How to tri-sect a given angle?
461392
If you know what the angle is, divide it by three and it is trisected.
394622
A triangle in which all 3 sides are of different lenghts, with different angles.
807541
Let A be the angle\n\nTurn the words into an equation:\n\n"The supplement" --> 180° - A\n"is" --> =\n"thirty less than" ---> - 30°\n"three times" ---> 3 x\n"the complement" --> 90° - A\n\n180° - A = 3(90° - A) - 30°\n\nDistribute the 3 through on the right:\n180° - A = 270° - 3A - 30°\n\nSimplify:\n180° - A = 240° - 3A\n\nAdd 3A to both sides:\n180° + 2A = 240°\n\nSubtract 180 from both sides:\n2A = 60°\n\nDivide both sides by 2:\nA = 30°\n\nSo the angle is 30° (the supplement is 150°, the complement is 60°)
389433
If you're asking how to do it: Just take the total number of degrees of the angle, divide by two, and there you go, you know the measurements of the 2 equal angles.\nanswered by Fading Away...\n\nAngle bisector Theorem.\n\nIf you are a Geometry student who knows constructions, locate a radi from your vertex and draw an arc. Choose another radii on that new arc that you just drew(That is more than half the arc) from the point where your newly drawn arc intersects the segment. Draw another arc. Repeat process on opposite side. Draw bisectore from vertex through intersection of two arcs.\n\nHope that helps!\nAnswered by Emily\n\n\n\nThis was asked and answered 13 hours ago, click this link.\nhttp://answers.yahoo.com/question/;_ylt=Al6hLACuaPLouhbP7BmwxcIjzKIX?qid=1006041024466
450833
join the 2 pointz frm da arc vit da radius.measure out da value of da angle forned here.then substitute da values 4 this formula:  2*pie*radius/da angle u measured\n\nhappy solvin!
265237
3s = -6\ndivide each side by 3\ns = -2
509207
Yes. You can use sine or cosine to determine the angle.\n\nSine x = Opposite / Hypotonuse\nSin x=5/13\nSin x=.38\nx=22.6*\n\nCosine x = Adjacent / Hypotonuse\nCos x=12/13\nCos x=.92\nx=22.6*
1106173
The direction (angle) of |u| and |v|?
1021808
Ok, you've got a good one here.\nFor a right triangle, height and base are the short sides; and area of the triange is (bh)/2. Since we know the area, and the length of the sides, we can plug it in to an equation, and find x. So...\nA = (bh)/2\n60 = ((5x)(3x-1))/2\n120 = (5x)(3x-1)\n120 = 15x^2 - 5x\n15x^2 - 5x -120 = 0\nDivide both sides by 5 to get\n3x^2 - x -24 = 0\nOk, there's a number of ways to do this, but the simplest way is to list the combinations of integers that will multiply to get -24 (the constant in the above equation). That gives us the combinations of (1, -24; 2, -12; 3, -8; 4, -6; 6, -4; 8, -3; 12, -2; and 24, -1). Now, we need to figure out the one combination that, when the second number is multiplied by three (the coefficient of x^2), you can add it to the first number to become (-1). It happens to be (8, -3) {-3*3=-9, -9+8 = -1}. There; you've factored the equation, leaving you with...\n(3x+8)(x-3)=0\nSince multiplication rules say that when ab=0, either a=0 or b=0, we can test by setting both numbers equal to zero; giving us...\n3x+8 = 0 ; x-3 = 0\nx = -8/3; x = 3\nSince x is an absolute length, and lengths can't be negative, we can disregard (-8/3) for now. That gives us x=3, and putting that back into the original lengths gives us sides of 15 and 8. To find the hypotenuse, we use Pythagorean Theorum: a^2 + b^2 = c^2, substituting in the known lengths gives us:\n15^2 + 8^2 = c^2\n225 + 64 = 289  = c^2\nc = √289 = 17\nSo, the hypotenuse of your triangle is 17.
826563
multiply the tangent of the given angle by the known side as long as it's not the sloping side (i.e. the hypotenuse). You will need a scientific calculator to do this. I am assuming of course that you are dealing with a right triangle. This will give you the other side of the triangle. Example:\n\nlets say you have side A of a right triangle and an angle of 30 degrees and A = 10. Take your scientific calculator press 30 and then the tan button depending on the type of caculator you have. Now multiply that result by 10 and you will get side B.\n\nSide C is found by A^2 + B^2 = C^2. take the square root for your answer.\n\nNow if the given angle is opposite the given side. Just calculate the angle of the given side by adding the given angle to 90 degrees and subtract the result from 180, and then multiply the tangent of the new new angle by the given side.\n\nTheoretically you can check the answer by multiplying the new side by the tangent of the new angle. You should in theory get the given side. But my trig is so rusty I may have forgotten more than  you will ever know. good luck.
816836
You need one more datum, either the third side or any angle.  If it were a right triangle, the right angle would comprise the third datum.
882048
Here's the answer you want:  With a compass set to the radius of the circle, start marking off radii around the circle, dividing it into 6 equal parts.  Connect any 3 alternate points and you'll have your triangle.\n\nNO MATH REQUIRED!
27599
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.
1021896
the squareroot of 2 over 2.  angle a must equal 45 degrees.\nnote- the answer may also appear as 1 over the squareroot of two.
721808
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.
388496
There are six ratios of the three sides of a right triangle. These ratios have been identified by name throughout the centuries. In the USA they are known as sine, cosine, tangent, cotangent, secant, and cosecant. Your calculator has buttons for three of these: [SIN], [COS], and [TAN]. Remember them with the non-word SohCahToa. \n\nThe tangent ratio of an angle is simply the ratio of the opposite side to the adjacent side of a right triangle. \n\nIf you know the angle, type \n\n[TAN] > angle > [ENTER].\n \nThe answer will be the tangent (ratio of the opposite side to the adjacent side).\n\nIf you know the opposite and adjacent sides and need to find the angle, type \n\n[2nd] > [TAN] > opposite > [/] > adjacent > [ENTER]. \n\nThe answer given will be the desired angle.\n\nCaution: \n(1) Different calculators require these buttons to be pushed in a different order.\n(2) If your angle is in degrees, then make sure your calculator's mode is in degrees.\n(3) Don't get the numbers backwards or you'll find the cotangent of the angle and not the tangent.
400701
what you want to do is draw an isosoles triangle with the 2 congruent angles the size you want. \n\nStart with a base with a certain length (use the compas to measure, say 4 inches).  Next determine the midpoint by marking arcs nboth above and below your line from each endpoint using the same compass diameter, then drawing a line through the two intersections.  You now have a right abngle at the midpoint.  You can now determine how tall your triangle must be using the sin table and use your compass to mark the point on the vertical line to complete the triangle with the angle you want.