The eccentric angles of the extremities of a chord of an ellipse (x^2/a^2)+(y^2/b^2 )are c1 and c2 .?
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The link below ought to provide some insight on this problem...
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If your curve is the upper half of an ellipse, then when you rotate it, you have half an ellipsoid.\n\nWikipedia has the formula for the surface are of an ellipsoid, which (surprize!) involve eliptic integrals.
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Let as suppose that your ellipse is rotated by 45 degrees to right (or by -45 degrees). The new coordinates are X = x cos (-45) - y sin (-45) and Y = x sin (-45) + y cos (-45) and they are satisfying X^2/a^2 + Y^2/b^2=1. There for (x + y)^2/2a^2 + (x - y)^2/2b^2 = 1 (in a few easy steps). The mayor axis over the x=y is given by |a| > |b| .
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The formula\n\nAx2 + Bxy + Cy2 + Dx + Ey + F = 0\n\nis a hyperbola if 4AC < B^2.  If 4AC > B^2, then it's an ellipse.\n\n4*1*19 = 76\n13^2 = 169\n\nSo, hyperbola.
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This seems like a homework problem, so I'll give a few hints.\n\nThe arclength is radius times the angle subtended.\n\nTo find the angle subtended, consider the isosceles triangle created inside the circle where the chord is the hypotenuse and the radius of the circle is the two other legs. \n\nSince the isosceles triangle created is not a right triangle you need to use a specific formula to find the angle "C" subtended by the chord.
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1. circle\n\n(x-h)^2 + (y-k)^2= r^2\n\n(h,k) -coordinates of the center\nr-radius\n\n2. hyperbola\n\n[(x-h)^2/a^2]  - [(y-k)^2/b^2] =1 - transverse axis is horizontal\n\n[(y-k)^2/z^2] - [(x-h)^2/b^2] = 1- transverse axis is vertical\n\n3. parabola\n\nThe standard form of the equation of a parabola with vertex (h,k) and directrix y=k-p is \n\n(x-h)^2 = 4p(y-k) - vertical axis\n\nFor directrix x=h-p, the equation is \n(y-k)^2 = 4p(x-h) - horizontal axis.\n\nThe focus lies on the axis p units from the vertex. The coordinates of the focus are as follows:\n\n(h,k+p)   vertical axis\n(h+p,k)   horizontal axis).\n\n4. ellipse\n\nThe standard form of the equation of an ellipse with center (h,k) and major and minor axes of lengths 2a and 2b, where a>b, is\n\n[(x-h)^2/a^2] + [(y-k)^2/b^2]=1 - Major axis is horizontal, or\n\n[(x-h)^2/b^2] + [(y-k)^2/a^2[=1 - Major axis is vertical.\nThe foci lie on the major axis, c units from the center, with c^2=a^2 - b^2.
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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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(x^2)/2 - 3(x^(3/2))/(3/2) + c = (x^2)/2 - 2x^(3/2) + c
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I got  the first part. just get the cosx by itself. then you wind up with the cosx = -(the square root of 2/2)\nwhich is 225+360n and 315 +360n(in degrees)\nSorry i can't help you on the other one. Good luck, just keep trying new things with it, would be my advice.
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Just use Cosinus Theorem: c^2=a^2+b^2-2abCos(gamma)\n1.\nc=sqrt(4+64-2*2*8*sqrt(3)/2))\nc=6.34\n\n2. Is not a right Triangle.. but anyway lets find the angle A:\n22^2=15^2+33^2-2*15*33*Cos(A)\nCos(A)=830/990\nA=ArcCos(83/99)\nA=33°1' 48.54"
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In rt triangle ABC with B= 90degrees (draw figure)      \nIf C is theta, cosC=7/8(ignoring sign)\ni.e., BC = 7, CA = 8 (in simplest terms)\nusing pythagoras theorem, AB= sqrt(15)\n \nsince C=theta is in the second quadrant tan C will be negative\n\nso tan C = AB / BC \n\n  tan C= - { sqrt(15) } / 7\n\ntan(theta) = -0.5533
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If it's a right triangle, use the Pythagorean Theorem.\n\na^2 + b^2 = c^2\n\n\nA and B are the legs (the ones that touch the right angle), C is the hypotenuse which is opposite the right angle.\n\n10^2 + x^2 = 26\n\n100 + x^2 = 676\n\nx^2 = 576\n\nx = 24
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The midpoint of A and B is (15+7)/2 = 11.\nIf C = 11 then AC =BC.\nTherefore AC < BC  for 11<C<15
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a=c*sin(30)=.5c since it is apposite the 30 degree angle\nb=sqrt(c^2-a^2)= .866c\nor b=c*cos(30)=.866c\n\na=b=c*sin(45)=.707c=
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angles are never isoceles... may be you mean the triangle? \nand also, what is C, where is it located? '
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Assuming that A, B, C are all on the same line.\n\nLarger circle:\nRadius = AC = 6\nArea = PI * r^2 = 36*PI\n\nSmaller circle:\nRadius = AC - AB = 6 - 2 = 4\nArea = PI * r^2 = 16*PI\n\nShaded area:\nArea = (Large A) - (Small A)\n     =  36*PI - 16*PI = 20*PI = 62.8 units^2
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do u mean angle or side? for side length you use the pithagorean theorem... a^2+b^2=c^2 ... yopu hace c, c=16, and a, a=9\nso ... 9^2+b^2=16^2......   b^2=16^2-9^2===175\nb= square root of 175 =13.2\nI