💡Okay, so I have this expression: (frac{(1+sqrt{1991})^{100} - (1-sqrt{1991})^{100}}{sqrt{1991}}). I need to figure out what kind of number this represents. The options are natural number, integer but not natural, rational but not integer, or irrational. Hmm, let me break this down step by step.First, I notice that the expression involves terms like ((1+sqrt{1991})^{100}) and ((1-sqrt{1991})^{100}). These look similar to binomial expansions. Maybe I can expand them using the binomial theorem. Let me recall that the binomial theorem states that ((a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n-k} b^k). So, applying this to both terms:[(1 + sqrt{1991})^{100} = sum_{k=0}^{100} binom{100}{k} (1)^{100 - k} (sqrt{1991})^k][(1 - sqrt{1991})^{100} = sum_{k=0}^{100} binom{100}{k} (1)^{100 - k} (-sqrt{1991})^k]Now, if I subtract these two expressions, the even-powered terms will cancel out because ((sqrt{1991})^k) and ((- sqrt{1991})^k) will be the same when (k) is even, but opposite when (k) is odd. So, subtracting them:[(1 + sqrt{1991})^{100} - (1 - sqrt{1991})^{100} = 2 sum_{k text{ odd}} binom{100}{k} (sqrt{1991})^k]This simplifies the expression to twice the sum of the odd-powered terms. Now, let's factor out (sqrt{1991}) from each term in the sum:[2 sum_{k text{ odd}} binom{100}{k} (sqrt{1991})^k = 2 sqrt{1991} sum_{k text{ odd}} binom{100}{k} (1991)^{(k-1)/2}]So, the original expression becomes:[frac{2 sqrt{1991} sum_{k text{ odd}} binom{100}{k} (1991)^{(k-1)/2}}{sqrt{1991}} = 2 sum_{k text{ odd}} binom{100}{k} (1991)^{(k-1)/2}]Now, looking at this sum, each term is of the form (binom{100}{k} (1991)^{(k-1)/2}). Since (k) is odd, ((k-1)/2) is an integer, so each term is an integer multiplied by a power of 1991, which is also an integer. Therefore, the entire sum is an integer, and multiplying by 2 keeps it an integer.But is this integer a natural number? Well, natural numbers are positive integers. Let's think about the terms in the sum. Each term is positive because (binom{100}{k}) is positive and (1991^{(k-1)/2}) is positive. So, the entire sum is positive, and multiplying by 2 keeps it positive. Therefore, the expression is a positive integer, which is a natural number.Wait, hold on. The options are (A) Natural number, (B) Integer but not a natural number, (C) Rational number but not an integer, (D) Irrational number. So, if it's a positive integer, it's a natural number, right? So, the answer should be (A).But wait, let me double-check. Maybe I made a mistake in assuming all terms are positive. Let me see:When I subtracted the two expansions, I got twice the sum of the odd terms. Each of these terms is positive because ((sqrt{1991})^k) is positive and the binomial coefficients are positive. So, yes, the entire expression is positive. Therefore, it's a natural number.But wait, another thought. The expression is (frac{(1+sqrt{1991})^{100} - (1-sqrt{1991})^{100}}{sqrt{1991}}). Since (1 + sqrt{1991}) is greater than 1, and (1 - sqrt{1991}) is negative, but raised to the 100th power, which is even, so it becomes positive. So, the numerator is a positive number minus a positive number, but since (1 + sqrt{1991}) is much larger than (1 - sqrt{1991}), the numerator is positive. Dividing by (sqrt{1991}), which is positive, keeps it positive.Therefore, the result is a positive integer, which is a natural number. So, the answer should be (A).But wait, another perspective. Maybe the expression is an integer but not necessarily a natural