answer:Let the side length of the inscribed square be x. In the diagram below, the triangles are drawn as blue equilateral triangles: asy size(100); pair A, B, C, D, E, F; B=(0,0); A=(0,10); D=(10,10); C=(10,0); real x = 5 -5/sqrt(3); pair E = (x,x); pair F = (10-x, 10-x); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); draw(E--F); draw((A+C)/2--(B+D)/2); label("A",A,NE); label("B",B,SW); label("C",C,SE); label("D",D,NW); label("E",E,NE); label("F",F,SE); label("x",(A+C)/2,NW); /asy As we can see, one side of the inscribed square has length x, and the other sides of the inscribed square have length x + x/2, the length of the overlap between the triangles. To maximize the area of the inscribed square, we need to draw it in a way that minimizes this overlap. By maximizing x (but not too much that it intersects any part of the blue triangle), we can minimize the overlap. Therefore, we draw x as long a possible, but not to the point of touching the blue triangle. The largest value of x can be determined by equating the side length of the square (i.e., 10) with the sum of the side length of the square minus 1/2 of the overlap and the side length of the overlap. Thus, x = 10 - (10/2). So, the largest x possible is 10 - (10/2) = 5. Therefore, the side length of the largest inscribed square is 5.,The largest possible square is the square with one vertex on the triangles' coincident vertices and with sides parallel to and coincident with those of the big square. There are two of them. We draw them in and label the diagram as shown: asy size(150); pair A, B, C, D, E, F; B=(0,0); A=(0,10); D=(10,10); C=(10,0); real x = 5 -5/sqrt(3); pair E = (x,x); pair F = (10-x, 10-x); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); draw(B--D,dashed); pair P=(0,x); pair Q=(x,0); draw(P--E--Q); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,NNE); label("F",F,SSW); label("P",P,W); label("Q",Q,S); draw((10,10-x)--(10-x,10-x)--(10-x,10)); draw(A--C,dashed); label("M",(5,5),W); /asy First, we find the side length of the equilateral triangle. M is the midpoint of EF; let MF=x, so AM=MC=xsqrt{3} and AC=2xsqrt{3}. AC is the diagonal of ABCD and thus has length 10sqrt{2}. So we have 2xsqrt{3}=10sqrt{2}. It follows that the side length of the triangle is 2x=frac{10sqrt{2}}{sqrt{3}}. Now, look at diagonal BD and notice that it is made up of twice the diagonal of the small square plus the side length of the triangle. Let the side length of the small square be y, so we have BD=BE+EF+FD=ysqrt{2}+frac{10sqrt{2}}{sqrt{3}}+ysqrt{2}=10sqrt{2}. Solving yields ysqrt{2}=5sqrt{2}-frac{5sqrt{2}}{sqrt{3}} so y=5-frac{5sqrt{3}}{3}., The side length of the largest square that can be inscribed in the space inside the square and outside of the triangles is 5(2 - sqrt(3)). Here's the reasoning: 1. The equilateral triangles have a height of 10sqrt(3)/2, so their base length is 10 - (10sqrt(3)/2). 2. The base of each equilateral triangle is shared by a right triangle with a hypotenuse of length 10 and a leg of length (10 - (10sqrt(3)/2)). 3. Using the Pythagorean theorem, we can find the other leg of this right triangle, which is the side length of the square we're looking for: sqrt(10^2 - (10 - (10sqrt(3)/2))^2). 4. Simplifying this expression, we get 5(2 - sqrt(3)). Therefore, the side length of the largest square that can be inscribed in the space inside the square and outside of the triangles is 5(2 - sqrt(3)).

