if two angles of one triangle are congruent to two angles of a second triangle, then the third angles of the triangles are congruent
199251
All you have to use is the fact that for any triangle, the measures of the three interior angles add to 180 degrees.  So if you have Triangles ABC and DEF, then you know:\nangleABC + angleBAC + angleBCA = 180 degrees, and\nangleDEF + angleEDF + angleEFD = 180 degrees.\n\nIf two angles are congruent, then they have the same measure in degrees.  So, if angleABC is congruent to angleDEF, and angleBAC is congruent to angleEDF, then:\nangleABC = angleDEF, and\nangleBAC = angleEDF.\n\nUsing substitution:\nangleABC + angleBAC + angleBCA = angleDEF + angleEDF + angleEFD\nangleDEF + angleBAC + angleBCA = angleDEF + angleEDF + angleEFD\nangleDEF + angleEDF + angleBCA = angleDEF + angleEDF + angleEFD.\n\nthen simplify by subtracting (angleDEF + angleEDF) from both sides:\n\nangleBCA = angleEFD.\n\ntherefore, the third angles are congruent.
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a triangle with angles 30, 60,90\nhas sides in ratio 1: (SR 3) : 2\nThe equilateral triangle can be formed from\ntwo 30,60,90 triangles.\neach of those can be formed from 3 smaller 30,60,90 triangles\nso the equilateral triangle can be formed from six 30,60,90 triangles.\nIf radius of circle is 1 unit,\neach of the six small 30,60,90 triangles has \nsides (1/2) ((SR 3)/2) 1\nso the equilateral triangle has side SR(3) units\nso if the equilateral triangle has side x, the circle has radius x /(SR(3)),\nand the circle circumference is 2 pi x /((SR 3))
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RHS cot a - tan b \n= cos a/sina - sinb/cosb \n= cosacosb - sinasinb/sinacosb\n= cos(a+b)/sinacosb\n= LHS (proven)\n\nThe idea is to prove that both sides are the same.
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26+64 = 90, and a triangle has 180 degrees, so the other angle must be 90.  A triangle with a 90 degree angle is a right-triangle.
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Two angles whose sum is 90 degrees.
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The problem is that 30, 31, and 63 are not the sides of a valid triangle.  For any triangle with sides a,b,c, you must have\na + b > c\nb + c > a\nc + a > b\nBut you have a + b < c, so the triangle isn't valid.
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sin(2A)=sin(A+A)=sinAcosA + sinAcosA=2sinAcosA
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this is difficult to do without drawing.  but here's the gist of it.\n\n1. <a + <b + <c + <d = 360  (definition of quadrilateral)\n2. <a + <b + <a + <b = 360  (since a = c and b = d)\n3. 2(a + b) = 360\n4. a + b = 180\n5. <a and <b are supplementary angles\n\nnow extend the line AB and include a new point X, so you have a picture that looks like:\n\nX\n\nA   D\n\n\nB   C\n\n<XAD and <BAD are supplementary since XAB is a line\n\nabove we proved that <ABC and <BAD are also supplementary\n\nso <XAD + <BAD = 180\nand <ABC + <BAD = 180\nwhich means <XAD = <ABC\n\nso now we can conclude that AD || BC\nbecause the corresponding angles <XAD and <ABC are congruent.\n\nyou can use this same logic to show that AB || DC\n\nhope this makes sense..
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Theorem 6-5\nSAS Inequality TheoremIf two sides of one triangle are congruent to two sides of another triangle, but the included angle of the first triangle is larger than the included angle of the second, then the third side of the first triangle is longer than the third side of the second triangle. not much, but might help.
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First we note that the two bases of the trapezoid are parallel and the segment joining the midpoints of the altitudes is parallel and equidistant from the bases. Therefore the vertical displacement of the midpoints of the bases from the midpoints of the altitudes is equal.\n\nIf we also contruct the segment joining the midpoints of the bases we see this intersects our other segment at its midpoint. We have now cut our quadrilateral into four triangles and it is an easy task to show that opposite triangles are congruent using SAS.
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(√3)² + 1 ² = 2²\nsin α = √3/2 α = 60º;  ß = 90 - α;  ß = 30º\n\nAnd your second question is\naⁿ = c² - b².  I'm going to assume that you wanted to put in a, b and c are all integers and a > 1.  Because if not, your theorem is not true and cannot be proven.  With that stipulation, this theorem passes the "intuitive" test.  It makes sense.\nWe know for certain that, for all even n, it must be true by the Pythagorean theorem.  So only odd n are of consequence to this theorem.\n\nI'm going to take a moment here and say that if you're trying to use this theorem as a FLT lemma, you must understand that it won't fly.  The lemma must use the carry the same stipulations as the successor theorem.  Said the other way, the corollary is bound by the restrictions of the theorem.\n\na=2, n=3 gives us c=3 and b=1.  8=9-1.  I took some time trying to find a counter-example and could not.  I did find some neat patterns emerging.\n\nThe proof to your theorem is simple: aⁿ = a²[(½a+½)² - (½a-½)²] because the expression in brackets will always be equal to a (a is odd prime; composite a can be simplified to prime a).\n\nI also found that 2ⁿ= [2^(n-2)+1]² - [2^(n-2)-1]².  So, in addition to all the solutions from the above expression, there will be another set of solutions for 2ⁿ; n is odd integer > 5.
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what are you supposed to prove?\n\nedit:\nyou mean why the three vertices of an equilateral triangle are concyclic? i am unable to make out theproblem
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1. All angles are equal\n2. All angles are less than 180° \n3. Size of each angle = (180 - 360/n)° where n = number of sides
13738
(A^3 + B^3)=(A+B)(A^2 - AB + B^2)
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since both sets of sides are parallel, it has to be at least a parallelogram (rules out trapezoid). And since two adjacent angles aren't the same, that rules out rectangle and square (both of which have all right angles). It can be either a rhombus or a more general parallelogram.
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h is perpendicular to a, so they form 2 right triangles.  The 2 sides are congruent, and h is congruent to itself. Thus, the 2 triangles are congruent by HL Theorem. \n\nUsing CPCTC and def'n of midpoint, the intersection of h and a is the midpoint of a, and using CPCTC again, the "upper angles" are congruent and are half the vertex angle of the isosceles triangle.\n\nLet x = angle between b and b'\n\nUsing trigonometric functions,\n\ntan (x/2) = (a/2)/h\ntan (x/2) = a/2h\nx/2 = arctan (a/2h)\nx = 2 arctan (a/2h)\n\nQED\n\nHope this helps^_^
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1 - A is enough because with the lengths of 2 sides you can use pythagoras' theorem to work out the length of the other