💡First, I recognize that the base of the hemisphere is a circle, and the area of a circle is given by the formula ( A = pi r^2 ). Given that the area of the base is 3, I can set up the equation:[pi r^2 = 3]To solve for ( r^2 ), I divide both sides of the equation by ( pi ):[r^2 = frac{3}{pi}]Next, I need to find the surface area of the hemisphere. The surface area of a hemisphere includes both the curved surface area and the base area. The curved surface area of a hemisphere is half the surface area of a full sphere, which is given by ( 2pi r^2 ). The base area is the area of the circle, which is ( pi r^2 ).Therefore, the total surface area of the hemisphere is:[text{Total Surface Area} = 2pi r^2 + pi r^2 = 3pi r^2]Substituting the value of ( r^2 ) from earlier:[text{Total Surface Area} = 3pi left( frac{3}{pi} right) = 9]So, the surface area of the hemisphere is 9.

💡First, I need to solve the inequality (2(x + 1) geq 5x - 3).I'll start by expanding the left side of the inequality:[2(x + 1) = 2x + 2]So, the inequality becomes:[2x + 2 geq 5x - 3]Next, I'll move all terms containing (x) to one side and the constant terms to the other side. Subtracting (2x) from both sides gives:[2 geq 3x - 3]Then, adding 3 to both sides:[5 geq 3x]Now, I'll solve for (x) by dividing both sides by 3:[x leq frac{5}{3}]Since we're looking for positive integer solutions, the only integer that satisfies (x leq frac{5}{3}) is:[x = 1]

💡Okay, so I need to factor the polynomial ( w^4 - 81 ) as far as possible, and the factors should be monic polynomials with real coefficients. Hmm, let me think about how to approach this.First, I remember that factoring polynomials often starts with looking for common factors or recognizing patterns. In this case, ( w^4 - 81 ) looks like a difference of squares because both ( w^4 ) and 81 are perfect squares. Specifically, ( w^4 ) is ( (w^2)^2 ) and 81 is ( 9^2 ). So, I can apply the difference of squares formula, which is ( a^2 - b^2 = (a - b)(a + b) ).Applying that here, I get:[w^4 - 81 = (w^2)^2 - 9^2 = (w^2 - 9)(w^2 + 9)]Alright, so now I have two factors: ( w^2 - 9 ) and ( w^2 + 9 ). Let me check if these can be factored further.Starting with ( w^2 - 9 ), that's another difference of squares because ( w^2 ) is ( w^2 ) and 9 is ( 3^2 ). So, applying the difference of squares formula again:[w^2 - 9 = (w - 3)(w + 3)]Great, so now ( w^2 - 9 ) is factored into linear terms. What about ( w^2 + 9 )? Hmm, that's a sum of squares, right? I remember that over real numbers, a sum of squares doesn't factor into real linear factors because it doesn't have real roots. But wait, can it be factored into quadratic terms with real coefficients?Let me think. If I try to factor ( w^2 + 9 ), I might consider using complex numbers, but the problem specifies real coefficients. So, since ( w^2 + 9 ) can't be factored further into real polynomials of lower degree, I guess that's as far as I can go for that part.Putting it all together, the original polynomial factors as:[w^4 - 81 = (w^2 - 9)(w^2 + 9) = (w - 3)(w + 3)(w^2 + 9)]So, the complete factorization into monic polynomials with real coefficients is ( (w - 3)(w + 3)(w^2 + 9) ).Wait, just to make sure I didn't miss anything, let me verify each step. First, recognizing ( w^4 - 81 ) as a difference of squares is correct. Then, factoring ( w^2 - 9 ) further is also correct because it's another difference of squares. And ( w^2 + 9 ) can't be factored further with real coefficients because it doesn't have real roots. So, yes, I think that's the most factored form possible under the given conditions.I don't think I made any mistakes here. It all seems to follow logically. So, I'm confident that the factorization is correct.**Final Answer**The polynomial ( w^4 - 81 ) factors completely as boxed{(w - 3)(w + 3)(w^2 + 9)}.

