prove the third angles theroem?
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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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Use the law of cosines.\n\n7^2=3^2 + 5^2 - 2(3)(5)cos(C)\n49=9 + 25 - 30cos(C)\n49=34 - 30cos(C)\n15= -30cos(C)\ncos(C)= -15/30\ncos(C)= -1/2\nC= 120 degrees.\nThe angle opposite the side of seven is 120 degrees.\n\nTo find the next angle, use the law of sines.\n\n7/sin(120) = 3/sin(A)\n3sin(120)= 7sin(A)\n(3sqrt3)/2= 7sin(A)\n(3sqrt3)/14= sin(A)\nA= 21.7867893\nThere are 21.7867893 degrees in the angle opposite the side of 3. \n\nTo find the final angle, add the two known ones and subtract their sum from 180.\n\n180 - (120 + 21.7867893) which equals 38.2132107 degrees.\n\nThere are 38.2132107 degrees in the angle opposite the side of 5.
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The easy way to remember this is to memorize the \n\n3 4 5 \n\nright triangle.  You will notice that any two multiples of the above will give you the other leg as a multiple.  The theorem is\n\nA^2 + B^2 = C^2,  so  3^2 + 4^2 = 5^2,  hence  9 + 16 = 25\n\nnow you have\n\n2*3 2*4 thus your other leg is 2*5\n\n6^2 + 8^2 = 10^2,  which is 36 + 64 = 100\n\nas for the angles, use the law of sines\n\nsin(90)/5 = sin(a)/3 = sin(b)/4\n\nso a = asin(sin(90)*3/5) = ~37 deg\n\nhence angles of 90 53 37
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There are several ways to prove two triangles are congruent (I believe this is what you are meaning by similar).\n\nProbably, several steps will need to take place before ultimately coming to one of three potentail final steps in a proof.\n\nThree final steps in proving the congruence of two triangles are:\n- ASA  (Angle Side Angle) - which means that the corresponding Angles, Side & Angles are equivalent\n- SSS (Side Side Side) - which means the three sides of the triangles are equivalent to the corresponding sides.\n- SAS (Side Angle Side) - which means the corresponding Sides and Angle of the two triangles are equivalent.\n\n*Note, there is no AAA theorem, as two triangles may have equivalent corresponding angles, but diff length sides.  These triangles would not be congruent.\n\nHere is an exerpt from a good site:\nHere is a quick summary of what each method of proof means:\n\n       SSS                SAS                ASA\n\n        +                    +                  +\n        /  &#92;                 /  &#92;                 /  &#92;\n     c/     &#92;a          c/     &#92;              /     &#92;\n     /        &#92;            /A     &#92;           /A   C &#92;\n   +-----------+   +-----------+  +-----------+\n          b                  b                  b\n\n        +                 +                +\n       /  &#92;               /  &#92;               /  &#92;\n    c/     &#92;a        c/    &#92;            /     &#92;\n     /       &#92;          /A    &#92;         /A  C &#92;\n    +------+    +--------+    +--------+\n          b                  b                  b\n\n    All three sides    Two sides and      Two angles and\n    of one triangle    the included       the included\n    agree with all     angle of one       side of one\n    three sides of     triangle agree     triangle agree\n    the other.         with the other.    with the other.
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which is angle 1 and which is angle 2?where is the diagram/description?\ninstead of saying angle 1 and angle 2 can you please name the angles?it will be very helpful
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A triangle has 180 degrees.  a right angle is 90 degrees.  a triangle has 3 angles.
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m<Z = 52\nThe three angles add to 180.\nm<X + m<Y + m<Z = 180\nThus,\nm<Y = 180 - 52 - 52\n= 76\nAlso, I assumed you meant that XY is congruent (or equal) to YZ.  Then m<X = m<Z.
461392
If you know what the angle is, divide it by three and it is trisected.
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Given: \n1) a right triangle\n2) hypotenuse = 2\n3) the other two angles are 70 and 20 degrees\n\nFind:\nLengths of the remaining sides\n\nNote: \nWe have to be more specific here.\n\nLet:\na = side opposite the 70 degree angle (side adj. 20 degrees)\nb = side opposite the 20 degree angle (side adj. 70 degrees)\nc = hypotenuse = 2\n\nSolution:\n\n1) For side a:\n\nsin 70 = opp./hyp. = a/c\nTherefore, a = csin70 = 2sin70 = 1.88 ANSWER\nOR\ncos 20 = adj./hyp. = a/c\nTherefore, a = ccos20 = 2cos20 = 1.88 ANSWER\n\n2) For side b:\n\ncos 70 = adj./hyp. = b/c\nTherefore, b = ccos70 = 2cos70 = 0.68 ANSWER\nOR\nsin 20 = opp./hyp. = b/c\nTherefore, b = csin20 = 2sin20 = 0.68 ANSWER\nOR\nPythagorean Theorem: a^2 + b^2 = c^2\nb^2 = c^2 - a^2\nb^2 = 2^2 - 1.88^2\nb^2 = 0.4656\nTherefore:\nb = 0.68 ANSWER
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If you are trying to make a triangle out of three equal lengths, all you do is devide 180 by 3.....and taddaaa you get your angles.\n\nIf does not matter how long each side is.
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the four angles add up to 360 degrees
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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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law of cosines : a^2 = b^2 + c^2 -2bc * cos(A)\nwhere a, b,c are side lengths and A is the angle oppisite side a.
