does a triagle with a hypotenuse have to be right, or can it just be with one longer side?
155406
By definition,hypotenuse is the side opposite the right angle. So it has to be a right angled triangle.
780884
As with many problems, it helps to draw a diagram.  I made a triangle, and then put in each of the altitudes, since they are my only known values.  With these in place, we now have 6 principal right triangles, of which we know the value one leg (the altitude).  In each of these triangles, the hypoteneuse is a complete side of the original triangle, and the angle opposite the known leg is also of the original triangle.  Personally, I chose to work with cosines and tnagents, but you could just as easily have used sines (though I think it might add a step or two).  Remember that the cosine of an angle in a right triangle is merely the ratio of the side opposite the angle to the hypoteneuse.  Don't worry that we don't know the value of any of these angles; I went through the entire problem without ever needing it.\n\nSo, since each angle, in its entirety, is in two of our right triangles, they give us two ratios to compare.  That is:\n\nIf the sides of the original triangle are x,y,z\ncos a1 = 20/x = 15/y\ncos a2 = 20/z = 12/y\ncos a3 = 12/x = 15/z\n\nWith a little cross multiplication and rearrangement of terms, we find that:\n12z = 15x = 20y.  \n\n(Your x, y, z may be in a different order depending on how you labeled your triangle, but if you manipulate your triangle a bit, you should be able to get it to match.  This will keep your coefficients identical to mine.  But even if you don't match it up, in the end you will get the same answer.  That's the beauty of math.)\n\nThe tangents are a little trickier, because they involve a fraction of the original side.  This involves the addition of a new variable on each, such that one portion of the line is "a", and the remainder is x - a; and then b and y - b; c and c - z.  The tangents then follow:\n\ntan a1 = 20/(y - b) = 15/a\ntan a2 = 12/(z - c) = 20/b\ntan a3 = 15/(x - a) = 12/c\n\nRearranging each gives us the three equations:\n20a = 15(y - b)\n12b = 20(z - c)\n15c = 12(x - a)\n\nFrom there, you can solve for either a, b, or c.  It's your choice.  I opted to solve for a.  Substituting the third equation into the second, and then that new equation into the first, we get:\n\n20a = 15(y - (20/12)(z - (12/15)(x - a)))\nwhich reduces to\n8a = 4x + 3y - 5z\n\nNow, side a is in a right triangle with hypoteneuse y and second leg 15, meaning that:\na^2 + 15^2 = y^2   or\na = sqrt(y^2 - 225)\n\nUsing this equation, plus the ratios we received from the cosine functions, we solve for y:\n\n8 * sqrt(y^2 - 225) = (16/3)y + 3y - (25/3)\n\nConveniently, the right half is equal to zero, so dividing by 8 and squaring:\n\ny^2 - 225 = 0\ny^2 = 225\ny = 15\n\nEnter this into our ratio:\n\n15x = 20y\n15x = 20*15\nx = 20\n\nand\n\n12z = 20y\n12z = 20*15\nz = 25\n\nSo finally we have the lengths of all three sides, and adding them together, the perimeter is\n15 + 20 + 25 = 60\n\nThis turns out to be a right triangle, but that is merely happenstance.  This process can be used to solve for any triangle with three known altitudes.\n\nHope this helps.
1002935
Its the Pythagorean Theorem. It applies to right angle triangles. \nIt states that a^2 + b^2 = c^2\nWhere a and b represent the lengths of legs of the triangle and c is the Hypotenuse.\nThe Hypotenuse is the side that is opposite the right angle (also the longest side). Whereas the legs are the two sides that meet at the right angle.
707890
If you drop a vertical line h representing the balloon's altitude, you have two right triangles.  Let the base of the 83 degree triangle be x.  Now here's where you run into a question and if the book shows a diagram, it might answer that question.  Are the stations looking toward a balloon in between them, or are they both looking in the same direction at a balloon very much farther away, and very much higher?  If it's between them, the base of the 57 degree triangle will be 15 km-x, but if it's on the same side of them both, that base will be 15 km+x.  Do you know the trick to remembering how the sides relate to the trig functions?  Ooh--aah--oh ah--ah oh--ha--ho.  You have to sound this out and it makes a distinctive but stupid sounding little chant.  For the sides opposite an angle, adjacent to the angle and the hypotenuse in a right triangle, sin=o/h, cos=a/h, tan=o/a, cot=a/o, sec=h/a, csc=h/o.  In this case we want the ratio of the height to the base, the opposite side being the height and the adjacent side being the base, so we want the tangents of the angles.  The altitude, opposite side to the angles, will be equal for both triangles.  Call it y.  x will be the base for the 83 degree triangle in both cases, but for the 57 degree triangle the base will be (15 km-x) if the balloon is between the stations or (15 km+x) if it's to the same side of both stations.  Tan 83=h/x and tan 57=h/(15-x) (between) or tan 57=h/(15+x) (same side).  Setting the heights of the triangles equal since the balloon can only be at one height: x tan 83=(15-x) tan 57 if between, x tan 83=(15+x) tan 57 if to the same side.  Solve for x.  Then plug x back into the earlier equations to get y, the altitude.  Work out the answer on your own, but I'll give you this: if the balloon is between the stations, x is between 2 and 3 and y is between 15 and 20.  If the balloon is to one side of both the stations, x is between 3 and 4 and y is between 25 and 30.