💡Okay, so I have this problem about an arithmetic sequence. Let me read it again: "Given an arithmetic sequence {a_n}, the sum of the first n terms is denoted as S_n. If a_4 = 4, then S_7 = ______." Hmm, alright, I need to find the sum of the first 7 terms given that the fourth term is 4.First, let me recall what an arithmetic sequence is. It's a sequence where each term after the first is obtained by adding a constant difference. So, if the first term is a_1 and the common difference is d, then the nth term a_n can be expressed as a_n = a_1 + (n-1)d. That makes sense.Given that a_4 = 4, I can plug that into the formula. So, a_4 = a_1 + (4-1)d = a_1 + 3d = 4. Okay, so that's one equation: a_1 + 3d = 4. I need another equation to solve for both a_1 and d, but maybe I don't need to find both? Let me think.The problem is asking for S_7, the sum of the first 7 terms. I remember that the sum of the first n terms of an arithmetic sequence is given by S_n = n/2 * (2a_1 + (n-1)d) or alternatively, S_n = n * (a_1 + a_n)/2. Both formulas are equivalent because a_n = a_1 + (n-1)d.So, for S_7, I can use either formula. Let me try the second one since it might be simpler: S_7 = 7 * (a_1 + a_7)/2. Hmm, but I don't know a_7 yet. Wait, but maybe I can express a_7 in terms of a_1 and d.Since a_n = a_1 + (n-1)d, then a_7 = a_1 + 6d. So, S_7 = 7 * (a_1 + a_1 + 6d)/2 = 7 * (2a_1 + 6d)/2 = 7 * (a_1 + 3d). Wait a minute, that's interesting. So, S_7 = 7*(a_1 + 3d). But from earlier, I know that a_1 + 3d = 4 because a_4 = 4. So, substituting that in, S_7 = 7*4 = 28.Oh, that's neat! So, I didn't even need to find a_1 or d individually. I just used the fact that a_4 is the average of a_1 and a_7, and since the sum is 7 times that average, it simplifies directly to 28.Let me double-check using the other formula for the sum to make sure I didn't make a mistake. The first formula is S_n = n/2 * (2a_1 + (n-1)d). So, plugging in n=7, we get S_7 = 7/2 * (2a_1 + 6d). Simplify that: 7/2 * 2*(a_1 + 3d) = 7*(a_1 + 3d). Again, since a_1 + 3d = 4, that gives S_7 = 7*4 = 28. Yep, same result.Just to be thorough, let me consider if there's another way to approach this. Maybe using the properties of arithmetic sequences. I remember that in an arithmetic sequence, the sum of terms equidistant from the ends is constant. So, for example, a_1 + a_7 = a_2 + a_6 = a_3 + a_5 = 2a_4. Since there are 7 terms, which is odd, the middle term is the 4th term, which is a_4. So, the sum S_7 would be 7 times the middle term, which is a_4. Therefore, S_7 = 7*4 = 28. That's another way to see it.So, all methods point to the same answer, 28. I think that's solid. I don't see any mistakes in my reasoning, and multiple approaches confirm the result. Therefore, I'm confident that S_7 is 28.

💡Alright, so I've got this problem where two identical triangles are positioned inside a square, and I need to find their angles. Hmm, okay, let me try to visualize this. Since it's a square, all sides are equal, and all angles are 90 degrees. Now, the triangles are identical, which means they have the same shape and size. First, I need to figure out how these triangles are placed inside the square. Are they placed in a way that they share a common side or maybe a common vertex? The problem mentions a diagram, but since I don't have it, I'll have to make some assumptions. Maybe the triangles are placed such that one vertex of each triangle is at a corner of the square, and the other vertices are somewhere along the sides or the opposite corners.Let me consider the square ABCD, with vertices A, B, C, D. Suppose one triangle is placed with vertex A and extends to some points on sides AB and AD. The other triangle might be placed similarly at another corner, say vertex C, extending to points on sides CB and CD. Since the triangles are identical, their corresponding sides and angles should be equal.Now, if I think about the angles of these triangles, they must add up to 180 degrees, as with any triangle. Since the square has right angles, some of the angles of the triangles might be related to those. Maybe one of the angles of the triangle is 45 degrees, which is half of 90 degrees, but I'm not sure yet.Let me try to break it down. If the triangles are identical and positioned inside the square, perhaps they are congruent triangles. Congruent triangles have all corresponding sides and angles equal. So, if I can figure out one angle, I can