number? No, because we've established it's positive. So, it's a natural number.Wait, but let me think about the specific value. Is it possible that the expression is zero? No, because (1 + sqrt{1991}) is much larger than (1 - sqrt{1991}), so their 100th powers would still have a significant difference. So, the numerator is positive, and dividing by a positive number keeps it positive. So, it's definitely a positive integer, hence a natural number.Therefore, I think the answer is (A).But wait, hold on. Let me check with a smaller exponent to see the pattern. Let's say instead of 100, let's take n=2.Compute (frac{(1+sqrt{1991})^2 - (1-sqrt{1991})^2}{sqrt{1991}}).First, compute numerator:[(1 + sqrt{1991})^2 = 1 + 2sqrt{1991} + 1991][(1 - sqrt{1991})^2 = 1 - 2sqrt{1991} + 1991]Subtracting them:[(1 + 2sqrt{1991} + 1991) - (1 - 2sqrt{1991} + 1991) = 4sqrt{1991}]Divide by (sqrt{1991}):[frac{4sqrt{1991}}{sqrt{1991}} = 4]Which is a natural number.Similarly, for n=1:[frac{(1+sqrt{1991}) - (1-sqrt{1991})}{sqrt{1991}} = frac{2sqrt{1991}}{sqrt{1991}} = 2]Also a natural number.For n=3:Compute (frac{(1+sqrt{1991})^3 - (1-sqrt{1991})^3}{sqrt{1991}}).First, expand:[(1 + sqrt{1991})^3 = 1 + 3sqrt{1991} + 3(1991) + (1991)^{3/2}][(1 - sqrt{1991})^3 = 1 - 3sqrt{1991} + 3(1991) - (1991)^{3/2}]Subtracting:[(1 + 3sqrt{1991} + 3(1991) + (1991)^{3/2}) - (1 - 3sqrt{1991} + 3(1991) - (1991)^{3/2}) = 6sqrt{1991} + 2(1991)^{3/2}]Divide by (sqrt{1991}):[frac{6sqrt{1991} + 2(1991)^{3/2}}{sqrt{1991}} = 6 + 2(1991)]Which is (6 + 3982 = 3988), a natural number.So, for n=1,2,3, the expression results in natural numbers. This suggests that for n=100, it's also a natural number.Therefore, I think the answer is (A) Natural number.But wait, let me consider the general case. Let’s define (a = 1 + sqrt{1991}) and (b = 1 - sqrt{1991}). Notice that (a + b = 2) and (ab = (1)^2 - (1991) = -1990). These are the roots of the quadratic equation (x^2 - 2x - 1990 = 0).Now, consider the sequence (S_n = frac{a^n - b^n}{sqrt{1991}}). We can analyze the properties of (S_n).From the recurrence relation, since (a) and (b) satisfy (x^2 = 2x + 1990), we can derive that (S_n) satisfies a linear recurrence relation.Let’s compute (S_0), (S_1), (S_2), etc., to find a pattern.For n=0:(S_0 = frac{a^0 - b^0}{sqrt{1991}} = frac{1 - 1}{sqrt{1991}} = 0)For n=1:(S_1 = frac{a - b}{sqrt{1991}} = frac{2sqrt{1991}}{sqrt{1991}} = 2)For n=2:As computed earlier, (S_2 = 4)For n=3:(S_3 = 3988)Wait, let me compute (S_3) properly.From earlier, (S_3 = 6 + 2(1991) = 6 + 3982 = 3988). So, yes.Now, let's see if there's a recurrence relation. Since (a) and (b) satisfy (x^2 = 2x + 1990), we can write:(a^{n} = 2a^{n-1} + 1990a^{n-2})Similarly,(b^{n} = 2b^{n-1} + 1990b^{n-2})Subtracting these:(a^{n} - b^{n} = 2(a^{n-1} - b^{n-1}) + 1990(a^{n-2} - b^{n-2}))Divide both sides by (sqrt{1991}):(frac{a^{n} - b^{n}}{sqrt{1991}} = 2frac{a^{n-1} - b^{n-1}}{sqrt{1991}} + 1990frac{a^{n-2} - b^{n-2}}{sqrt{1991}})Thus,(S_n = 2S_{n-1} + 1990S_{n-2})With initial conditions:(S_0 = 0)(S_1 = 2)So, this recurrence relation generates integer values for (S_n), starting from (S_0 = 0) and (S_1 = 2). Therefore, (S_n) is always an integer for all (n geq 0).Moreover, since (S_n) is defined as (frac{a^n - b^n}{sqrt{1991}}), and (a > 1), (b < 0), but since (n=100) is even, (b^{100}) is positive, but much smaller than (a^{100}), so (S_{100}) is positive.Therefore, (S_{100}) is a positive integer, which is a natural number.Hence, the answer is (A) Natural number.