660313
the smaller 1 being right u get the hypotenuse as 5..\n\nthe bigger triangle n the smaller 1 are similar...so sides are proportional...(the sides r acually duble of tht of the smaller 1) so the perpendicluar is 6... n the hypotenuse of the biggger triangle is 10..(duble of the smaller 1)...\n\nso perimeter is 8+6+10=24\n\nperimeter of the shaded region is indeed 19..4+6+5=19..got it?
653474
Pick an angale, not the 90 degree angle and call it Theta.  The short side that forms this angle is called the x side of the triangle and the long side that forms this angle is the hypotenuse or 'r'.  The third side is called the opposite side or 'y'.\n\n\n.... /|\n..../.|\n.../..|\n../...|\n./....|\n/___|  Bottom is x, slanting side is r, and right side is \n          the y side.  The angle at the bottom left is Theta.\n\n\nSin(Theta) will = y/r\nCos(Theta) will = x/r\nand\nTan(Theta) will = y/x side.\n\nThis all presumes you don't already know 2 of the sides.  If you know 2 sides already, you use the pythagorean thm.\n\nExampe: Theta = 30 degrees, r = 10 feet, what are the others?\n\nSin(Theta) = y/r  and Sin(30 degrees) happens to be 1/2.\n\nso, 1/2 = y/10.  Solving for y you get y=5.
984494
sine opposite/hypotenuse\ncosine adjacent/hypotenuse\ntangent opposite/adjacent
1176396
sin=opposite/hypotenuse\ncos=adjacent/hypotenuse\ntan=opposite/adjacent
922431
Use pythagorean formula to solve:  a^2 + b^2 = c^2, where a and b are the legs of the triangle and c is the hypotenuese.\n\nIt's a little unclear what you mean by "right side is x" -- not sure if you mean it's one of the legs or the hypotenuese.  I will assume you mean hypoteneuse.\n\nSubstitute 51 and 45 for a and b in the formula to solve for c (or x as it's being called in your problem):\n\n51^2 + 45^2 = x^2\n\n2601 + 2025 = x^2\n\nx^2 = 4626\n\nx = √4626 = 3√514\n\n.
769451
yes ofcoure it is meant to for an example i can tell you\n\n\nthe trigonometry is used in aviation\n\nu know how to calculate distance say a flight from \n\n30 deg north and 45 deg east is landing at 21 south and 36 west \n\n\n\nhere they use right angle  triangle the phythogers theorem ,\n\nmark the two location in a chart scaled say one deg per square,\nthen you can joint the lines to make a right angle triangle,\nno with no of degres in adjacent and opposite side of triange u can calculate the hypotunuse ( sum of suares of other two sides is the sum of squares of hypotenuse).\n\n\nthis is an ex: u have every thing use in one or more way in actual  applications.
1084262
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
461392
If you know what the angle is, divide it by three and it is trisected.
724499
It is a tough Q - coz it is NOT right angle...\nThe best solution would be to name the angles, LHS Triangle as angle x, y z,,,,,     RHS triangle as  x', y', z'\n\nThe actual proof will be long, but the basic idea is\nSum of the Angles of a Tri = 180\nand the angles in straight line = 180\n\nx + y + z =180\n\nx' + y' + z' = 180\n\nz + y' = 180\n\ny' = 180-z                                                 \n\nx = x'\n\nx' + 180-z + z' = 180\nx + z' = z\n\nTherefore, z is larger Angle than x  (or x')\n\nSide opposite to larger angle is larger
80712
If you mean 1 cm and 5 cm, then the answer would be\n\n3 cm\n4 cm\nand a 5 cm hypotenuse.
865033
okay. so. there's a special set of rules for a 30-60-90 triangle.\n\nthe sides are proportional to each other by the ratio \n\nx : v/( 3)x : 2x\n\nwhere the v/ is the square root.\n\nthat means that the greatest angle corresponds to the greatest length.\n\nbecause the longest side is 7.5 meters, that means that is 2x.\n\ndivide by two for x, and that is the 30* length. (because its the smallest angle, therefore is opposite the smallest length)
1186712
arctan x should be interpreted as "the angle whose tangent is x". So draw a right triangle and include an angle whose tangent is x. Then figure out the sine of that angle in terms of x and you're done.\n\nEasiest way to do this is to let the side opposite the angle be x and the side adjacent to the angle be 1. Then the hypotenuse will have length sqrt(1+x^2).
273778
The lengths of a right triangle are a^2 + b^2 = c^2 (Pythagorean theorem).  For 6, 7, 8, this would mean that 36 + 49 = 64, which is false.  For 10, 15, 20 you would need 100+225 = 400, also false.\n\nThe one common ration that works is 3, 4, 5  (9+16 = 25).  Multiples also work, such as 15, 20, 25 (225 + 400 = 625).