find the others.Wait, maybe the triangles are right-angled triangles? If that's the case, one of their angles is 90 degrees, and the other two angles add up to 90 degrees. But since they are identical and placed inside a square, the other angles might be equal, making them 45 degrees each. That would make sense because 45 degrees is half of 90 degrees, and it would fit neatly inside the square.But hold on, if both triangles are right-angled and have angles of 45 degrees, wouldn't they each occupy a corner of the square, creating a sort of diagonal division? That might leave some space in the middle of the square, but the problem says they are positioned inside the square, not necessarily covering the entire area.Alternatively, maybe the triangles are not right-angled. Perhaps they are isosceles triangles with two equal sides and two equal angles. If that's the case, the base angles would be equal, and the vertex angle would be different. Given the symmetry of the square, it's possible that the triangles are arranged symmetrically, which would support the idea of them being isosceles.Let me think about the possible angles. If the triangles are isosceles and positioned inside the square, maybe their base is along one side of the square, and the two equal sides extend towards the opposite sides. In that case, the base angles would be equal, and the vertex angle would be different.Suppose the base of each triangle is along the bottom side of the square, from corner A to some point M on the top side. Then, the two equal sides would go from M to corners B and D. If that's the case, then the triangle would have a base angle at A and two equal angles at M and some other point.Wait, this is getting a bit confusing. Maybe I should try to sketch it out mentally. Imagine the square with points A, B, C, D. If I place a triangle with vertex A and extend it to points on sides AB and AD, and another triangle with vertex C extending to points on sides CB and CD, these triangles might intersect somewhere inside the square.If the triangles are identical, their points of intersection should be symmetrical. Maybe they intersect at the center of the square. If that's the case, then the angles at the center would be equal for both triangles.Hmm, okay, let's consider that the triangles intersect at the center of the square. The center point would then be a common vertex for both triangles. Since the square's center is equidistant from all sides, the triangles would be mirror images of each other across the center.In that case, each triangle would have one vertex at a corner of the square, one at the center, and one somewhere along a side. The angles at the corner and the center would determine the shape of the triangle.Let me try to calculate the angles. If the triangle has vertices at corner A, center O, and some point M on side AB, then the sides AO and OM would be equal if the triangle is identical to the one at corner C. Wait, no, AO is a diagonal from corner A to center O, and OM would be from center O to point M on side AB.Actually, AO is half the length of the diagonal of the square. If the square has side length 'a', then the diagonal is a√2, so AO is (a√2)/2 = a/√2. The length of OM would depend on where M is located on side AB.If the triangles are identical, then the lengths from O to M and from O to the corresponding point on side AD should be equal. That would mean that point M is the midpoint of side AB, right? Because if M is the midpoint, then OM would be equal to the distance from O to the midpoint of AD.Wait, but if M is the midpoint of AB, then AM is a/2. Then, triangle AOM would have sides AO = a/√2, OM = a/2, and AM = a/2. Let me check if this satisfies the triangle inequality.AO = a/√2 ≈ 0.707aOM = a/2 = 0.5aAM = a/2 = 0.5aSo, AO < OM + AM: 0.707a < 0.5a + 0.5a = a, which is true.OM < AO + AM: 0.5a < 0.707a + 0.5a = 1.207a, which is true.AM < AO + OM: 0.5a < 0.707a + 0.5a = 1.207a, which is true.Okay, so triangle AOM is a valid triangle. Now, let's find its angles.Using the Law of Cosines to find angle at A:cos(angle at A) = (AO² + AM² - OM²) / (2 * AO * AM)Plugging in the values:AO² = (a/√2)² = a²/2AM² = (a/2)² = a²/4OM² = (a/2)² = a²/4So,cos(angle at A) = (a²/2 + a²/4 - a²/4) / (2 * (a/√2) * (a/2))= (a²/2) / (2 * (a²)/(2√2))= (a²/2) / (a²/√2)= (1/2) / (1/√2)= (1/2) * (√2/1)= √2/2 ≈ 0.7071So, angle at A is arccos(√2/2) which is 45 degrees.Okay, so angle at A is 