💡First, I need to determine the side length of the square. The perimeter of the square is 64, so each side is 64 divided by 4, which equals 16.Next, I'll calculate the area of the square by squaring the side length: 16 multiplied by 16 equals 256.Now, I'll find the area of the triangle. The formula for the area of a triangle is one-half the base multiplied by the height. Here, the height is 32, and the base is represented by ( x ). So, the area of the triangle is ( frac{1}{2} times 32 times x = 16x ).Since the areas of the square and the triangle are equal, I can set up the equation ( 16x = 256 ).Finally, I'll solve for ( x ) by dividing both sides of the equation by 16, which gives ( x = 16 ).

💡First, identify the total number of cards in the box: 1 Ace, 2 Kings, and 2 Queens, making a total of 5 cards.Next, calculate the total number of ways to draw two cards from these five using the combination formula:[binom{5}{2} = frac{5 times 4}{2 times 1} = 10]Now, determine the number of ways to draw two cards of the same rank. Since there are 2 Kings and 2 Queens:[binom{2}{2} = 1 text{ way for Kings}][binom{2}{2} = 1 text{ way for Queens}]Total ways to draw two cards of the same rank:[1 + 1 = 2]Calculate the probability of drawing two cards of the same rank:[frac{2}{10} = frac{1}{5}]Finally, find the probability of drawing two cards of different ranks by subtracting the probability of drawing the same rank from 1:[P(text{different ranks}) = 1 - frac{1}{5} = frac{4}{5}]

💡Okay, so I have this geometry problem here, and I need to figure out which statement is correct among options A, B, C, and D. Let me take it step by step because I want to make sure I understand each part correctly.First, let's parse the problem. We have two lines, a and b, and two planes, α and β. The question is about the relationships between these lines and planes in terms of parallelism and perpendicularity. The options are:A: If a is parallel to α and b is parallel to α, then a is parallel to b.B: If a is perpendicular to α and α is perpendicular to β, then a is parallel to β.C: If a is parallel to α and b is perpendicular to α, then a is perpendicular to b.D: If a is parallel to α and α is perpendicular to β, then a is perpendicular to β.I need to determine which of these statements is correct. Let me go through each option one by one.Starting with option A: If a is parallel to α and b is parallel to α, then a is parallel to b.Hmm, okay. So, if two lines are both parallel to the same plane, does that mean they are parallel to each other? I'm not sure. Let me visualize this. Imagine a plane α, like a flat tabletop. If line a is parallel to this tabletop, it means it's either lying on a parallel plane above or below, or it's just running alongside it without intersecting. Similarly, line b is also parallel to the same tabletop. Now, can these two lines a and b be parallel? Yes, they can. But can they also be skew or intersecting? Wait, if they're both parallel to the same plane, they could be in different directions. For example, if I have two lines on a different plane that's parallel to α, they can be parallel, but they could also be intersecting if they lie on the same plane. Wait, no, if they're both parallel to α, they don't necessarily lie on the same plane. They could be on different parallel planes, so they could be skew lines as well. So, just because both lines are parallel to the same plane doesn't necessarily mean they are parallel to each other. They could be parallel, intersecting, or skew. Therefore, option A is not necessarily true. So, A is incorrect.Moving on to option B: If a is perpendicular to α and α is perpendicular to β, then a is parallel to β.Alright, so line a is perpendicular to plane α, and plane α is perpendicular to plane β. Does this mean that line a is parallel to plane β? Let me think. If a is perpendicular