Triangles in the Cartesian Plane?
928297
Send me an email, I will give you a PDF file with details
1155045
You are dealing with a triangle.  It is not lying flat on the ground, but all three lines are in the same plane.  One side is 350 ft, another is 400 ft, and the angle between them is 50 degrees.  This sounds like a Law of Cosines problem to me.\n\nCalling the unknown side x,\nx^2 = 350^2 + 400^2 - (2)(350)(400)(cos 50)\nLooks like 320.2 ft to me.
890883
1)One leg of a right triangle is 12 cm long. The diference between the other two sides is 2 cm.\nFind the sides of the triangle. (Pythagorean Theorem)\n\nWhat we can take from this problem is that one LEG is 12 cm long, while the other LEG is unknown (we will call it X) and the HYPOTENUSE -- the longest side of a right triangle -- is also unknown (we will designate it X+2, since it differs 2 cm from the unknown leg).\n\nSetting up Pythagorean Theorem as always:\n\n "Leg squared" + "Leg squared" = "Hypotenuse squared"\n\n  12^2 + x^2 = (x+2)^2\n\nExpanding:\n\n  144 + x^2 = x^2 + 4x + 4\n\nCollecting like terms:\n\n  4x = 140\n\nDivide by 4:\n\n   x = 35\n\nThe unknown leg is 35 cm, while the hypotenuse is 37.  Thus, we can also call the triplet (12,35,37) a "Pythagorean Triple" (of which there are infinitely many!).\n\n\n2) The hypotenuse of a right triangle is 58 cm long. The diference between the other two sides\nis 2 cm. Find the sides of the triangle. (Pythagorean Theorem\n\nIn this case, we know the hypotenuse, but do not know the lengths of the legs.  For convenience, we will again use the expressions X and (X+2) for each side's length.\n\nBy P.T.:\n\n x^2 + (x+2)^2 = 58^2\n\nExpanding:\n\n x^2 + x^2 + 4x + 4 = 3364\n\nCollect like terms:\n\n 2 x^2 + 4x - 3360 = 0\n\nDivide by the GCF (2):\n\n  x^2 + 2x - 1680 = 0\n\nCan we factor this trinomial easily?  We could, if we knew integer factors of -1680 (one positive, one negative) that combine to make 2.\n\nLuckily, -1680 can be factored into -2*2*2*2*3*5*7, and these can be grouped into two sets:\n      2*3*7 = 42\n -2*2*2*5 = -40\n\nFactor:\n\n (x+42) (x-40) = 0\n\nAlthough two solutions come from these factors (-42 and 40), we may only consider the positive answer, since lengths of triangle's legs are always positive!\n\n Therefore, x=40 (Leg 1), and Leg 2 is 42 cm.
337862
go do your homework by yerself next time.\n\nBut becos im feeling nice today...\n\na triangle = 1 triangle , duh\na quadilateral can be divided into two triangles with vertices (points) all at the points of the quadilateral\n5 sided --> 3 triangles\n6 sided --> 4 triangles\n....\n12 sided --> 10 triangles\n1 triangle -->180 degrees\n10 triangles --> 1800 degrees
712617
No, if u draw two right triangles of a common( extended base ) snd then try drawing another triangle ( any type ) by extending the perpendiculars, its impossible as the perpendicualrs are parallel to each other.
463801
The cosine laws apply to triangles which are NOT right angled. I think you have maybe read that wrong:\n\nhttp://www.themathpage.com/aTrig/law-of-sines.htm\n\ngives you a full guide to using the cosine laws to solve problems like these.
683962
that is possibel only if your triangle abc is equilateral or isosceles. otherwise, that wouldn't happen:)
735084
a normal triangle has 3 sides. asking q just for fun?
721808
If you are trying to make a triangle out of three equal lengths, all you do is devide 180 by 3.....and taddaaa you get your angles.\n\nIf does not matter how long each side is.
797707
See the picture:\n\nhttp://mudandmuck.com/str2/sin&cos.jpg\n\nThe two triangles in the picture are equivalent.  (Same angles, and one side of each we know to be the same (hypotenuse=1), so all the sides have the same absolute length, only their signs are different.)
219885
triangle- 3side\nquadrilateral- 4side\npentagon- 5side\nhexagon- 6side\nheptagon- 7side\noctagon- 8side\nnonagon- 9side\ndecagon- 10side\ndodecagon- 12side\nicosagon- 20side
39543
There are seven different (even if rotated, they're still different) ways to join the points together. Multiplied by 5, this gives 35 triangles total.
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.
987754
Actually the only way it would be possible is if you use Elliptic or Riemannian geometry, and would require a manifold.  A representation of such a construct would be similar to drawing a 4 dimensional shape in 3 or even 2 dimensions.\n\nEuclidian geometry doesn't allow triangles with a sum of all angles to be greater than 180 degrees.
722736
the angles are just unknown if i get you correctly.\nthere are sin, cos, and tan proportions you can work out to find the rest. i think you only need three of the measurements to work out the rest of the triangle, like 2 sides and an angle or 2 angles and a side.  not positive.
142784
an isoceles triangle pointing up\nand another pointing down right on top of it
882787
let \nc=hypotenuse\na,b = legs\n\ntherefore:\nc=25\na=2b-10\n\nform pythagorean theorem;\n\na^2+b^2 = c^2\n(2b - 10)^2 + b^2 = 25^2\n4b^2 - 40b + 100 + b^2 = 625\n5b^2 - 40b - 525 = 0\n\ndividing both sides by 5\nb^2 - 8b - 105 = 0\n\nfactoring the left side\n\n(b - 15)(b + 7) = 0\n\nb + 7 = 0\nb = -7 absurd (no side can be negative)\n\nb - 15 = 0\nb = 15\n\nsolve for a:\n\na = 2b - 10\na = 2*15 - 10\na = 20\n\ntherefor the two legs of the right triangle are\n20 cm and 15 cm\n\nchecking:\nc^2 = 20^2 + 15^2\nc^2 = 400 + 225\nc^2 = 625\nc = 25 cm.\n\n\nthat's it