Find the hypothesis of an obtuse and acute triangle?
463801
The cosine laws apply to triangles which are NOT right angled. I think you have maybe read that wrong:\n\nhttp://www.themathpage.com/aTrig/law-of-sines.htm\n\ngives you a full guide to using the cosine laws to solve problems like these.
1085012
Its been a long time since I took Geometry but maybe I can give you a small pointer.\n\nTry dividing the rectangle into 3 segments lengthwise and 3 segments widthwise\n\nA--B--C--D\nE--F--G--H\nJ---K--L--M\nN--O--P--Q\n\nDraw the diagonal through NKGD\n\nProve that the segments NK, KG, and DG are equal (prove that the triangles are the same)\n\nNow draw the segments from the vertex Q to the mid point of AD and AN.  Prove that the mid point of AD is the same as the mid point of BC and the same for AN and EJ.\n\nNow we have to prove that the line segment we drew in the last step instersect points K and G. You can probably do this with similar triangles (the triangle NQ and mid point of EJ is similar to JK and mid point of EJ therefore the point K is in the triangle)\n\nSo if you prove that the lines drawn from Q to the midpoints intersect K and G and we proved that NK KG and GD are congruent we have proved the original hypothesis.\n\nYou can turn this gabably gook into a real mathematical proof.\n\nGood Luck
583757
Your statement is incorrect.  It reads:\n\n"In the image below, there are right angles. How would I prove that they're right angles without a protractor in any triangle?"\n\nActually, it should read:\n\n"In the image below, there are some angles."\n\nAnything else is pure conjecture.  (That means a guess.)\n\nIf you would like us to say something about those angles, you need to tell us how the figure was constructed.  After you tell us that, and ask your question more clearly, your answer will be obtained in short order.\n\nedit...\n\nYou have added some text in your extra text file that doesn't help at all:\n\n"I know that these green points are right angles..."\n\nSo you think a point can also be an angle.  It will be a long time before you get someone (who knows and understands geometry) to agree with that.\n\nIf you start over and explain what you want to know, in correct terms, you may have a chance at getting an answer.
1002935
Its the Pythagorean Theorem. It applies to right angle triangles. \nIt states that a^2 + b^2 = c^2\nWhere a and b represent the lengths of legs of the triangle and c is the Hypotenuse.\nThe Hypotenuse is the side that is opposite the right angle (also the longest side). Whereas the legs are the two sides that meet at the right angle.
422360
A triangle is 180 degrees... so 28 + 58 = 86. so now solve this: 180-86=....94!! Tah Dah!
381599
i think its a triangle
217181
Are you looking for someone to do your homework?\nGeometry is very simple. There are three basic shapes, The sphere, Triangle and the Square. Learn that a circle is nothing more than a shadow of a sphere and that shadow is but a line. The angle is your point of view. To solve for X you must know two of the three. X is only unknown for a very short time. There are 10 dimensions. Length, width, height, area, volume, weight, mass, velocity, time, and infinity. Any math problem may be solved if two or more dimensions are known. The lowest common denominator is two and if two of anything is known the third dimension will be also known.\nie. a triangle with equal sides is composed of 3, 60 degree angles, 60 + 60 + 60 = 180 degrees. Knowing that half of a circle which is half of a sphere multiply by the common denominator of 2 and 360 appears which is a sign for infinity, through in any of the dimensions given  such as volume, mass, velocity, etc. any problem can be solved. Linear problems are solved with the triangle also. For every side there is the other. Make the triangle square or draw a square and divided it in two by a linear line from the opposing corner. Solve the problem L x W x H and you have area. Divide the area of the square by two and you have the area of the triangle. Pi is infinity, in math it is 3.14 and then some. Pi is also the Greek word for the Egyption (word) hieroglift of the lazy 8 which is a symbol for infinity also, it is a mineral we call magnetite. Magnetite in crystal form is in the shape of tetrahedron which we know as a pyramid times 2. The numeral 8 is an Aribic number which means 4 + 4. It is not of any quincedence that a pyramid has four faces. The mystery of the pyramids has been solved by Paul Einstein. Geometry is fun if you know how it came to be.