to α, that means it's orthogonal to every line in α. Now, if α is perpendicular to β, that means the dihedral angle between α and β is 90 degrees. So, how does line a relate to plane β? If a is perpendicular to α, and α is perpendicular to β, then a could be parallel to β, but it might also lie within β or intersect β at some angle. Wait, actually, if a is perpendicular to α, and α is perpendicular to β, then a must be parallel to β. Because if α and β are perpendicular, their normals are also perpendicular. Since a is along the normal of α, it would be parallel to the normal of β, hence parallel to β. Hmm, is that right? Wait, no. If a is perpendicular to α, then a is parallel to the normal vector of α. If α is perpendicular to β, then the normal vector of α is parallel to the normal vector of β. Therefore, a, being parallel to the normal of α, is also parallel to the normal of β, which means a is parallel to β. So, actually, option B might be correct. Wait, but hold on. If a is perpendicular to α, and α is perpendicular to β, does that necessarily mean a is parallel to β? Or could a lie within β? Because if a is perpendicular to α, and α is perpendicular to β, then a could lie within β as well. For example, imagine α is the xy-plane, β is the xz-plane, which is perpendicular to α. Then, a line a that is perpendicular to α would be along the z-axis. Is the z-axis parallel to the xz-plane? No, the z-axis lies within the xz-plane. So, in this case, a is not parallel to β; it's actually lying within β. So, in this example, a is not parallel to β. Therefore, option B is not necessarily true because a could lie within β or be parallel to β. So, B is incorrect.Okay, moving on to option C: If a is parallel to α and b is perpendicular to α, then a is perpendicular to b.Alright, so line a is parallel to plane α, and line b is perpendicular to plane α. Does this imply that a is perpendicular to b? Let me think. If a is parallel to α, that means a doesn't intersect α and lies in a direction that's parallel to α. On the other hand, b is perpendicular to α, so it's orthogonal to every line in α. Since a is parallel to α, it's essentially lying in a direction that's within α or parallel to it. Therefore, b, being perpendicular to α, must be perpendicular to a as well. Because if b is perpendicular to the entire plane, it's perpendicular to every line in that plane, including those parallel to the plane. So, yes, a and b must be perpendicular. Let me test this with an example. Suppose α is the xy-plane. Then, a line a parallel to α could be, say, along the x-axis in some plane above the xy-plane. Line b, being perpendicular to α, would be along the z-axis. The x-axis and z-axis are indeed perpendicular. So, in this case, a and b are perpendicular. Another example: if a is along the y-axis in a plane above the xy-plane, and b is along the z-axis, they are still perpendicular. So, yes, it seems that option C is correct.Just to be thorough, let's look at option D: If a is parallel to α and α is perpendicular to β, then a is perpendicular to β.So, line a is parallel to plane α, and plane α is perpendicular to plane β. Does this mean that a is perpendicular to β? Let me visualize. If α is perpendicular to β, their dihedral angle is 90 degrees. Line a is parallel to α, so it's either lying on α or on a plane parallel to α. Now, does that make a perpendicular to β? Not necessarily. For example, take α as the xy-plane and β as the xz-plane, which is perpendicular to α. Line a could be along the x-axis, which is parallel to α (the xy-plane). Is the x-axis perpendicular to β (the xz-plane)? No, the x-axis lies within the xz-plane, so it's not perpendicular. Therefore, a is not necessarily perpendicular to β. It could lie within β or be at some angle. So, option D is incorrect.So, after going through each option, I think option C is the correct one. Let me just recap:- A: Incorrect because two lines parallel to the same plane can be parallel, intersecting, or skew.- B: Incorrect because a line perpendicular to α could lie within β if α and β are perpendicular.- C: Correct because a line parallel to α and another line perpendicular to α must be perpendicular to each other.- D: Incorrect because a line parallel to α doesn't have to be perpendicular to β even if α and β are perpendicular.Therefore, the correct answer is option C.**Final Answer**boxed{C}

