how a angle changing begin's?
397231
BECAUSE OF THE TORQUE THAT IS EXERTED TO THE BODY.\n \n WE HAVE \n          TORQUE = MOMENT OF INERTIA * ANGULAR ACCELERATION
1033465
Angle a=159 and the value of y=55.33.\n-\nSimple  supplementary angles  angle a =180-21=159.\n\n159=3y-7\n159+7=3y\n166=3y\ny=166/3\ny=55.33
806989
The easiest way I see to do this is to rotate it first so its axis of symmetry is vertical (or horizontal), do the calculations, and rotate it back.\n\nThe transformation for rotation is best described as\nx → xcosθ - ysinθ\ny → xsinθ + ycosθ where θ is the angle of rotation\n\nSo rotate by 45° anticlockwise whence axis becomes vertical:\n\n(10, 0) \nx → xcos45° - ysin45° = 10cos45° - 0 = 5√2\ny → xsin45° + ycos45° = 10sin45° = 5√2\n\n(-2,0)  \nx → xcos45° - ysin45° = -2cos45° = -√2\ny → xsin45° + ycos45° = -2sin45° = -√2\n\n(12,4)  \nx → x cos45° - ysin45° = 12cos45° - 4sin45° = 4√2\ny → xsin45° + ycos45° = 12sin45° + 4cos45° = 8√2\n\nSo now we find the equation of the parabola with a vertical exis through (5√2, 5√2), (-2√2, -2√2) and (4√2, 8√2)\n\nWell the parabola is of the form y = ax² + bx + c\n\nWhen x = 5√2, y = 5√2\nSo 50a + 5√2b + c = 5√2 ................. (1)\n\nWhen x = -√2, y = -√2\nSo 2a - √2b + c = -√2................. ...... (2)\n\nAnd when x = 4√2, y = 8√2\nSo 32a + 4√2b + c = 8√2 ................. (3)\n\nEquation (1) + 5 x equation (2)\n\n60a + 6c = 0\nie 10a + c = 0 .................... ........... (4)\n\nEquation (3) + 4 x equation (2)\n\n40a + 5c = 4√2 ................ ........... (5)\n\nEquation (5) - 5 x equation (4)\n\n-10a = 4√2\na = -0.4√2\nThus c = 4√2\nand b = 4.2\n\nSo the rotated parabola is y = -0.4√2x² + 4.2x + 4√2  ← ***\n\nNow we need to unrotate this parabola and to do that we need to rotate it BACK by an angle of 45° ie an angle of -45°\n\nUnder this rototation\n\nx → xcos(-45°) - ysin(-45°) = x/√2 + y/√2\ny → xsin(-45°) + ycos(-45°) = -x/√2 + y/√2\n\nNow y = -0.4√2x² + 4.2x + 4√2\nThus -x/√2 + y/√2 = -0.4√2(x/√2 + y/√2)² + 4.2(x/√2 + y/√2) + 4√2\n= -0.4√2(x²/2 + 2xy/2 + y²/2) + 4.2x/√2 + 4.2y/√2 + 4√2\n= -0.2√2x² - 0.4√2xy - 0.2√2y² + 4.2x/√2 + 4.2y/√2 + 4√2\nMultiply by √2\n-x + y = -0.4x² - 0.8xy - 0.4y² + 4.2x + 4.2y + 8\n\nMultiply by 10 and collect terms:\n\n4x² + 8xy + 4y² - 52x - 32y - 80 = 0\nie x² + 2xy + y² - 13x - 8y - 20 = 0\n\nno sorry made another .... whoops!!!\nIt makes no sense ... it is correct up to line marked ← ***
830426
The crux of this problem is for you to realize that the given angles are "complementary pairs", which add up to 90 degrees.\n\nYou are expected to know that the "co-function" of angle A gives the same value as the "function" of angle A's complement.\n\nSpecifically:\n\n cos(A) = sin(90-A)\n\nA graphical way of looking at this statement is that "cosine and sine are the same function, but offset by 90 degrees!"  We call that "out of phase by pi/2 or 90 degrees."\n\nThe same works for tangent and its co-function:\n\n tan(A) = cot(90-A)\n\nSo the answer to the second problem is 0.\n\nIf one of the fuctions in the first problem had been sin, the quotient would have been 1.  However, that's not the case, and I leave the upper problem as an exercise for another Answerer.\n\n Hope this helps!
291453
It shifts it.  If y = mx + b, b is the y-intercept (where the graph crosses the y-axis), and changing b would move the entire line up or down along the y axis.\n\nTry graphing these two lines by plotting points\n\ny = x\ny = x+3\n\nSee how the lines are parallel, but one is slightly higher than the other?
