there another way to see this without going through all the coordinate algebra? Maybe using vectors or some geometric interpretation.Let me think about vectors. Let’s denote vectors **a**, **b**, **c**, **d** as the position vectors of points A, B, C, D respectively. Then, the distances squared can be expressed using dot products:AC^2 = |**c** - **a**|^2 = (**c** - **a**) · (**c** - **a**)Similarly for the others. So, the inequality becomes:|**c** - **a**|^2 + |**d** - **b**|^2 + |**d** - **a**|^2 + |**c** - **b**|^2 ≥ |**b** - **a**|^2 + |**d** - **c**|^2Let me expand each term:|**c** - **a**|^2 = |**c**|^2 + |**a**|^2 - 2 **a** · **c**|**d** - **b**|^2 = |**d**|^2 + |**b**|^2 - 2 **b** · **d**|**d** - **a**|^2 = |**d**|^2 + |**a**|^2 - 2 **a** · **d**|**c** - **b**|^2 = |**c**|^2 + |**b**|^2 - 2 **b** · **c**Adding these up:= (|**c**|^2 + |**a**|^2 - 2 **a** · **c**) + (|**d**|^2 + |**b**|^2 - 2 **b** · **d**) + (|**d**|^2 + |**a**|^2 - 2 **a** · **d**) + (|**c**|^2 + |**b**|^2 - 2 **b** · **c**)Combine like terms:= 2|**c**|^2 + 2|**a**|^2 + 2|**d**|^2 + 2|**b**|^2 - 2(**a** · **c** + **b** · **d** + **a** · **d** + **b** · **c**)Now, the right-hand side:|**b** - **a**|^2 + |**d** - **c**|^2= (|**b**|^2 + |**a**|^2 - 2 **a** · **b**) + (|**d**|^2 + |**c**|^2 - 2 **c** · **d**)So, the inequality is:2|**c**|^2 + 2|**a**|^2 + 2|**d**|^2 + 2|**b**|^2 - 2(**a** · **c** + **b** · **d** + **a** · **d** + **b** · **c**) ≥ |**b**|^2 + |**a**|^2 - 2 **a** · **b** + |**d**|^2 + |**c**|^2 - 2 **c** · **d**Subtracting the RHS from both sides:[2|**c**|^2 + 2|**a**|^2 + 2|**d**|^2 + 2|**b**|^2 - 2(**a** · **c** + **b** · **d** + **a** · **d** + **b** · **c**)] - [|**b**|^2 + |**a**|^2 - 2 **a** · **b** + |**d**|^2 + |**c**|^2 - 2 **c** · **d**] ≥ 0Simplify term by term:2|**c**|^2 - |**c**|^2 = |**c**|^22|**a**|^2 - |**a**|^2 = |**a**|^22|**d**|^2 - |**d**|^2 = |**d**|^22|**b**|^2 - |**b**|^2 = |**b**|^2-2(**a** · **c** + **b** · **d** + **a** · **d** + **b** · **c**) - (-2 **a** · **b** - 2 **c** · **d**)= -2**a** · **c** - 2**b** · **d** - 2**a** · **d** - 2**b** · **c** + 2**a** · **b** + 2**c** · **d**So, combining all terms:|**c**|^2 + |**a**|^2 + |**d**|^2 + |**b**|^2 - 2**a** · **c** - 2**b** · **d** - 2**a** · **d** - 2**b** · **c** + 2**a** · **b** + 2**c** · **d** ≥ 0Hmm, this looks similar to the coordinate approach. Maybe I can factor this expression somehow. Let me try grouping terms:= (|**a**|^2 + |**b**|^2 + |**c**|^2 + |**d**|^2) + (-2**a** · **c** - 2**a** · **d** - 2**b** · **c** - 2**b** · **d**) + (2**a** · **b** + 2**c** · **d**)Let me see if I can write this as a sum of squares. Maybe consider (**a** + **b** - **c** - **d**) · (**a** + **b** - **c** - **d**). Let's compute that:(**a** + **b** - **c** - **d**) · (**a** + **b** - **c** - **d**) = |**a**|^2 + |**b**|^2 + |**c**|^2 + |**d**|^2 + 2**a** · **b** - 2**a** · **c** - 2**a** · **d** - 2**b** · **c** - 2**b** · **d** + 2**c** · **d**Wait, that's exactly the expression we have! So,(**a** + **b** - **c** - **d**) · (**a** + **b** - **c** - **d**) = |**a** + **b** - **c** - **d**|^2Which is always non-negative because it's a squared magnitude. Therefore, the inequality holds.So, both the coordinate approach and the vector approach lead us to the conclusion that the left-hand side minus the right-hand side is a sum of squares, which is non-negative. Hence, the original inequality is true.Alternatively, is there a geometric interpretation without coordinates or vectors? Maybe using the triangle inequality or some properties of quadrilaterals?Let me think about quadrilaterals. In a quadrilateral, the sum of the squares of the sides can be related to the sum of the squares of the diagonals. Wait, there's a formula for that. In any quadrilateral, the sum of the squares of all six edges is equal to the sum of the squares of the three diagonals plus four times the square of the line connecting the midpoints. But I'm not sure if that's directly applicable here.Wait, actually, in any quadrilateral, the following holds:AC^2 + BD^2 + AB^2 + CD^2 = AD^2 + BC^2 + 2MN^2Where MN is the segment connecting the midpoints of the diagonals. But in our case, the inequality is:AC^2 + BD^2 + AD^2 + BC^2 ≥ AB^2 + CD^2Which seems similar but not exactly the same. Maybe if I rearrange the terms:AC^2 + BD^2 + AD^2 + BC^2 - AB^2 - CD^2 ≥ 0From the quadrilateral formula, we have:AC^2 + BD^2 + AB^2 + CD^2 = AD^2 + BC^2 + 2MN^2So, rearranging:AC^2 + BD^2 + AD^2 + BC^2 = AB^2 + CD^2 + 2MN^2Therefore,AC^2 + BD^2 + AD^2 + BC^2 - AB^2 - CD^2 = 2MN^2 ≥ 0Which shows that the left-hand side is equal to twice the square of the midline, which is always non-negative. Hence, the inequality holds.So, that's another way to see it, using properties of quadrilaterals. The difference between the sums of the squares of certain sides and diagonals equals twice the square of the midline, which is non-negative.Therefore, regardless of the approach—coordinate, vector, or geometric—the inequality holds true.**Final Answer**boxed{AC^2 + BD^2 + AD^2 + BC^2 geq AB^2 + CD^2}