💡Okay, so I have this geometry problem about a square ABCD. Let me try to visualize it first. The square has points E and F on sides AD and BC respectively, such that EF is parallel to AB. Then there are points G and H on sides AB and DC respectively, such that GH is parallel to AD. These lines EF and GH intersect at point K. The area of rectangle KFC H is twice the area of rectangle AGKE. I need to find the measure of angle FAH.Hmm, okay. Let me try to draw this in my mind. Square ABCD, so all sides are equal, all angles are 90 degrees. Let me label the square: A is the top-left corner, B is the top-right, C is the bottom-right, and D is the bottom-left. So, sides AB, BC, CD, DA.Points E and F are on AD and BC respectively, and EF is parallel to AB. Since AB is the top side, which is horizontal, EF must also be horizontal. Similarly, points G and H are on AB and DC respectively, and GH is parallel to AD, which is vertical, so GH must be vertical.So, EF is a horizontal line inside the square, and GH is a vertical line inside the square. They intersect at point K. So, K is somewhere inside the square where these two lines cross.Now, the areas of rectangles KFC H and AGKE are given. KFC H is a rectangle, and AGKE is another rectangle. The area of KFC H is twice that of AGKE. I need to find the measure of angle FAH.Let me try to assign some variables to make this more concrete. Let me assume the square has side length 1 for simplicity. So, each side is 1 unit long.Let me denote AE as x. Since E is on AD, which has length 1, then ED would be 1 - x. Similarly, since EF is parallel to AB, which is the top side, EF must be a horizontal line. Therefore, F is on BC, so BF would be equal to AE, right? Because EF is parallel to AB, so the distance from E to A is the same as the distance from F to B. So, BF = x, which would make FC = 1 - x.Similarly, let me denote AG as y. Since G is on AB, which is the top side, then GB would be 1 - y. GH is parallel to AD, which is vertical, so GH is a vertical line. Therefore, H is on DC, and since GH is vertical, the distance from G to A is the same as the distance from H to D. So, DH = AG = y, which makes HC = 1 - y.Now, since EF is horizontal and GH is vertical, their intersection K divides the square into four regions: two rectangles AGKE and KFC H, and two other regions which are triangles or something else.Wait, but the problem says KFC H and AGKE are rectangles. So, KFC H is a rectangle with sides FC and HC, which are both 1 - x and 1 - y. Similarly, AGKE is a rectangle with sides AG and AE, which are y and x. So, the area of KFC H is (1 - x)(1 - y), and the area of AGKE is x * y.Given that the area of KFC H is twice the area of AGKE, so:(1 - x)(1 - y) = 2 * (x * y)Let me write that equation:(1 - x)(1 - y) = 2xyExpanding the left side:1 - x - y + xy = 2xyBring all terms to one side:1 - x - y + xy - 2xy = 0Simplify:1 - x - y - xy = 0Hmm, so:1 = x + y + xyOkay, that's an equation relating x and y. I need another equation to solve for x and y, but maybe I can find a relationship from the geometry.Wait, angle FAH is the angle between lines AF and AH. So, point F is on BC, point A is the corner, and point H is on DC. So, AF is a diagonal from A to F, and AH is a diagonal from A to H.I need to find the measure of angle FAH. Maybe I can find the coordinates of points