answer:The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with 12 equilateral pentagons) in which the 20 vertices have all been truncated to form 20 equilateral triangles with common vertices. The resulting solid has then p=12 smaller equilateral pentagons and t=20 equilateral triangles yielding a total of t+p=F=32 faces. In each vertex, T=2 triangles and P=2 pentagons are concurrent. Now, the number of edges E can be obtained if we count the number of sides that each triangle and pentagon contributes: E=frac{3t+5p}{2}, (the factor 2 in the denominator is because we are counting twice each edge, since two adjacent faces share one edge). Thus, E=60. Finally, using Euler's formula we have V=E-30=30. In summary, the solution to the problem is 100P+10T+V=250.,We have a convex polyhedron with 32 faces, each of which is either a triangle or a pentagon, so the total of triangles and pentagons sums up to 32. Furthermore, we know that around each vertex, T triangular faces and P pentagonal faces meet. This implies that T+P=32. Now, Euler's formula for a convex polyhedron states that V-E+F=2. In this scenario, F=T+P which means that the number of faces equals the sum of triangles and pentagons (because our polyhedron only has triangles and pentagons for faces). So, we can rewrite Euler's formula as V-E+(T+P)=2. But, we also know that T+P=32 so we can substitute T+P with 32 in the previous formula. Thus, we have V-E+32=2. Dividing both sides of the equation by 2, we get V-E=30. From here, we are asked to find the value of 100P+10T+V. As seen, we already know the value of T+P or V-E, so we only need to find T and P separately. We can notice a relationship between each vertex of the polyhedron and the faces surrounding it. If a vertex is surrounded by T triangles and P pentagons, then it means that edge linking each two consecutive faces surrounding it is shared by one triangle and one pentagon. If we sum the number of edges for all the faces connected to a vertex, the result is T+2P. This is because each face has three edges, and each pair of consecutive faces has one edge in common. Since there are T triangles and P pentagons around the vertex, and each pentagon has 5 edges, while each triangle has 3, the total number of edges at the vertex can be calculated as: E = (T+2P) cdot 3/2. Now, substituting this in V-E=30, we get V = 30 + (T+2P) cdot 3/2. Multiplying both sides by 2, we obtain 2V = 60 + (T+2P). We also know that T+P=32, so we can substitute it in the above equation to get 2V = 60 + (T+2P). Substituting, we have 2V = 60 + 32K = 30 + (T+2P) cdot 3/2. Subtracting 60 from both sides, we have 2V - 30 = (T+2P) cdot 3/2, which gives us V = (T+2P) cdot 3/4. So, if we plug in the value of V, we can find the values of T and P. Plugging in V in the new equation, we get V = (T+2P) cdot 3/4. Substituting what we found in the equation above, and since we want to determine the value of 100P+10T+V, we can factor out the T and P terms from each side to isolate the coefficients: (T+2P) cdot 3/4 = 100P+10T+V. Multiplying both sides by 4/3, we get: (T+2P) cdot frac{4}{3} cdot frac{4}{3}= 100P+10T+V. Dividing both sides by frac{4}{3}, and simplifying the equation on the left-hand side by distributing the frac{4}{3}, gives: (T+2P) (4/3 cdot 4/3) = frac{10}{3}(T+2P). Canceling out (T+2P) from both sides, we have: frac{10}{3} = frac{100}{9}. Dividing everything by frac{10}{3}, we get: P = 3. Since we know that T+P=32, we can now find T: T+3 = 32. Solving for T, we get T=29. Plugging these values into the equation 100P+10T+V, we get: 100(3)+10(29)+V = 300+290+V = 28., We can use Euler's formula to find an equation involving V, E, F, T, and P. We are given that F=T+P=32. Since every edge is shared by exactly two faces, we have 2E=3T+5P because each triangle has three edges and each pentagon has five. Now we can find V-E+F=V-(3T+5P)/2+32=2. Solving for V, we get V=2(3T+5P)/2-32=3T+5P-32. Now we want to find 100P+10T+V=100P+10T+(3T+5P-32). Combining terms, we get 100P+10T+V=105P+13T-32. Note that 105P+13T is a multiple of 5 and 13, so it can never equal 32. Therefore, no such convex polyhedron exists, and the question becomes invalid.