860005
I finished the problem.  The answer is s = 0.48517m.  It was hard, although now that I've worked it, it doesn't seem as difficult anymore.  Had to do some serious thinking though.\n\nThe collar A will tend to slide <down> the bar CD.  The vector CD (from C to D) is [-0.2, -0.3, 0.25] with magnitude 0.43875 and direction cosines [-0.45584, -0.68376, 0.56980].  The weight of collar A, directed downward, is 8g = 78.4N and is represented as [0, -78.4, 0].\n\nWe don't have to worry about the collar being supported by bar CD (the normal force).  The component of the collar directed down the bar is the inner (scalar, dot) product of the weight vector and the direction cosines.  This force is 78.4 x .68376 = 53.60706N.\n\nLet E be the point on the bar CD such that BE is perpendicular to CD.  Then BEA is a right triangle, and by the Pythagorean Theorem, the force along BE is sqrt(150^2 - 53.60706^2) = 140.09384N.  (I'm using numbers stored in the calculator and showing five decimal digits.  All this can be rounded off at the end.)\n\nSince the force triangle BEA is similar to the distance triangle BEA, we just have to get one of the legs to find our distance s = AB.  We will get the distance BE.\n\nTriangle BEA is in the plane defined by the points B, C, and D.  Vector CB = [-0.4, 0.2, 0.15].  The vector cross product CB x CD (see above) is normal to the plane defined by points B, C, and D.  CB x CD = [0.095, 0.07, 0.16].\n\nNow, the cross product CD x (CB x CD) lies in the plane defined by B, C, and D; is perpendicular to CD; and is parallel to the line segment BE.\n\nCD x (CB x CD) = [-0.0655, 0.05575, 0.0145] with magnitude 0.08723 and direction cosines [-0.75091, 0.63914, 0.16623].\n\nNow we can get parametric equations of the lines CEAD and BE, and find the point of intersection E.  Starting with point C and using vector CD, the first line is\n\nx = 0.4 - 0.2t\ny = 0.3 - 0.3t\nz = 0 + 0.25t\n\nStarting from point B and using the direction cosines from vector BE, the second line is\n\nx = 0 + 0.75091u\ny = 0.5 - 0.63914u\nz = 0.15 - 0.16623u\n\nSetting like terms equal to each other, we have three equations in two unknowns.  The three equations are not linearly independent.  Solving the x and y equations for t and u, we get\n\nt = 0.29871 and u = 0.45313\n\n(and these results are confirmed in the z equations).  Plugging these results into the x, y, and z equations, we find that the intersection point E is (0.34026, 0.21039, 0.07468).\n\nWe know the coordinates of point B, so by the Pythagorean Theorem, the distance BE is 0.45313m.\n\nComparing the force and distance vectors in similar triangles BEA, we have\n\ns/150 = 0.45313 / 140.09384\n\nand the length s of the cable AB is 0.48517m.  (Answer)\n\nA difficult, but fun, problem.
1015799
Use the second derivative test, not sure exactly what it is still but it should be easy to find on the internet.
516766
I'm not giving you the answer, but Cos is adjacent over hypotenuse.  Also, if you wanted the question answered correctly, us answering people might want to know where all of the letters are in relation to the triangle.
570600
Ok, first just so you get the figure right, draw dots n name them like i have n at the places like m showing them below\n      A                B\n      \n    E                   F     \n\n   D                      C             G\n\nNow, join AB, D-C-G, EF , A-E-D and B-F-C  also.. join\n A-F-G .. GET IT?? and you'll have AE equal to ED as well as BF equal to CF as the median of the trapezium joins the mid=points of the non-parallel sides...\n\nNOW IN triangle BFA and triangle CFG ,you have\nBF congruent to CF ----   Proved tht F midpoint of BC\nangle BFA congruent to angle CFG--vertically opposite angles\n and as AB is parallel to CD (opposite sides of a trapezoid) it is also parallel to CG ( construction was to join D,C and G , remember??\n\ntherefore, abgle ABF congruent to angle FCG --- alternate angles for parallel lines AB and CG ..