F and H, then compute the vectors AF and AH, and then find the angle between them.Since I assumed the square has side length 1, let me assign coordinates:Let me place point A at (0,1), B at (1,1), C at (1,0), and D at (0,0). So, side AB is from (0,1) to (1,1), BC is from (1,1) to (1,0), CD is from (1,0) to (0,0), and DA is from (0,0) to (0,1).Point E is on AD, which is from (0,0) to (0,1). Since AE = x, point E is at (0, x). Similarly, point F is on BC, which is from (1,1) to (1,0). Since BF = x, point F is at (1, 1 - x).Point G is on AB, which is from (0,1) to (1,1). Since AG = y, point G is at (y, 1). Point H is on DC, which is from (1,0) to (0,0). Since DH = y, point H is at (y, 0).Now, lines EF and GH intersect at point K. Let me find the coordinates of K.First, equation of line EF: it's a horizontal line through E (0, x) and F (1, 1 - x). Wait, but EF is parallel to AB, which is horizontal, so EF is a horizontal line at y = x? Wait, no, because E is at (0, x) and F is at (1, 1 - x). Wait, that can't be horizontal unless x = 1 - x, which would mean x = 0.5. But that might not necessarily be the case.Wait, hold on, maybe I made a mistake. If EF is parallel to AB, which is horizontal, then EF must be a horizontal line. So, the y-coordinate of E and F must be the same. But E is on AD at (0, x) and F is on BC at (1, 1 - x). So, for EF to be horizontal, the y-coordinate must be the same, so x = 1 - x, which gives x = 0.5.Wait, that seems important. So, if EF is parallel to AB, which is horizontal, then E and F must be at the same height. So, E is at (0, x) and F is at (1, x). So, BF = 1 - x, which would mean FC = x. Wait, that contradicts my earlier thought.Wait, maybe I messed up the labeling. Let me clarify.If EF is parallel to AB, which is horizontal, then E and F must lie on the same horizontal line. So, E is on AD, which is vertical from (0,0) to (0,1). So, E is at (0, e) for some e between 0 and 1. Similarly, F is on BC, which is vertical from (1,1) to (1,0). So, F is at (1, f) for some f between 0 and 1.Since EF is horizontal, the y-coordinate of E and F must be equal, so e = f. So, E is (0, e) and F is (1, e). So, AE = e, so ED = 1 - e. Similarly, BF = 1 - e, so FC = e.Similarly, GH is parallel to AD, which is vertical. So, GH is a vertical line. G is on AB, which is from (0,1) to (1,1). So, G is at (g, 1) for some g between 0 and 1. H is on DC, which is from (1,0) to (0,0). So, H is at (h, 0) for some h between 0 and 1.Since GH is vertical, the x-coordinate of G and H must be equal, so g = h. So, G is (g, 1) and H is (g, 0). So, AG = g, so GB = 1 - g. Similarly, DH = g, so HC = 1 - g.So, now, lines EF and GH intersect at K. Let's find the coordinates of K.Line EF is the horizontal line y = e, from (0, e) to (1, e).Line GH is the vertical line x = g, from (g, 1) to (g, 0).So, their intersection K is at (g, e).So, point K is at (g, e).Now, the rectangles AGKE and KFC H.Rectangle AGKE has vertices at A (0,1), G (g,1), K (g,e), and E (0,e). So, its sides are AG = g, and AE = e. So, area is g * e.Rectangle KFC H has vertices at K (g,e), F (1,e), C (1,0), and H (g,0). So, its sides are FC = 1 - e, and HC = 1 - g. So, area is (1 - e)(1 - g).Given that area of KFC H is twice the area of AGKE:(1 - e)(1 - g) = 2 * (g * e)So, expanding:1 - e - g + e g = 2 e gBring all terms to one side:1 - e - g + e g - 2 e g = 0Simplify:1 - e - g - e g = 0So,1 = e + g + e gThat's the same equation as before, just with e and g instead of x and y.So, 1 = e + g + e g.I need another equation to solve for e and g, but perhaps I can find a relationship from the geometry.Wait, angle FAH is the angle between AF and AH. Let me find the coordinates of points F and H.Point F is at (1, e), and point H is at (g, 0). Point A is at (0,1).So, vector AF is from A (0,1) to F (1, e), which is (1 - 0, e - 1) = (1, e - 1).Vector AH is from A (0,1) to