45 degrees. Now, let's find angle at O.Again, using the Law of Cosines:cos(angle at O) = (AO² + OM² - AM²) / (2 * AO * OM)Plugging in the values:AO² = a²/2OM² = a²/4AM² = a²/4So,cos(angle at O) = (a²/2 + a²/4 - a²/4) / (2 * (a/√2) * (a/2))= (a²/2) / (2 * (a²)/(2√2))= (a²/2) / (a²/√2)= (1/2) / (1/√2)= √2/2 ≈ 0.7071So, angle at O is also 45 degrees.Wait, that can't be right because the sum of angles in a triangle is 180 degrees. If two angles are 45 degrees each, the third angle would be 90 degrees. But in our case, the triangle AOM has sides AO, OM, and AM with lengths a/√2, a/2, and a/2. So, if two angles are 45 degrees, the third angle would indeed be 90 degrees.But in our calculation, both angles at A and O came out to be 45 degrees, which would make angle at M equal to 90 degrees. But in reality, point M is on side AB, so angle at M should be the angle between sides AM and OM.Wait, maybe I made a mistake in assigning the sides. Let me double-check.In triangle AOM, vertex A is at corner A, vertex O is at the center, and vertex M is on side AB.So, sides:AO: from A to O, length a/√2OM: from O to M, length a/2AM: from A to M, length a/2So, sides AO, OM, and AM are as I mentioned.Using the Law of Cosines for angle at A:cos(angle at A) = (AO² + AM² - OM²) / (2 * AO * AM)= (a²/2 + a²/4 - a²/4) / (2 * (a/√2) * (a/2))= (a²/2) / (a²/√2)= (1/2) / (1/√2)= √2/2So, angle at A is 45 degrees.Similarly, for angle at O:cos(angle at O) = (AO² + OM² - AM²) / (2 * AO * OM)= (a²/2 + a²/4 - a²/4) / (2 * (a/√2) * (a/2))= (a²/2) / (a²/√2)= √2/2So, angle at O is also 45 degrees.Therefore, angle at M is 180 - 45 - 45 = 90 degrees.Wait, but point M is on side AB, so angle at M is the angle between AM and OM. If angle at M is 90 degrees, that would mean that OM is perpendicular to AM. But since AM is along side AB, which is horizontal, OM would have to be vertical. However, O is the center of the square, so OM would be a line from the center to the midpoint of AB, which is indeed vertical if AB is the top side.Wait, no, if AB is the top side, then OM would be going from center O to midpoint M of AB, which is horizontal. Wait, no, in a square, the center is equidistant from all sides, so if AB is the top side, then the midpoint M of AB is directly above the center O. So, the line OM would be vertical.But in that case, if OM is vertical and AM is along AB, which is horizontal, then the angle between AM and OM at point M would indeed be 90 degrees. So, that makes sense.Therefore, triangle AOM has angles of 45 degrees at A, 45 degrees at O, and 90 degrees at M.But wait, the problem says there are two identical triangles inside the square. If triangle AOM is one of them, then the other triangle would be triangle COM, where C is the opposite corner. But triangle COM would be congruent to triangle AOM, so it would have the same angles: 45 degrees at C, 45 degrees at O, and 90 degrees at M'.But in this case, the two triangles would overlap at the center O, and their right angles would be at points M and M' on sides AB and CD respectively.However, the problem states that the two triangles are positioned inside the square, but it doesn't specify whether they overlap or not. If they overlap, then the angles would still be 45 degrees, 45 degrees, and 90 degrees.But maybe the triangles are placed differently. Perhaps they are not right-angled. Maybe they are equilateral triangles, but that wouldn't fit inside a square neatly. Or maybe they are acute or obtuse triangles.Wait, another thought. If the two triangles are placed such that their bases are along the diagonals of the square, then their angles could be different. For example, if each triangle has a vertex at a corner and extends along the diagonal to the opposite corner, but that would make them larger than the square, which doesn't make sense.Alternatively, maybe the triangles are placed with their bases along the sides of the square and their apexes meeting at the center. In that case, each triangle would have a base angle at the corner, a base angle at the midpoint of the side, and a vertex angle at the center.If that's the case, then the base angles would be equal, and the vertex angle would be different. Let's calculate those angles.Assuming the square has side length 'a', the distance from a corner to the center is a√2/2, as before. The distance from the center to the midpoint of a side is a/2.So, if we have a triangle with vertices at