find the indicated arc or angle??  wz=??
492887
Your image isn't there...sorry.
93089
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.
826437
(x/a)²+(y/b)² = 1 (← Note my variation! I am using a², b² in place of your a, b)\n\nx = acos t , y = bsint is its parametric equation, a, b >0\n\nThen if T1(acost, bsint), (0 ≤ t ≤ π/2),  is the point on the ellipse in the 1st quadrant, the point on the ellipse vertically below it is\nT2 (acost, -bsint).\nThen the point horizontally to the left of that on the ellipse is\nT3 (-acost, - bsint)\nThe final point is vertically above on the ellipse namely\nT4 (acost, -bsint) which is also horizontally to the left of the point T1\n\nSo perimeter P = |T1T2| + |T2T3| + |T3T4| + |T4T1|\n= 2bsint + 2acost + 2bsint + 2acost\n= 4(bsint + acost)\n\ndP/dt = 4(bcost - asint)\n= 0 for stationary points\nSo asint = bcost\nie sint/cost = b/a\nSo t = arctan(b/a) (0 ≤ t ≤ π/2) (ie a,b>0)\n\nd²P/dt² = -4(acost + bsint)\n< 0 for (0 ≤ t ≤ π/2)\nand therefore P is maximum when t = arctan(b/a) (0 ≤ t ≤ π/2)\n\nIf t = arctan(b/a) then tan t = b/a \nThus cost = a/[√(a² + b²)] and sint = b/[√(a² + b²)]\n\nThus T1 ≡(a²/[√(a² + b²)], b²/[√(a² + b²)]), \nT2 ≡(a²/[√(a² + b²)], -b²/[√(a² + b²)]),\nT3 ≡(-a²/[√(a² + b²)], -b²/[√(a² + b²)]),\nT4 ≡(-a²/[√(a² + b²)], b²/[√(a² + b²)])
844799
Wallstream is correct -- start with a right triangle.\n\nLabel the interior angles a, b and c, with c being the 90-degree angle.\n\nNow notice that a and b are complementary angles that add up to 90-degrees.\n\nLet x = sina.  Since the angles are complementary, we also have that x = cosb.\n\nSo we can rewrite as follows:\n\nIf x = sina, then a = arcsinx\nIf x = cosb, then b = arccosx\n\nBut a+b = 90-degrees (aka pi/2).\n\nSo arcsinx + arccosx = pi/2\n\nHope that helped!
854468
Induri Kartic is almost right but she has missed a certain steps.\n\nlet x = tan A , y= tan B z = tan C (A B C are acute angles) \n\nnow tan(A+B+C) = tan (A+D) using B+C = D\n= (tan A + tan D)/(1- tan A tan D)\n= tan A + (tan B+tan C)/(1- tan B tan C)/denominator say a\n\nnumerator = tan A + tan B + tan C - tan A tan B tan C / some thing say b\nwe have not ecaluated ab \ntan (A+B+C) = (x+y+z-xyz)/ab = 0\nso evaluation of a b was not required\n\nso A+B+C is  pi (as A+B+C < 3pi/2)\n\nnow tan 3(A+B+C) = 0\nso tan 3A + tan 3B + tan 3C = tan 3A \n\ntan 3A = (3tan A -tan^3 A)/(1-3 tan^2 A) = (3x-x^3)/(1-3x^2)\nputting values we get the resultant immediately
510334
x = a*sin(ang)\ny = b*cos(ang)\n\nang goes from 0 to 360 degress, a and b determine how oval it is (if a=b it is a circle)
486759
You're trying to find the vertical angle, angle of site and site.  (But your overall goal is to find the elevation of the guns)\n\nRange and charge
945543
A short element of the curve is dL=sqrt(dr^2+(r*dt)^2), r*dt elementary length of the arc with radius r and polar angle t (your theta), dr elementary increase of r for dt.\nThus dL=sqrt(r’^2+r^2)*dt.\nIf  r=a*(1-cost), then dL=a*sqrt((sint)^2+(1-cost)^2)=a*sqrt(2)*sqrt(1-cost)*dt. \nFortunately 1-cost=2*(sin(t/2))^2 or dL=2a*sin(t/2)*dt, L=-4a*cos(t/2) for t=0 until 2pi, thus L=-4*a*(-1-1)=8*a – Oh! check it poor kid!\nIf r=(sin(t/2))^2=0.5*(1-cost), then a=0.5 for t=0 until pi. It must be done likewise. Thus L=-4*0.5*(0-1)=2
726975
Use sin(arcsinx) = x...\n\ncos(arcsinB) = sqrt(1-sin^2(arcsinB))\n= sqrt(1-B^2)\n\nSimilarly, sin(arccosA) = sqrt(1-A^2).\n\nUse the cos(x+y) equality to compute the last one...