H (g, 0), which is (g - 0, 0 - 1) = (g, -1).The angle between vectors AF and AH can be found using the dot product formula:cos(theta) = (AF . AH) / (|AF| |AH|)Compute AF . AH:= (1)(g) + (e - 1)(-1)= g - (e - 1)= g - e + 1Compute |AF|:= sqrt(1^2 + (e - 1)^2)= sqrt(1 + (1 - e)^2)Compute |AH|:= sqrt(g^2 + (-1)^2)= sqrt(g^2 + 1)So,cos(theta) = (g - e + 1) / (sqrt(1 + (1 - e)^2) * sqrt(g^2 + 1))I need to find theta, which is angle FAH.But I don't know e and g yet. I only have the equation 1 = e + g + e g.I need another equation. Maybe from the areas or some other geometric consideration.Wait, perhaps I can express one variable in terms of the other.From 1 = e + g + e g, let me solve for g in terms of e.1 = e + g + e g1 - e = g (1 + e)So,g = (1 - e) / (1 + e)Similarly, I can express e in terms of g.But let me see if I can find another relationship.Wait, maybe from the coordinates of K, which is (g, e). Since K is inside the square, g and e are between 0 and 1.Alternatively, perhaps I can use similar triangles or something else.Wait, let me think about the coordinates.Point K is at (g, e). So, from point K, we can consider the lines to F and H.Wait, maybe not. Alternatively, perhaps I can use the fact that AF and AH are vectors, and the angle between them is determined by their slopes.Wait, the slope of AF is (e - 1)/1 = e - 1.The slope of AH is (-1)/g.The angle between two lines with slopes m1 and m2 is given by:tan(theta) = |(m2 - m1)/(1 + m1 m2)|So, tan(theta) = |( (-1/g) - (e - 1) ) / (1 + (e - 1)(-1/g))|Simplify numerator:= (-1/g - e + 1)= (1 - e) - 1/gDenominator:= 1 - (e - 1)/g= (g - (e - 1))/gSo,tan(theta) = | [ (1 - e) - 1/g ] / [ (g - e + 1)/g ] |Simplify numerator:= (1 - e) - 1/g = (g(1 - e) - 1)/gDenominator:= (g - e + 1)/gSo,tan(theta) = | [ (g(1 - e) - 1)/g ] / [ (g - e + 1)/g ] | = | (g(1 - e) - 1) / (g - e + 1) |Simplify numerator:g(1 - e) - 1 = g - g e - 1Denominator:g - e + 1So,tan(theta) = | (g - g e - 1) / (g - e + 1) |But from earlier, we have 1 = e + g + e g, so e + g + e g = 1.Let me see if I can substitute.From 1 = e + g + e g, we can write e g = 1 - e - g.So, numerator:g - g e - 1 = g - (1 - e - g) - 1 = g - 1 + e + g - 1 = 2g + e - 2Wait, let me check that:Wait, e g = 1 - e - g, so g e = 1 - e - g.So, numerator:g - g e - 1 = g - (1 - e - g) - 1 = g - 1 + e + g - 1 = 2g + e - 2Denominator:g - e + 1So,tan(theta) = | (2g + e - 2) / (g - e + 1) |Hmm, that seems complicated. Maybe I can find another way.Wait, let me go back to the equation 1 = e + g + e g.Let me solve for e in terms of g.1 = e + g + e g1 - g = e(1 + g)So,e = (1 - g)/(1 + g)Similarly, I can express e in terms of g.So, e = (1 - g)/(1 + g)Now, let me substitute e into the expression for tan(theta).From earlier, tan(theta) = | (2g + e - 2) / (g - e + 1) |Substitute e = (1 - g)/(1 + g):First, compute numerator:2g + e - 2 = 2g + (1 - g)/(1 + g) - 2Let me combine terms:= 2g - 2 + (1 - g)/(1 + g)= 2(g - 1) + (1 - g)/(1 + g)Factor out (1 - g):= (1 - g)[ -2/(1 - g) + 1/(1 + g) ]Wait, maybe not. Alternatively, let me get a common denominator.= [2g - 2](1 + g) + (1 - g) all over (1 + g)= [ (2g - 2)(1 + g) + (1 - g) ] / (1 + g)Expand numerator:(2g - 2)(1 + g) = 2g(1 + g) - 2(1 + g) = 2g + 2g^2 - 2 - 2g = 2g^2 - 2So, numerator:2g^2 - 2 + 1 - g = 2g^2 - g -1So, numerator is (2g^2 - g -1)/(1 + g)Denominator of tan(theta):g - e + 1 = g - (1 - g)/(1 + g) + 1Again, get common denominator:= [g(1 + g) - (1 - g) + (1 + g)] / (1 + g)Expand numerator:g(1 + g) = g + g^2- (1 - g) = -1 + g+ (1 + g) = 1 + gSo, total numerator:g + g^2 -1 + g +1 + g = g^2 + 3gSo, denominator is (g^2 + 3g)/(1 + g)So, tan(theta) = | [ (2g^2 - g -1)/(1 + g) ] / [ (g^2 + 3g)/(1 + g) ] | = | (2g^2 - g -1)/(g^2 + 3g) |Simplify:tan(theta) = | (2g^2 - g -1)/(g(g + 3)) |Hmm, this is getting complicated. Maybe I can factor the numerator.2g^2 - g -1. Let me try to factor it.Looking