corner A, midpoint M of side AB, and center O, then the sides are:AM = a/2AO = a√2/2OM = a/2Wait, that's the same triangle as before, triangle AOM. So, it still has angles of 45 degrees at A and O, and 90 degrees at M.Therefore, regardless of how I place the triangles, if they are identical and positioned inside the square with vertices at the corners and midpoints, their angles would be 45 degrees, 45 degrees, and 90 degrees.But the problem says "two identical triangles are positioned inside a square," and it doesn't specify that they form a particular shape or cover the entire square. So, perhaps the triangles are arranged differently.Another possibility is that the two triangles together form the square when combined. That is, each triangle is half of the square, making them right-angled triangles with angles 45 degrees, 45 degrees, and 90 degrees.But the problem says "two identical triangles are positioned inside a square," not necessarily that they form the square. So, they could be smaller triangles inside the square.Wait, maybe the triangles are placed such that each has a vertex at a corner of the square and extends to the midpoints of the adjacent sides. For example, triangle ABD, where D is the midpoint of side AB, and triangle CBD, where D is the midpoint of side CB. But in that case, the triangles would overlap at the center.Alternatively, perhaps the triangles are placed with their bases along the sides of the square and their apexes meeting at the center, but not necessarily at the midpoints.Wait, let me think differently. Suppose the two triangles are placed such that each has a vertex at two adjacent corners of the square and their apexes somewhere inside the square. For example, one triangle has vertices at A, B, and some point M inside, and the other triangle has vertices at C, D, and the same point M inside. If these triangles are identical, then point M must be positioned such that the triangles are congruent.In this case, the triangles would share the common vertex M inside the square, and their other vertices would be at the corners A, B and C, D respectively.Since the triangles are identical, the distances from M to A, B, C, and D must be equal in some way. But since A and B are adjacent corners, and C and D are adjacent corners opposite to A and B, the distances from M to A and M to B must be equal, and similarly, the distances from M to C and M to D must be equal.But for the triangles to be identical, the distances from M to A and M to C must be equal, and the distances from M to B and M to D must be equal. This would imply that M is at the center of the square.So, if M is at the center, then the triangles would be congruent, with sides MA = MC and MB = MD.In this case, each triangle would have sides MA, MB, and AB for triangle ABM, and sides MC, MD, and CD for triangle CDM.Since MA = MC and MB = MD, and AB = CD, the triangles would indeed be congruent.Now, let's find the angles of triangle ABM.Since M is the center, MA = MB = a√2/2, where 'a' is the side length of the square.Side AB is of length 'a'.So, triangle ABM has sides:AB = aAM = a√2/2BM = a√2/2So, it's an isosceles triangle with two sides equal to a√2/2 and base a.Let's find the angles.Using the Law of Cosines to find angle at M:cos(angle at M) = (AM² + BM² - AB²) / (2 * AM * BM)= ((a√2/2)² + (a√2/2)² - a²) / (2 * (a√2/2) * (a√2/2))= ( (2a²/4) + (2a²/4) - a² ) / (2 * (2a²/4))= ( (a²/2 + a²/2 - a²) ) / (2 * (a²/2))= ( (a² - a²) ) / (a²)= 0 / a²= 0So, angle at M is arccos(0) = 90 degrees.Now, let's find angle at A.Using the Law of Cosines:cos(angle at A) = (AB² + AM² - BM²) / (2 * AB * AM)= (a² + (a√2/2)² - (a√2/2)²) / (2 * a * (a√2/2))= (a² + 0) / (2 * a * (a√2/2))= a² / (a²√2)= 1/√2 ≈ 0.7071So, angle at A is arccos(1/√2) = 45 degrees.Similarly, angle at B would also be 45 degrees, since the triangle is isosceles.Therefore, triangle ABM has angles of 90 degrees at M and 45 degrees at A and B.But wait, the problem says "two identical triangles are positioned inside a square." If each triangle has angles of 90 degrees, 45 degrees, and 45 degrees, then both triangles would have the same angles.However, in this configuration, the two triangles together would form the entire square, with their hypotenuses along the diagonals. But the problem says they are positioned inside the square, not necessarily forming the square.Alternatively, maybe the triangles are placed such that they don't