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If you are talking about trisecting an angle using a compass and straight edge, then it is not and unsolved problem, it is problem for which there is no solution.\n\nThere is an area of abstract algebra called Galois theory. One of the things that you learn in Galois theory is that there exist certain angles that cannot be trisected using the kind of techniques that you would use to bisect an angle or line. Flat out can't be done. I recall, but don't quote me on this, that 60 degrees is one example.\n\nAnother thing that comes from Galois theory is that there is no formula, that will give you the roots of a 5th degree polynomial. You are probably familiar with the quadratic equation which gives you the roots of a second degree polynomial. There are formulas that you can use to get the roots of 3rd and 4th degree polynomials. But there can't be a formula that will work for a general 5th degree polynomial.\n\nI know more than one high school student who wasted a lot of time trying to devise a method for trisecting angles when he/she heard it was a great unsolved problem. Their teachers neglected to tell them that it was not only a great unsolved problem, but that it was a great unsolvable problem.
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64 proofs\nexplanation\nProof #1\nThis is probably the most famous of all proofs of the Pythagorean proposition. It's the first of Euclid's two proofs (I.47). The underlying configuration became known under a variety of names, the Bride's Chair likely being the most popular.\n\n    \n\nThe proof has been illustrated by an award winning Java applet written by Jim Morey. I include it on a separate page with Jim's kind permission.\n\nFirst of all, ABF = AEC by SAS. This is because, AE = AB, AF = AC, and\n\n  BAF = BAC + CAF = CAB + BAE = CAE.\n \n\nABF has base AF and the altitude from B equal to AC. Its area therefore equals half that of square on the side AC. On the other hand, AEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel to AE. Thus the area of AEC equals half that of the rectangle AELM. Which says that the area AC2 of the square on side AC equals the area of the rectangle AELM.\n\nSimilarly, the are BC2 of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB.\n\nThe configuration at hand admits numerous variations. B. F. Yanney and J. A. Calderhead (Am Math Monthly, v.4, n 6/7, (1987), 168-170 published several proofs based on the following diagrams\n\n   \n\n\n\nProof #2\n    \n\nWe start with two squares with sides a and b, respectively, placed side by side. The total area of the two squares is a2+b2.\n\n    \n\nThe construction did not start with a triangle but now we draw two of them, both with sides a and b and hypotenuse c. Note that the segment common to the two squares has been removed. At this point we therefore have two triangles and a strange looking shape.\n\n    \n\nAs a last step, we rotate the triangles 90o, each around its top vertex. The right one is rotated clockwise whereas the left triangle is rotated counterclockwise. Obviously the resulting shape is a square with the side c and area c2.\n\n(A variant of this proof is found in an extant manuscript by Thâbit ibn Qurra located in the library of Aya Sofya Musium in Turkey, registered under the number 4832. [R. Shloming, Thâbit ibn Qurra and the Pythagorean Theorem, Mathematics Teacher 63 (Oct., 1970), 519-528]. ibn Qurra's diagram is similar to that in proof #27. The proof itself starts with noting the presence of four equal right triangles surrounding a strangenly looking shape as in the current proof #2. These four triangles correspond in pairs to the starting and ending positions of the rotated triangles in the current proof. This same configuration could be observed in a proof by tesselation.)\n\n\n\nProof #3\n    \n\nNow we start with four copies of the same triangle. Three of these have been rotated 90o, 180o, and 270o, respectively. Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c.\n\n    \n\nThe square has a square hole with the side (a-b). Summing up its area (a-b)2 and 2ab, the area of the four triangles (4·ab/2), we get\n\n  c2 = (a-b)2+2ab \n  = a2-2ab+b2+2ab \n  = a2+b2 \n \n\n\n\nProof #4\n    \n\nThe fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side (a+b) and a hole with the side c. We can compute the area of the big square in two ways. Thus\n\n  (a + b)2 = 4·ab/2 + c2  \n\nsimplifying which we get the needed identity.\n\n\n\nProof #5\n\nThis proof, discovered by President J.A. Garfield in 1876 [Pappas], is a variation on the previous one. But this time we draw no squares at all. The key now is the formula for the area of a trapezoid - half sum of the bases times the altitude - (a+b)/2·(a+b). Looking at the picture another way, this also can be computed as the sum of areas of the three triangles - ab/2 + ab/2 + c·c/2. As before, simplifications yield a2+b2=c2.\n\nTwo copies of the same trapezoid can be combined in two ways by attac
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Let v1=C-B , v2=A-C , v3=B-A be the three displacement vectors making up the sides of the triangle. Define f1,f2,f3 to be the three forces X,Y,Z each rotated 90 degrees counterclockwise. Then f1 is parallel to v1, and so on. That is:\n\nf1 = s1*v1\nf2 = s2*v2\nf3 = s3*v3\n\nwhere s1,s2,s3 are three scalars. It's only necessary to show that s1=s2=s3 to prove that the three forces are proportional to the three opposite side lengths (|f1|/|v1|=|f2|/|v2|=|f3|/|v3|). This is easily done as follows:\n\nSince the forces are in equilibrium (f1 + f2 + f3 = 0):\n\ns1*v1 + s2*v2 + s3*v3 = 0\n\nMultiplying (v1+v2+v3)=0 by s1 and subtracting from the above equation:\n\n(s2-s1)*v2+(s3-s1)*v3 = 0\n\nSince v2 and v3 are linearly independent both coefficients must be zero\n\n-->\n\ns1 = s2 = s3\nQED.
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theorm of Pythagorus states that if one of the angle of triangle is right angle{90degrees} than the hypotenuese square{the side opposite to right angle} is egual to the sum of squares of the sides forming the right angle \ni.e. (Hyp)square =one side square + second side square \n a2=b2+c2