for two numbers a and b such that a * b = 2*(-1) = -2 and a + b = -1.Hmm, 1 and -2: 1*(-2) = -2, 1 + (-2) = -1. So, split the middle term:2g^2 + g - 2g -1 = g(2g +1) -1(2g +1) = (g -1)(2g +1)So, 2g^2 - g -1 = (2g +1)(g -1)So, tan(theta) = | (2g +1)(g -1) / (g(g + 3)) |So,tan(theta) = | (2g +1)(g -1) / (g(g + 3)) |Hmm, interesting. Now, I can write this as:tan(theta) = | (2g +1)(g -1) / (g(g + 3)) |But I don't know the value of g yet. I need another equation to solve for g.Wait, from earlier, we have e = (1 - g)/(1 + g). So, e is expressed in terms of g.Is there another relationship? Maybe from the areas or something else.Wait, perhaps I can use the fact that the areas are related, but we already used that to get 1 = e + g + e g.Alternatively, maybe I can consider the coordinates of point K and see if there's a relationship.Point K is at (g, e). So, in the square, it's somewhere inside. Maybe I can consider the triangles formed by K.Wait, another approach: since we're dealing with a square and parallel lines, maybe there's some similarity or congruence.Alternatively, perhaps I can assign specific values to e and g that satisfy 1 = e + g + e g and see what happens.Let me try to solve for g.From 1 = e + g + e g and e = (1 - g)/(1 + g), let's substitute e into the equation.Wait, but that's how we got e in terms of g. So, maybe I can express tan(theta) in terms of g and then find a relationship.Alternatively, perhaps I can assume that angle FAH is 45 degrees and see if that works.If angle FAH is 45 degrees, then tan(theta) = 1.So, set tan(theta) = 1:| (2g +1)(g -1) / (g(g + 3)) | = 1So,(2g +1)(g -1) / (g(g + 3)) = ±1Let me consider positive case first:(2g +1)(g -1) = g(g + 3)Expand left side:2g(g) + 2g(-1) +1(g) +1(-1) = 2g^2 -2g +g -1 = 2g^2 -g -1Right side:g^2 + 3gSo,2g^2 -g -1 = g^2 + 3gBring all terms to left:2g^2 -g -1 -g^2 -3g = 0Simplify:g^2 -4g -1 = 0Solve quadratic:g = [4 ± sqrt(16 +4)] / 2 = [4 ± sqrt(20)] / 2 = [4 ± 2*sqrt(5)] / 2 = 2 ± sqrt(5)Since g must be between 0 and 1, because it's a point on AB which has length 1, so 2 - sqrt(5) is approximately 2 - 2.236 = -0.236, which is negative, so invalid. 2 + sqrt(5) is about 4.236, which is greater than 1, so also invalid.So, no solution in this case.Now, consider the negative case:(2g +1)(g -1) = -g(g + 3)Expand left side:2g^2 -g -1Right side:- g^2 -3gSo,2g^2 -g -1 = -g^2 -3gBring all terms to left:2g^2 -g -1 +g^2 +3g = 0Simplify:3g^2 +2g -1 = 0Solve quadratic:g = [-2 ± sqrt(4 +12)] / 6 = [-2 ± sqrt(16)] /6 = [-2 ±4]/6So,g = (-2 +4)/6 = 2/6 = 1/3org = (-2 -4)/6 = -6/6 = -1Again, g must be between 0 and 1, so g = 1/3 is valid.So, g = 1/3.Then, from e = (1 - g)/(1 + g) = (1 - 1/3)/(1 +1/3) = (2/3)/(4/3) = (2/3)*(3/4) = 1/2.So, e = 1/2.So, e = 1/2, g = 1/3.So, now, let's check if tan(theta) = 1.From earlier,tan(theta) = | (2g +1)(g -1) / (g(g + 3)) |Plug in g = 1/3:Numerator:(2*(1/3) +1)(1/3 -1) = (2/3 +1)(-2/3) = (5/3)(-2/3) = -10/9Denominator:(1/3)(1/3 +3) = (1/3)(10/3) = 10/9So,tan(theta) = | (-10/9) / (10/9) | = | -1 | = 1So, tan(theta) = 1, which implies theta = 45 degrees.So, angle FAH is 45 degrees.Therefore, the measure of angle FAH is 45 degrees.

💡First, I need to determine A's salary based on the information provided.I know that the combined salary of A and B is 7000. Let's denote A's salary as ( A ) and B's salary as ( B ). This gives me the equation:[ A + B = 7000 ]Next, I'm told that A spends 95% of his salary, which means he saves 5% of it. Similarly, B spends 85% of his salary, saving 15% of it. Since their savings are equal, I can set up another equation:[ 0.05A = 0.15B ]To simplify this, I'll divide both sides by 0.05:[ A = 3B ]Now, I can substitute ( A = 3B ) back into the first equation:[ 3B + B = 7000 ][ 4B = 7000 ][ B = 1750 ]With B's salary determined, I can now find A's salary:[ A = 3 times 1750 = 5250 ]Therefore, A's salary is 5250.