overlap and don't form the entire square. Perhaps they are smaller triangles inside the square, each occupying a corner.Wait, another thought. Suppose each triangle is placed in a corner of the square, with one vertex at the corner and the other two vertices along the adjacent sides. If the triangles are identical, their positions would be symmetrical.Let me consider triangle AED, where E is a point along side AB and D is a point along side AD. Similarly, triangle BFC, where F is a point along side BC and C is a point along side CD. If these triangles are identical, then AE = BF and ED = FC.In this case, the angles of the triangles would depend on the positions of E and F.But without specific information about where E and F are located, it's hard to determine the exact angles. However, if the triangles are identical and placed symmetrically, their angles would be equal.Assuming that the triangles are right-angled, with the right angle at the corner of the square, then their other angles would be acute angles adding up to 90 degrees. If they are isosceles right-angled triangles, then the other two angles would be 45 degrees each.But if they are not isosceles, the angles could be different. For example, one angle could be 30 degrees, and the other 60 degrees, making the triangle a 30-60-90 triangle.However, since the problem doesn't provide specific information about the positions or side lengths, I have to make assumptions based on typical configurations.Given that the square has right angles, and the triangles are identical and placed inside, the most straightforward assumption is that they are isosceles right-angled triangles with angles of 45 degrees, 45 degrees, and 90 degrees.But earlier, I considered triangles with angles 45, 45, and 90 degrees, but the problem specifies two triangles inside the square, not necessarily forming the square. So, perhaps they are smaller triangles with different angles.Wait, another approach. Let's consider that the two triangles are placed such that each has a vertex at the midpoint of a side and extends to two adjacent corners. For example, triangle AMB, where M is the midpoint of side AB, and triangle CMD, where M is the midpoint of side CD.In this case, each triangle would have sides:For triangle AMB:- AM = MB = a/2- AB = aSo, triangle AMB is an isosceles triangle with two sides of length a/2 and base a.Using the Law of Cosines to find the angles:First, angle at M:cos(angle at M) = (AM² + BM² - AB²) / (2 * AM * BM)= ((a/2)² + (a/2)² - a²) / (2 * (a/2) * (a/2))= (a²/4 + a²/4 - a²) / (a²/2)= (a²/2 - a²) / (a²/2)= (-a²/2) / (a²/2)= -1So, angle at M is arccos(-1) = 180 degrees, which doesn't make sense because it would mean the points are colinear, which they are not.Wait, that can't be right. If M is the midpoint of AB, then triangle AMB would actually be a straight line, not a triangle. So, that configuration doesn't work.Therefore, the triangles must be placed differently.Perhaps each triangle has vertices at two adjacent corners and a point inside the square. For example, triangle ABD, where D is inside the square, and triangle BCD, where D is the same point inside the square.In this case, the two triangles share the common vertex D inside the square.Since the triangles are identical, the distances from D to A, B, and C must be equal in some way.But without specific information about the position of D, it's hard to determine the angles.Alternatively, maybe the triangles are placed such that each has a vertex at a corner and extends to the midpoints of the opposite sides.For example, triangle AEC, where E is the midpoint of side BC, and triangle AFD, where F is the midpoint of side CD.But again, without specific information, it's difficult to determine the angles.Given the lack of a diagram, I have to rely on common configurations. The most likely scenario is that the two triangles are congruent right-angled triangles with angles of 45 degrees, 45 degrees, and 90 degrees, positioned at opposite corners of the square.Therefore, their angles are 45 degrees, 45 degrees, and 90 degrees.But wait, earlier I considered triangles with angles 45, 45, and 90 degrees, but the problem specifies two identical triangles inside the square, not necessarily right-angled.Another possibility is that the triangles are equilateral triangles, but that wouldn't fit inside a square neatly because equilateral triangles have 60-degree angles, which