\n\nSO YOU NOW HAVE triangle ABF congruent to triangle CFG --- Angle-Side-Angle Test of congruency...\n\ntherefore, AF congruent to FG --- corresponding sides for congruent triangles\n\nNOW, IN THE BIG TRIANGLE ADG,\n  E is the midpoint of side AD - by construction\n  F is the midpoint of AG  --- Proved AF congruent to FG\n \ntherefore... EF parallel to DG --- By the Midpoint Theorem which states that-- A line passing through the midpoints of any two sides of the triangle is parallel to the third side (and has half the length of the third side )\n\nSO WE'VE NOW PROVED EF // DG\n .\n. . , EF IS ALSO // TO DC --- D,C AND F LIE ON THE SAME LINE.. I.E. THEY ARE COLLINEAR\n\n SO, EF//DC\nBUT DC IS PARALLEL TO AB ---- OPPOSITE SIDES OF THE TRAPEZIUM ABCD\n \n .\n. . , EF IS ALSO PARALLEL TO AB\n\n .\n. ., WE CAN PROVE THAT ----> AB // EF // DC  (REASON- both AB and DC are parallel to the same linE EF)  \n                     \n\n .\n. . WE HAVE PROVED THAT THE MEDIAN OF THE TRAPEZIUM (in this case GE)  IS PARALLEL TO BOTH THE BASES (i.e AB and CD )
485045
Let a(P,Q) be the direction angle of the vector going from point P to Q measured counterclockwise from AB (so that a(A,B)=0). Then the direction angles of the bisectors going through B and C respectively are:\n\na(B,C)/2\n(a(B,C)-pi+a(A,C))/2\n\nand the angle between the two (A1)\n\nA1 = a(B,C)/2 - (a(B,C)-pi+a(A,C))/2 = pi/2 - a(A,C)/2\n\nIn terms of the original angle at A (which is a(A,C)) this is:\n\nA1 = pi/2 - A/2\n\nThat is, the recursion relation is:\n\nA_(n+1) = pi/2 - A_n/2\n\nSubstituting A_n = pi/3 + X_n, gives:\n\nX_(n+1) = - (1/2)*X_n\n\nSo clearly X_n = (-1/2)^n * Constant\n-->\nA_n = pi/3 + (-1/2)^n * Constant\n\nSince A_0 = A , the constant is: A-pi/3\n\n--> answer 4\n\nTo guess the answer without doing any work, just note that A_0 must be A, so only 3 and 4 are eligible. It's easy to see A_n can't be a constant just by sketching a triangle with a very obtuse angle at A.
865033
okay. so. there's a special set of rules for a 30-60-90 triangle.\n\nthe sides are proportional to each other by the ratio \n\nx : v/( 3)x : 2x\n\nwhere the v/ is the square root.\n\nthat means that the greatest angle corresponds to the greatest length.\n\nbecause the longest side is 7.5 meters, that means that is 2x.\n\ndivide by two for x, and that is the 30* length. (because its the smallest angle, therefore is opposite the smallest length)
787829
SOHCAHTOA\nsinθ = O/H = a/c\ncosθ = A/H = b/c\ntanθ = O/A = a/b
735406
Hint: the three side are equal long.\n\nTh
447231
Aha, geometry!\n\nDraw a diagram and follow the steps below.\n\nThe trick is to show that the triangles DBE and \ntriangle DAF are similar.\n\nProof That triangles DBE and triangle DAF are similar.\n\nLet <ABC> donote angle ABC.\nWe need to show any two corresponding angles of these traingles are equal.\n\nFirst angle,\n<FDA> = <BDE>  because D,B,F collinear and D,E,A collinear.\n\nSecond angle,\n<DAF> = <DAC> because A,F,C collinear\n      = (pi) - <ACD> - <ADC> angles in triangle ACD\n      = (pi) - <ACD> - <ADC>  ...(1)\n\nNow, separately  manipulate,<ACD> and <ADC>.\n\n<ACD> = <ACB> + <BCD> splitting <ACD> with line CB\n      = <ACB> + <CAB>, tangent property <BCD> = <CAB> \n      = (pi) - <CBA>  angles in triangle ABC\n      = <ABE>  because C,B,E are all colinear\n\nNow, similarly\n\n<ADC> = <ADB> + <BDC> splitting <ADC> with line DB\n      = <ADB> + <BAD>, tangent property <BDC> = <BAD> \n      = (pi) - <ABD>  angles in triangle ABD\n         \nSubstituiting these two info into (1), we get,  \n\n<DAF> = (pi) - <ACD> - <ADC>\n      = (pi) - <ABE> - [ (pi) - <ABD> ]\n      = <ABD> - <ABE>\n      = <BDE> because BE splits <ABD> \n\nDone. In showing the triangle DAF and triangle DBE are similar.\n\nHence, by ratio for similar triangles,\n\n        DA/DF = DB/DE\n\nOr Rewriting we get\n \n      DA * DE = DB * DF\n\nProved.\n\nNot so difficult, was it. hehehe.\n\nRegards
1068661
Perimeter of the triangle multiplied by the height of the prism.