1 More Hypotenuse Question?
962823
Can you give angle measures ?  Cant do much without those or some relationship between the other sides.
921650
The area of a triangle is half base times height.\n\nCould you properly define the length of the base, height and hypotenuse?
748059
First use the Pythagorean Theorem.\n\nThe hypotenuse is 20. One of the legs has length x. Call the length of the other leg y.\n\nx^2 + y^2 = 20^2\nx^2 + y^2 = 400\n\nNow solve for y in terms of x.\n\ny^2 = 400 - x^2\ny = sqrt(400 - x^2)\n\nNow, since the triangle is right, the two legs are the altitude (or height) and the base.\n\nA = (1/2)*b*h\nA = (1/2)*x*y\nA = (1/2)*x*sqrt(400-x^2)\n\nThere's the area in terms of x!
803546
Use the pythagorean Therym(?)\nThat is, well, here is an example:\nIf one leg of a right triangle is 6 units, and the other leg is 8 units, what is the legnth of the hypotenuse?\nthat is:\nleg 1=a, leg 2=b, and the hypotenuse = c\nThe hypotenuse always = c, it is the side parallel to the right angle of the triangle.\nthe pythagorean therym(?) is a^2(a squared)+b^2(b squared)=c^2(c squared)\na=6, b=8, and c=?\n6^2+8^2=c^2\n36 + 64= c^2\n100= c^2\nc=10^2\nso the missing side would b 10 units, just use that example for your problem, and remember, if you are trying to find the legnth of a leg, just plug in the one leg length that you have (this would be "a") and the hypotenuse(this would be "c") and subtract the square of "a" from "c". find the square root, and vua la, there you have it. hope that helps
886349
SOHCAHTOA \nSOH =sine=Opposite /Hypotenuse\nCAH cos=adjacent/hypotenuse\nTOA tan=opposite/adjacent\nIn finding the adjacent,we use CAH\ncos19=adjacent/13.8\nadjacent=cos19x13.8\nadjacent=13.048
289962
every one that passes by (-1,6). there are infinite equations. You must give some more clues.\n\n Let me explan myself: \nif your equation were x=-1 (completely horizontal)\nif your equation were y=6 (completely vertical)\nand so on f you have a crcle, a hperbolic, a sinoidal, a straight line with some inclination.... there are just a lot!!!!\n\nhope i help\n\nthanks for the details\n\nJust substitute x=-1 on each equaton, and the one that results on y=6 is the one you are looking for.
28789
Errata corrige to the last answer: the pythagorean triple is 15,20,25 and not 12,15,25.\n\nHope this helps
701134
a and b can be either one and not the hypotenuse.\nc can ONLY be the hypotenuse. \nSome triples:\n3-4-5   5-12-13   7-24-25   9-40-41\n\nHere how you find them\nstart from 3-4-5\n3*1+1 = 4, 4+1 = 5\nSo first set = 3-4-5\nNext odd number\n5*2+2 = 12, 12+1 = 13\nSo second set = 5-12-13\nNext odd number\n7*3+3 = 24, 24+1 = 25\nSo third set = 7-24-25\nNext odd number\n9*4+4 = 40, 40+1=41\nSo fourth set = 9-40-41\n\nAND SO ON.....
406591
Switch your calculator from degrees (DEG) to radians (RAD). That should do the trick. \nHere are some quick trig rules you need to know:\nIn radians just remember that pi is 180 degrees. Pi/3 is 60 degrees, pi/2 is 90, pi/4 is 45 degrees and so on. \nIn a unit circle, the hypotenuse is always 1.\nSin is always the far side away from the angle (the vertical side). We call that the opposite side.\nCos is always the adjacent side (or horizontal side).\nIn a right triangle with angles 60 and 30, the shortest side is always 1/2 the hypotenuse, so in a unit triangle, it is just 1/2. You can use the pathagorean theorem to find that the longer side is the square root of (3/4), or (sqrt(3))/2.\nIn a right triangle with angles 45 and 45, the opposite and adjacent sides are equal, and we can use the pathagorean theorem to find that the sides are the square root of (1/2), or 1/(sqrt(2)).\n\n1. Sin 1/3 pi. This is the same as sin pi/3 which is a 60 degree angle. If you draw the triangle on a unit circle you would see that the opposite side is the longer side of a 90/60/30 triangle. We have seen above that the length is (sqrt(3))/2.\n\n2. Cos pi. Easy. pi is 180 degrees, so draw the unit circle. You see that it is just a striaght horizontal line going the other way so the answer is -1.\n\n3. cos pi/4 is a 45 degree angle. We know from above that both sides have to be 1/(sqrt(2)).\n\n4. sin pi/4. Same as #3.\n\n27. What is 315 degrees? Well draw your unit circle. Go, 90, then to 180, then to 270, then go 45 more degrees and you will end up at 315 degrees. Draw your triangle. We know that the hypotenuse is 1, so the coordinates will simply be, (cos 315, sin 315). Cos 315 is just a 90/45/45 triangle in the bottom right quadrant so the adjacent side is positive. The opposite is negative, so you end up with coordinates (1/(sqrt(2)), -1/(sqrt(2))). Do the same to solve the last question youeself.