don't align with the square's 90-degree angles.Alternatively, the triangles could be acute or obtuse triangles with angles that complement the square's angles.Wait, perhaps the triangles are placed such that their bases are along the sides of the square, and their apexes meet at points inside the square, creating angles that are not right angles.For example, if each triangle has a base along the top and bottom sides of the square and their apexes meet at the center, then the triangles would be congruent and their angles would depend on the height from the base to the apex.But in this case, the triangles would overlap at the center, and their angles would be determined by the proportions of the square.However, without specific measurements, it's hard to calculate the exact angles.Given all these possibilities, the most straightforward answer, assuming the triangles are right-angled and isosceles, is that their angles are 45 degrees, 45 degrees, and 90 degrees.But earlier, I considered that if the triangles are placed with their vertices at the corners and midpoints, their angles would be 45, 45, and 90 degrees. However, if they are placed differently, the angles could be different.Wait, another thought. If the two triangles are placed such that each has a vertex at a corner of the square and extends to the midpoints of the opposite sides, then their angles would be different.For example, triangle AEC, where E is the midpoint of side BC, and triangle AFD, where F is the midpoint of side CD.In this case, the triangles would have sides:For triangle AEC:- AE: from A to E, which is a diagonal from corner A to midpoint of BC.- EC: from E to C, which is half the side length.- AC: the diagonal of the square.Wait, but AC is the diagonal of the square, which is longer than AE and EC.Wait, no, if E is the midpoint of BC, then EC is a/2, and AE is the distance from A to E.Using coordinates might help. Let's assign coordinates to the square:Let’s place the square ABCD with A at (0,0), B at (a,0), C at (a,a), and D at (0,a).Then, the midpoint E of BC would be at (a, a/2).So, triangle AEC has vertices at A(0,0), E(a, a/2), and C(a,a).Let's calculate the lengths of the sides:AE: distance from A to E = √[(a - 0)² + (a/2 - 0)²] = √(a² + a²/4) = √(5a²/4) = (a√5)/2EC: distance from E to C = distance from (a, a/2) to (a,a) = √[(a - a)² + (a - a/2)²] = √[0 + (a/2)²] = a/2AC: distance from A to C = diagonal of the square = a√2So, triangle AEC has sides of length (a√5)/2, a/2, and a√2.Now, let's find the angles using the Law of Cosines.First, let's find angle at A.cos(angle at A) = (AE² + AC² - EC²) / (2 * AE * AC)= [(5a²/4) + (2a²) - (a²/4)] / (2 * (a√5/2) * (a√2))= [(5a²/4 + 8a²/4 - a²/4)] / (2 * (a²√10)/2)= [(12a²/4)] / (a²√10)= [3a²] / (a²√10)= 3/√10 ≈ 0.9487So, angle at A is arccos(3/√10) ≈ 18.43 degrees.Next, angle at E.cos(angle at E) = (AE² + EC² - AC²) / (2 * AE * EC)= [(5a²/4) + (a²/4) - (2a²)] / (2 * (a√5/2) * (a/2))= [(6a²/4 - 8a²/4)] / (2 * (a²√5)/4)= [(-2a²/4)] / (a²√5/2)= (-a²/2) / (a²√5/2)= -1/√5 ≈ -0.4472So, angle at E is arccos(-1/√5) ≈ 116.57 degrees.Finally, angle at C.Since the sum of angles in a triangle is 180 degrees, angle at C = 180 - 18.43 - 116.57 ≈ 45 degrees.Wait, that's interesting. So, triangle AEC has angles approximately 18.43 degrees at A, 116.57 degrees at E, and 45 degrees at C.But the problem states that the two triangles are identical. If triangle AEC is one of them, the other triangle would have to be congruent to it. However, in this configuration, the other triangle would be triangle AFD, which would have the same angles.But in this case, the angles are not all equal, so the triangles are not equiangular, but they are congruent in terms of side lengths and angles.However, the problem asks for the angles of the triangles, not specifying whether they are all equal or not. So, if the triangles are congruent with angles approximately 18.43 degrees, 116.57 degrees, and 45 degrees, then that would be the answer.But this seems more complicated than necessary, and the problem likely expects a simpler answer, such as 45 degrees, 45 degrees, and 90 degrees.Given that, and considering the initial assumption that the triangles are right-angled and isosceles, I think the intended answer is that the triangles have angles of 45 degrees, 45 degrees, and 90 degrees.Therefore, the angles of the two identical triangles are 45 degrees, 45 degrees, and 90 degrees.

