What is the longest side of the right triangle that is across from the right triangle?
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Like in pythagorean therom? Hypotenuse
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When you are trying to find out various measurements of a right triangle (angle measurements or lengths of sides). Try this mnemonic (memory device):\n\nSaddle Our Horses\nCanter Away Happily\nTo Other Adventures
155406
By definition,hypotenuse is the side opposite the right angle. So it has to be a right angled triangle.
715239
Triangle, quadrilateral, pentagon, hexagon, heptagon, octagon, nonagon, decagon
779641
Ahh, a triangle. Since you know the two long sides of the triangle and the angle between them, use the Lore of Co-signs to get the third side (Law of Cosines).
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This is a neat problem, and I am sure that there are many ways to solve it. I will be interested in seeing if the different ways to solve it end up producing answers that are at all similar.\n\nHere is what I did. I'll start with part (b), since I found it to be easier.\n\nIf x > 7, then the largest angle is the angle opposite the side having length x. This is the included angle of the sides having length 4 and 7.\n\nUse the law of cosines to relate the lengths of the two shorter sides and their included angle to the side having length x.\n\nc^2 = a^2 + b^2 - 2abcos(theta)\nx^2 = 4^2 + 7^2 - 2(4)(7)cos(theta)\n\nx^2 = 16 + 49 - 56cos(theta)\nx^2 - 65 = -56cos(theta)\ncos(theta) = (65 - x^2)/56\n\nOr, theta = arccos[(65 - x^2)/56]\n\nIn this case, x must be greater than 7 and less than 11.\n\nIf x is less than 7, then the largest angle is the angle opposite the side having length 7. This is the included angle of the sides having lengths 4 and x.\n\nMy first idea didn't work, so let's try something that could be even uglier :D\n\nCall the included angle of the sides of length 4 and 7 alpha.\n\nUsing the law of cosines, \n\nc^2 = a^2 + b^2 - 2abcos(alpha)\nx^2 = 4^2 + 7^2 - 2(4)(7)cos(alpha)\n\nx^2 = 16 + 49 - 56cos(alpha)\nx^2 - 65 = -56cos(alpha)\ncos(alpha) = (65 - x^2)/56\n\nLet's derive sin(alpha) from cos(alpha).\n\nSince sin^2(alpha) + cos^2(alpha) = 1,\n\nsin(alpha) = sqrt[1 - cos^2(alpha)]\n=sqrt{1 - [(65 - x^2)/56]^2}\n\nAt this point, you're probably wondering what this has to do with theta.\n\nNow that we know sin(alpha), we can use the law of sines on two of the sides of the triangle, and the angles opposite them.\n\na/sin(A) = b/sin(B) = c/sin(C)\n\nIn this case, theta is still the angle opposite the side having length 7. alpha is the angle opposite the side having length x.\n\nSo:\n\nx/sin(alpha) = 7/sin(theta)\n\nor\n\nsin(theta) = 7sin(alpha)/x\n\nWe know sin(alpha):\n\nsin(theta) = 7*sqrt{1 - [(65 - x^2)/56]^2}/x\n\nSo theta equals arcsin [7*sqrt{1 - [(65 - x^2)/56]^2}/x]\n\nThat's not very pretty.\n\nIn this case, x must be greater than 3 and less than 7.
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Side side side, Side side angle, not sure for hl, angle side angle, Angle angle angle, Angle angle side, and angle angle side. It has to do with triangles, and see if they are congruent.\n\nEDIT: HL could that be hypotenuse and length?
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the shortest side will be opposite to the smallest angle in that triangle. So if the shortest length of one triangle is the connected line, the shortest length of the other triangle is the shortest overall.
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Do you mean a right triangle, if so the answer is 13. 5^2 + 12^2 = 13^2. If not, it depent on the angles.
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You have to use the Pythagorean Theorem.  It says the lengths of the two short sides of a right triangle squared, then added together, equal the length of the long side squared.  Picture a triangle on your sheet of graph paper.  One leg goes from point P straight left to 2,4.  The second goes from 2,4 straight down to Q.  The third connects P and Q and is the longest of the three legs.  The length of the first leg is 2.  The length of the second is 5 (or -5 if you like; it doesn't matter since we're fixing to square it).  The lengths squared are 25 and 4.  So the distance between P and Q squared is 29.  The distance is root(29), which is 5.385.  That's your answer.
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There are two names for a seven sided shape, heptagon and septagon.  Both answers are equally correct
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That would be a triangle....
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It really depends upon your problem.\n\nIf you are asked to find the sine, cosine, and tangent then for any right triangle:\n\nsine = opposite side/hypotenuse  = y/r\n\ncosine = adjacent side/hypotenuse = x/r\n\ntangent = opposite side/adjacent side = y/x\n\nwhere opposite side is the side opposite the angle in consideration.\n\nwhere adjacent is the side closest to the angle.\n\nwhere hypotenuse is the longest segment in the triangle.
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use pythagoras theorm,if left hand side is equal to right hand side then it is a right triangle.\n\nsince 8 is the biggest no here that means the length of hypotenues is 8. \n\nlet side 1 be a=4\nlet side 2 be b=4 sqrt(3)\nlet hypotenuese be c=8\n\naccording to pythagoras theorm,\n\n(a)*(a)+(b)*(b)=(c)*(c)\n\nL.H.S\n\n=4*4+4sqrt(3)*4sqrt(3)\n=16+48\n=64\n\nR.H.S\n=8*8\n=64\n\nsince L.H.S=R.H.S\n\ntherefore, it is a right triangle\n\nhope u understand it\nif u hv any prob u can contact me
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I think it's the left side. I don't know if this is exactly what you're looking for, but check out the website below.
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there isn't any, but the longest that i know of are:\npar 3: 250\npar 4: 560\nPar 5: 650
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it's the Pythagorean Theorem\n\nIn any right triangle, the area of the square whose side is the hypotenuse (the side of a right triangle opposite the right angle) is equal to the sum of the areas of the squares on the other two sides.