answer:The triangle is shown below: asy pair A,B,C; A = (0,0); B = (10,0); C = (10,15); draw(A--B--C--A); draw(rightanglemark(C,B,A,26)); label("A",A,SW); label("B",B,SE); label("C",C,N); /asy We have sin A = frac{BC}{AC} and cos A = frac{AB}{AC}, so 2sin A = 3cos A gives us 2cdot frac{BC}{AC} = 3cdotfrac{AB}{AC}. Multiplying both sides by AC gives 2BC = 3AB, so AB= frac23 BC. The Pythagorean Theorem gives us AB^2 + BC^2 = AC^2. Substituting AB = frac23BC gives left(frac23BCright)^2 + BC^2 = AC^2.Simplifying the left side gives frac{13}{9}BC^2 = AC^2, so frac{BC^2}{AC^2} = frac{9}{13}, which means sin A = frac{BC}{AC} = sqrt{frac{9}{13}} = frac{sqrt{9}}{sqrt{13}} = frac{3}{sqrt{13}} = frac{3sqrt{13}}{13}}.We also could have noted that (sin A)^2 + (cos A)^2 = 1 for any angle A, so 2sin A = 3cos A gives us cos A = frac23 sin A and (sin A)^2 + left(frac23sin Aright)^2 = 1, which gives frac{13}{9}(sin A)^2= 1. Therefore, we have (sin A)^2 = frac{9}{13}. Since A is an acute angle, we have sin A > 0, so (sin A)^2 = frac{9}{13} gives us sin A = sqrt{frac{9}{13}} = frac{sqrt{9}}{sqrt{13}} = frac{3}{sqrt{13}} = boxed{frac{3sqrt{13}}{13}., Since angle B = 90^circ, we can apply the trigonometric identity for the sine of the complementary angle: sin A = cos(90^circ - A). We are given 2sin A = 3cos A, so we can rewrite this equation as follows: frac{2}{3} = frac{sin A}{cos A} = tan A. Now, recall the inverse tangent function: A = arctan{tan A}. Thus, we have A = arctanleft(frac{2}{3}right). Now, recall that sin(arctan x) = frac{x}{sqrt{x^2 + 1}}. We apply this formula to find the answer: sin A = sinleft(arctanleft(frac{2}{3}right)right) = boxed{frac{2}{sqrt{13}}}.,We have two equations: 1. 2sin A = 3cos A 2. sin^2 A + cos^2 A = 1 We can solve these two equations simultaneously to find sin A. From eq. 1 2sin A - 3cos A = 0, multiply both sides by cos A to eliminate any possibility of dividing by zero (i.e., assuming cos A neq 0): 2sin A cos A - 3cos^2 A = 0 Now substitute sin^2 A + cos^2 A = 1 from eq. 2 to transform our equation into: 2sin A (sin A + cos A) - 3cos^2 A = 0 Divide the equation by cos A to simplify: 2(sin^2 A + sin A cos A) = 3cos A Now we will use trigonometric identities, namely the sum of squares identity, sin^2 A + cos^2 A = 1, and the product of sin and cos, cos^2 A - sin^2 A = cos 2A = 1 - 2sin^2 A, which also gives us -1 + 2sin^2 A = cos 2A. We also know the double angle identity for cosine, cos 2A = 2cos^2 A - 1, so: -1 + 2sin^2 A = 2cos^2 A - 1 sin^2 A = 2cos^2 A - 2cos^2 A + 1 sin^2 A = cos^2 A sin Acos A = 0 This means sin A = 0 or cos A = 0. If cos A = 0, then angle A would be frac{pi}{2} or frac{3pi}{2}, which are not possible in a right triangle. Thus, cos A = sin A = 0. Therefore, sin A equals 0.

answer:We can determine the possible values of n by considering the angle measures in the triangle. We are given that: angle A = 2n + 12 angle B = 3n - 3 angle C = 180 - (angle A + angle B) = 180 - (2n + 12 + 3n - 3) = 180 - 5n + 9 Since angle A > angle B > angle C, we can write the following inequalities: 2n + 12 > 3n - 3 3n - 3 > 180 - 5n + 9 Solving the first inequality, we get: n < 15 Solving the second inequality, we get: 8n > 186 n > 23.25 Since n must be an integer, we have: n geq 24 Combining both inequalities, we find that the possible values of n are: 24 leq n < 15 However, this inequality contains no valid integer solutions for n. Therefore, there are no possible values of n that satisfy the given conditions.,The sides of the triangle must satisfy the triangle inequality, so AB + AC > BC, AB + BC > AC, and AC + BC > AB. Substituting the side lengths, these inequalities turn into begin{align*} (3n - 3) + (2n + 7) &> 2n + 12, (3n - 3) + (2n + 12) &> 2n + 7, (2n + 7) + (2n + 12) &> 3n - 3, end{align*} which give us n > 8/3, n > -2/3, and n > -22, respectively. However, we also want angle A > angle B > angle C, which means that BC > AC and AC > AB. These inequalities turn into 2n + 12 > 2n + 7 (which is always satisfied), and 2n + 7 > 3n - 3, which gives us n < 10. Hence, n must satisfy n > 8/3 and n < 10, which means 3 le n le 9. The number of positive integers in this interval is 9 - 3 + 1 = 7.,This seems to be a geometry problem, which I am unable to solve without knowing the specific angles and their measurements. Please provide more information or a different question.