💡Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. The problem gives me an equation: (cos A)/a + (cos B)/b = (2c cos C)/(ab). I need to find the value of angle C first, and then, given a=2 and c=√5, find the length of side b.Starting with part (1). The equation given is (cos A)/a + (cos B)/b = (2c cos C)/(ab). Hmm, maybe I can manipulate this equation to find a relationship involving angle C.First, let me write down the given equation:(cos A)/a + (cos B)/b = (2c cos C)/(ab)To make it easier, I can multiply both sides by ab to eliminate the denominators. Let's do that:ab*(cos A)/a + ab*(cos B)/b = ab*(2c cos C)/(ab)Simplifying each term:b cos A + a cos B = 2c cos COkay, so now I have b cos A + a cos B = 2c cos C. Hmm, this seems familiar. I remember in triangles, there's a formula related to the projection of sides onto each other. Specifically, in any triangle, the length of a side can be expressed as the sum of the projections of the other two sides onto it.Wait, actually, I think the formula is a = b cos C + c cos B, and similarly for the other sides. Let me verify that. Yes, in triangle ABC, each side is equal to the sum of the other two sides multiplied by the cosine of the included angle. So, a = b cos C + c cos B, b = a cos C + c cos A, and c = a cos B + b cos A.Looking back at my equation: b cos A + a cos B = 2c cos C. Hmm, if I compare this with the formula for c, which is c = a cos B + b cos A, then my equation becomes c = 2c cos C.Wait, that's interesting. So, from the given equation, after simplifying, I get c = 2c cos C. Let me write that down:c = 2c cos CAssuming c ≠ 0, which it can't be in a triangle, I can divide both sides by c:1 = 2 cos CSo, cos C = 1/2. Therefore, angle C must be the angle whose cosine is 1/2. I remember that cos(60°) = 1/2, and in radians, that's π/3. So, angle C is π/3 radians or 60 degrees.Okay, that seems straightforward. So, part (1) is solved, and angle C is π/3.Moving on to part (2). Now, I'm given that a = 2 and c = √5. I need to find the length of side b. Since I know angle C is π/3, I can use the Law of Cosines here. The Law of Cosines states that c² = a² + b² - 2ab cos C.Let me plug in the known values into this formula. So, c is √5, so c² is (√5)² = 5. a is 2, so a² is 4. Angle C is π/3, and cos(π/3) is 1/2. So, substituting these into the equation:5 = 4 + b² - 2*2*b*(1/2)Simplify the equation step by step. First, calculate 2*2*b*(1/2). Let's see, 2*2 is 4, times b is 4b, times 1/2 is 2b. So, the equation becomes:5 = 4 + b² - 2bNow, let's rearrange this equation to form a quadratic equation. Subtract 5 from both sides:0 = 4 + b² - 2b - 5Simplify the constants:4 - 5 = -1, so:0 = b² - 2b - 1Which can be written as:b² - 2b - 1 = 0Now, I need to solve this quadratic equation for b. The quadratic equation is of the form ax² + bx + c = 0, so here, a = 1, b = -2, c = -1. Using the quadratic formula:b = [-B ± √(B² - 4AC)]/(2A)Plugging in the values:b = [2 ± √((-2)² - 4*1*(-1))]/(2*1)Calculate the discriminant:(-2)² = 4, and 4*1*(-1) = -4, so:√(4 - (-4)) = √(4 + 4) = √8 = 2√2So, the solutions are:b = [2 ± 2√2]/2Simplify by dividing numerator and denominator by 2:b = 1 ± √2Now, since b is a length of a side in a triangle, it must be positive. So, we discard the negative solution:b = 1 + √2Therefore, the length of side b is 1 + √2.Wait, let me just double-check my steps to make sure I didn't make any mistakes. Starting from the Law of Cosines:c² = a² + b² - 2ab cos CPlugging in:5 = 4 + b² - 2*2*b*(1/2)Simplify:5 = 4 + b² - 2bSubtract 5:0 = b² - 2b - 1Quadratic equation:b = [2 ± √(4 + 4)]/2 = [2 ± √8]/2 = [2 ± 2√2]/2 = 1 ± √2Yes, that seems correct. Since b must be positive, 1 + √2 is the only valid solution.So, summarizing:1. Angle C is π/3 radians.2. The length of side b is 1 + √2.I think that's it. I don't see any errors in my reasoning.**Final Answer**(1) The value of angle ( C ) is (boxed{dfrac{pi}{3}}).(2) The magnitude of ( b ) is (boxed{1 + sqrt{2}}).