You have a triangle RSTProve SinR^2 + CosR^2 = 1 Given ST Perp RT PLZ ANSWER FAST Answr = 10pnts?
472052
ok so i assume you know the basic rules of trigonomerty for a triangle.\ni.e sinR = opposite/hypotenuse and cosR= adjacent/hypotenuse.\n\ndraw the triangle on a piece of paper.\nYou will see that sin R = ST/SR\n\n(where ST is the edge running from S to T and similarly SR is the edge running from Sto R)\n\nalso cos R= RT/SR\n\nso sinR^2 + cosR^2 = (ST)^2/(SR^2) + (RT)^2/(SR^2)\n\n= [(ST)^2 + (RT)^2]/(SR^2)\n\nNow we use pythagoras' theorem. I assume u know this? The sum of the squares of the two shorter sides of the triangle equals the square of the longer side (the hypotenuse)\n\nso (ST^2) + (RT)^2 = SR^2\n\nSO WE NOW HAVE\nsinR^2 + cosR^2 = SR^2/SR^2 = 1\n\nthere you go\n\n\nCANT E-MAIL YOU BACK WESTILE BECAUSE IT SAYS YOUR E-MAIL IS NOT VERIFIED
516766
I'm not giving you the answer, but Cos is adjacent over hypotenuse.  Also, if you wanted the question answered correctly, us answering people might want to know where all of the letters are in relation to the triangle.
729006
3sin(x) - 2[cos(x)]^2 = 0\n\nNow [cos(x)]^2 = 1 - [sin(x)]^2 from the rule [sin(x)]^2 + [cos(x)]^2 =1\n-->\n\n3sin(x) - 2{1 - [sin(x)]^2} = 0\n2[sin(x)]^2 + 3sin(x) - 2 = 0\n\nsubstitute sin(x) = t\n\n2t^2 + 3t - 2 = 0\n(2t -1) (t + 2) = 0\n\nSolving ...t = -2, t = 0.5\n\nSubstituting back for sin(x)\n\nSin(x) = -2, 0.5\n\nSin(x) = -2 is not in the range since -1<sin(x)<1\n\nTherefore sin(x) = 0.5\n\nx = 30 degrees, 150 degrees for 0<x<360 degrees
791344
R = [ a^2 + b^2 -2*a*b* cos(theta), where  theta is the angle between a and b.\nIf a and b are fixed,\nR is max when  cos(theta) is min.   \nMin value of cos(thete) = -1  at  thete = 180 degree
258058
you know that sin^2(t)+cos^2(t)=1. so cos^2(t) is 144/169. but because it is on the positive part of the axis, cos(t)=12/13.
109605
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"
497748
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
562730
cos2t=cos^2t-sin^2t or 2cos^2t-1 or 1-2sin^2t\n\nso you can arrive at something like 2cost-sect (remember 1/cost=sect). that's the simplest answer that you can get.
551883
The answer is C\nwell, one side's length of a triangle should be shorter than the other 2's plus (2 sides plus should be longer than one side)\nthen: 3x+4+8x+2>6x-1\n      8x+2+6x-1>3x-4\n      6x-1+3x+4>8x+2\n--> x>-7/5\n    x>3/11\n    x>-1\nThus, the answer should be x>3/11\n:)
722996
Drop a perpendicular to side of length 10 from the vertex where the side of length 8 ends.\n\nA right angle triangle is formed,\n\nthe perpendicular is 8*sin 60 = 4sqrt(3)\n\nArea of ||gm is 10*4*sqrt(3) = 69.282 approx.
820147
sin(2a)=2*sin(a)*cos(a)\ncos(2a)=cos^2(a)-sin^2(a)
459338
Go to http://mathworld.wolfram.com/Cosine.html
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1. 5m^2 - 15mp - 2mr - 6pr <-- Impossible to factor by grouping. It needs to be either, exactly one of the signs needs to be reversed.\n\n2. 6-3x-2y+xy\n(6 - 3x) - (2y - xy)\n3(2 - x) - y(2 - x)\n(3 - y)(2 - x)
129693
1) square both sides\n\n2) then make trig substitutions to get same 'variable'. Get sin^2[x] throughout equation. \n\ncos^2[x] + sin^2[x] = 1, so\n\n1 - cos^2[x] = sin^2[x].\n\n3) sin[x] cos[x] = ...\n\n4) Set equal to zero\n\n5) Solve.
910165
cos2A = 2cos²A - 1 → cos²A = ½(cos2A + 1)\n\ncos2A = 1 - 2sin²A → sin²A = ½(1 - cos2A)
539310
After 2.5 hrs, Boat A has gone 50 km, and Boat B 80 km. Let the distances of A and B be two sides of a triangle, a 70 degree angle connecting them.\n\nThus we have a triangle with two known sides and one known angle.\n  \nLabel it A, B and C starting from the bottom left angle. Label each angle's opposite side its respective lowercase (a,b, and c). Fill in for b and c, and angle A. We don't know if this is a right triangle or not, so we're going to need to use the law of cosines:\n\na^2 = b^2 + c^2 - 2*b*c*cos(a)\nSubstitute for each known variable:\na^2 = (50)^2 + (80)^2 - 2(50)(80)cos(70 degrees)\na^2 = 2500 + 6400 - 8000 * cos (70 degrees)\na^2 = 8900 - 8000 * .34202\na^2 = 6163.84\na = 78.5101\n\nThus, using the law of cosines, the distance (a) from the two boats is approximately 78.5101 km. It is not exact as there is no exact value for cos (70 degrees).
507417
r/SinR = s/SinS\nSinS = 24*(Sin(38)/20) = 0.73879....\nAngle S = Sin^-1 (0.73879.... = 47.628 degrees\nAngle T = (180-(38+47.628))94.372\nUse same procedure to find side t.