199013
add 2 other angle in degree and substract from 180\n ie. c=180 -(a+b)\nwhere a ,b are known angles \nand c is unknown angle
1030296
for it to be a triangle the angles have to measure to 180, so if you have a 90 degree andlge off one, you could subtract 90 from that, take the remaining 90 and work with that, since you know your need a 1:1 of the legs , then you know you need an equal angle.. angle has to equal 90/2... and thats where i get stumped...\n      gl
228730
You can describe rotations and the effect they have on each point in the figure mathematically by means of simple trigonometric equations which can be condensed into a matrix. Imagine (or better, draw) a point on the plane in coordinate system X0,Y0, located at X0=1, Y0=0. Draw a line to the point from the origin. Now draw another coordinate system X1,Y1 with the same origin as X0,Y0 but rotated CW by an angle Theta. (Make Theta small, say 30 degrees.) What are the coordinates of our point in the new system? The answer is a function of Theta. We find that the point is above the X1 axis and nearer the Y1 axis. Specifically, X1 = X0*cosine(Theta) and Y1 = X0*sine(Theta). \nNow if you repeat this with a new point at X0=0, Y0=1, you find that X1 = -Y0*sine(Theta) and Y1 = Y0*cosine(Theta).\nNow the good part. You can put these four equations together to find X1,Y1 for any value of X0,Y0 (not just points located on one axis or the other). The result is\nX1=X0*cosine(Theta) - Y0*sine(Theta)\nY1=X0*sine(Theta) + Y0*cosine(Theta)\nNext you form a rotation matrix R from the Theta-dependent parts of these equations:\nR=[cosine(Theta)___ -sine(Theta)]\n__[sine(Theta) ___ cosine(Theta)] \nNow when you multiply any vector P0, representing any point in X0,Y0 coordinates, by this matrix R using the rules of matrix multiplication, you are actually using the above two equations, and the result is a vector P1 which is the same point represented in X1,Y1 coordinates. Expressed compactly, P1 = R*P0. P0 is said to be transformed by R to P1.\nOf course, this is a 2-dimensional transformation, and R is a 2-by-2 matrix. Sophisticated 3-D computer graphics use a 3-by-3 matrix implementing very similar equations to do 3-D transformations. This is how you can see convincing 3-D models of objects in rotation.
27065
I've been studying this stuff recently, and I'm pretty sure this is how it goes:\n\nK is now (-2,1) and D is now (-3,-2).\n\nFor future reference on determining rotations, simply turn the graph counter-clockwise one time for 90 degrees, upside down for 180 degrees, and for 270 degrees, just turn the graph clockwise once to see what the new cooridnates are.
958527
You need to know the angle between "x" and "y", call it "a"\nLet initial position of "x" be "x0"\nLet the time from start of "x" be "t"\nLet "z" be the distance between "x" and "y" at time "t"\n\nThen:  \ny = 38 + 2t\nx = x0 + 11t   \n\nAnd from Pythagoras:\nz^2 = x^2 + y^2 - 2xycos(a)\n\nHopefully that gets you started....
165372
S= Radius * Alpha Angle\nAlpha Angle needs to be in radians\n\nSo, Radius= S/Alpha Angle (in Radians)
819957
Start with a graph of:\nf(x) = x\n\nThat's a straight line going through the origin with a slope of 1.  It should look like a 45 degree angle if you draw it correctly.\n\nNow to transform this to:\nf(x) = | x |\n\nYou have to think of each value smaller than 0.  Rather than -1, for example, you'll have 1.  Rather than -2, you'll have 2, etc.  This is like "mirroring" the graph that is below the x-axis so that it is above the x-axis.\n\nThis results in a "V" shape.  A slope of -1 down to the origin, then a slope of 1 from the origin.\n\nFinally:\nf(x) = | x | - 2\n\nFor every value that results, you now subtract 2.  For example, x = 0, gives -2, rather than 0.  This is true for every point x, making every point y smaller by 2.  It would look like you shifted the whole graph down 2 units.\n\nSo it would be "V" like before, but it would go down at a slope of -1, cross the x-axis at -2, and continue to (0, -2).  Then it would slope up (m=1) through the x-axis at 2 and continue upward.\n\nYou can go to the first link and type 'abs(x) - 2' in the box.  The resulting graph is shown as the 2nd link.
443769
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.
653474
Pick an angale, not the 90 degree angle and call it Theta.  The short side that forms this angle is called the x side of the triangle and the long side that forms this angle is the hypotenuse or 'r'.  The third side is called the opposite side or 'y'.\n\n\n.... /|\n..../.|\n.../..|\n../...|\n./....|\n/___|  Bottom is x, slanting side is r, and right side is \n          the y side.  The angle at the bottom left is Theta.\n\n\nSin(Theta) will = y/r\nCos(Theta) will = x/r\nand\nTan(Theta) will = y/x side.\n\nThis all presumes you don't already know 2 of the sides.  If you know 2 sides already, you use the pythagorean thm.\n\nExampe: Theta = 30 degrees, r = 10 feet, what are the others?\n\nSin(Theta) = y/r  and Sin(30 degrees) happens to be 1/2.\n\nso, 1/2 = y/10.  Solving for y you get y=5.
951198
^ Parasite alert LOL! ^\n\n30 s for C to move from A to B, maybe? I'll assume that. Also, x seems to be a length, not an angle. Maybe a) is asking for the rate of change of z (called z')? We know the angular rate of C about O (y') is pi/30 (answer to b)), and triangle CAO is isosceles, so z'=-0.5*pi/30. \nFor c), circumferential velocity is pi*r/30. Angle CAO is, conveniently, 45 deg, so x' = sqrt(0.5)*circumferential velocity.\nFor d), the area of triangle OAC is .5*AO*altitude, and altitude=r*sin(y). This is a maximum when y (not x) =pi/2.\nSorry, I know you wanted  no answers, but I couldn't see how/where to stop short of most of the answers.
765889
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
367687
an angle that is inside a circle or any other figure