821495
a) Assume the following:\n\nAngles a= 10 deg, b= 80 Deg, c=90 deg \nHypotenuse length = 2.5 m\nheight = 2.5 * cos 80 Deg= .434 m\nlength = 2.5 * sin 80 Deg= 2.462 m\nacceleration = "g" * Sin 10 Degree
984494
sine opposite/hypotenuse\ncosine adjacent/hypotenuse\ntangent opposite/adjacent
42660
One. \n\nThis relation is really just Pythagoras' theorem, hiding in a new form. Recall:\n\nH^2 = A^2 + B^2 \n\nThat is, the square of the hypotenuse H is the sum of the squares of the other two sides A and B, for a right triangle with sides A, B and hypotenuse H.  \n\nOf course, A=H*cosx and B=H*sinx for some angle x, so that \n\nH^2= (H^2)*(sinx*sinx + cosx*cosx)\n\nleading to the result:\n\n1=sinx*sinx + cosx*cosx\n\nThis result also is nicely connected to the simple fact that \n\nx - x = 0 \n\nbecause \n\nexp(i*x)=cosx + i*sinx and\nexp(-i*x)=cosx -i*sinx\n\nwhere i=sqrt(-1). \n\nThen multiplying these two, \n\nsinx*sinx + cosx*cosx=exp(i*x)*exp(-i*x)=\nexp(i*[x-x])=exp(0)=1
1104008
This forms a right triangle.  You can use the pythagorean theorem to find L.\nThe hypotenuse is 3L - 2, and the legs are L and 2L + 2\n(3L-2)^2 = L^2 + (2L+2)^2\n9L^2 - 12L + 4 = L^2 + 4L^2 + 8L + 4\n9L^2 - 12L + 4 = 5L^2 + 8L + 4\nSubtract 4 from each side.\n9L^2 - 12L = 5L^2 + 8L \nMove everything to one side.\n4L^2 - 20L = 0\nFactor\n4L(L-5) = 0\nL = 0 or L = 5\nThe height can't be zero; therefore, L = 5\ntan(x) = L/(2L+2)\ntan(x) = 5/(10+2)\nx = arctan(5/12)\nx = 22.619864948\nx = 23
633404
ANSWER: \n34 metres\n\nEXPLANATION:\nLet one leg be 'x' units long. \nThen hypotanuse = 2 metres + 2*(length of leg) = (2 + 2x) metres.\nThe other leg is 30 metres.\nIn right angled triangle using Pythagorus' theorem\n(Hypotanuse)^2 = (One leg)^2 + (Another leg)^2\nor (2 + 2x)^2 = x^2 + 30^2\nor 4 + 8x + 4x^2 - x^2 - 30^2 = 0\nor 3x^2 + 8x - 896 = 0\nor 3x^2 + 56x - 48x - 896 = 0\nor x(3x + 56) -16x(3x + 56) = 0\nor (x - 16)(3x + 56) = 0\n\nTherefore x = 16 [as x = -56/3 is inadmissible ('x' being length of a side have to be positive)]\nHence the hypotanuse = 2 + 2*16 = 34 metres
488283
I think that what you really want here is the tangent ratio. You know the slope of the line, right (hint, it is right there in the equation)? You know that the tangent ratio is a over b (where a and b are the non-hypotenuse sides of the triangle), right? You should be able to use that to figure out the tangent of the angle made by the line. Now all you have to do is figure out the angle that has that number as a tangent. That would be the "tan -1" (inverse tangent) function on your calculator.\n\nIf you are working from trig tables rather than a calculator, find the number in the second column of the tangent table that is closest to your tangent number then look in the first column for the angle.\n\nIf you are required to use sin and cosine remember your Pythagorean theorem and use it to find the hypotenuse.\n\nASIDE TO Drakestone: Your solution only finds the sine of the angle, not the angle itself. You need to apply "sin -1" to your answer to get the final answer.\n\n\nASIDE TO eternal...: You are wrong too. graph y=5x+3 and you will see that it is way too steep to be an 11 degree angle. You mixed up your opposite and adjacent values in calculating your tangent.
759539
Where c is the hypotenuse, and a and b are legs of the triangle:\nc² = a² + b² --- Pythagoren Therom\nc² = (12)² + (20)² --- Replace what you know...\nc² = 144 + 400 --- Find 12² and 20²...\nc² = 544 --- Add them together...\nc = √544 --- Square root of 544 equals c\nc ≈ 23.32"
1105810
a² + b² = c²\n\nWhere c is the hypotenuse.\n\nDoes (5 x 5) + (12 x 12) = (13 x 13) ?\n\n25 + 144 = 169, so yes it is right.