706522
It's often handy to draw yourself out a little sketch.  From it you can see the location of the given point.\n\nTo identify the angle the given point makes with the x-axis and the y-axis use trigonometry.  Switch to radians.\n tan Θ = opp./ adj.\n Θ = tan^-1 opp./ adj.\n Θ = tan^-1 (-0∙4425./ -0∙8968)\n Θ = tan^-1 (-0∙493 210 52)\n Θ =  0∙458 370 613 rad.\n    This is the value of the angle below the x-axis.  You need to include the angle above the x-axis, which is 180º or π radians.\n Θ =  0∙458 370 613 rad. + π rad.\n Θ =  (0∙458 370 613 + 3∙14) rad.\n Θ =  3∙598 370 613 rad.\n Θ ≈  3∙6 rad.
1139471
Yeah where's angle 1 and 2.\nBut remember the angle opposite the longest side has the greastest measurement.
699032
wx - x =\n\nx(w - 1)\n\n- - - - - - -s-
723523
Int tanx ln(cosx)dx\n\nwrite tanx=sinx/cosx\n\nsubstitute y = cos(x)\ndy=-sinx dx\n\n-Int ln(y)/y dy\n\nsub z = ln(y)\ndz=ln(y)/y dy\n\n-Int z dz = -z^2/2 = -(1/2)ln^2(y) = -(1/2)ln^2(cos(x)) + C
343495
To find arc length, you need the angle that the arc subtends and the radius of the circle.  To find the angle, you'd have to do some geometry...\n\nImagine a triangle with sides 10, 10, and 5.  This is pretty much a chord on your circle (5) and the radii from the center to the ends of the chord (10).  If you draw a line from the center perpendicular to the chord, it bisects the chord.  This gives you two right triangles, with hypotenuse 10, side 2.5 (half the chord), and side unknown (the segment from the center to the chord).  To find the angle at the center formed by the radius and the unknown, you can use the sine function:\n\nangle = arcsin(opposite/hypotenuse)\nangle = arcsin(2.5/10)\nangle = 14.48 degrees or 0.25 radians\n\nThe total angle is then 0.5 radians, since the above only found half of that total angle.\n\nThe formula for arc length is\n\narc length = radius * angle\n= 10 cm * 0.5 rad\n= 5 cm\n\nAnd this kinda makes sense - if the total circumference is 20pi, and the arc is 28.96 degrees, then the arc length can also be found by\n\narc length = angle * circumference/360\n= 28.96 * 20pi/360\n= 5 cm
1106173
The direction (angle) of |u| and |v|?
830100
'x= rcost\ny=rsint\n\nand r= sqrt(x^2+y^2)\nt=arctan(y/x)\nso\nif x=2, and y=2, then r= sqrqt(4+4)=sqrt(8)\nand t=arctan(2/2) = arctan(1) = pi/4
826982
If the point is in the yz-plane, it has x=0. In other words, its projection on the xy-plane lies on the y-axis. But that means that the angle theta is *not* zero but either Pi/2 (when y>0) or 3Pi/2 (when y<0).