Lets say i have a non-right triangle and i need to find the hypothesis.\n\nAcute Triangle\nBase = 2\nVertical Side = 4\nAngle = 75, which is opposite of the hypothesis\nHypothesis = x\n\nObtuse\nBase = 2\nVertical Side = 4\nAngle = 110\nHypothesis = x\n\nI've tried using cos(x) or c^2 = a^2 + b^2 - 2ab cos(C), however that only applies to right triangles.
209753
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.
310677
if the triangle is a regular triangle\n\na = (1/2)bh\n\na = (1/2)(6)(6 * sin60)\n\na = (1/2)(6)(3√3)\n\na = 9√3 = 15.588\n\n\nfor a regular polygon, \narea = (n*t^2) / (4*tan(π/n))\n\nwhere n = number of sides & t = length of side
189358
Initial case n = 3\n180(3-2) = 180 which is known to be true for a triangle\n\nAssumption: Angle sum of n sided polygon = 180(n-2)\nProve: Angle sum of n+1 sided polygon = 180((n+1)-2) = 180(n-1)\n\nNow, take any n+1 sided polygon, and split it into an n sided polygon and a triangle by drawing a line between 2 vertices separated by a single vertex in between. The total angle sum of the n+1 polygon will be equal to the angle sum of the n sided polygon plus the triangle.\n\nUsing the assumption, the angle sum of the n-sided polygon is 180(n-2). The total angle sum therefore is\n\n180(n - 2) + 180 = 180(n - 2 + 1) = 180(n - 1)    QED\n\nTherefore since it is true for n = 3, and if it is true for n it is also true for n+1, by induction it is true for all n >= 3.
636895
The  Pythagorean Theorem gives the length of the hypotenuse (the longest side) of a right triangle ( one of the angles is 90 degrees )\na^2 + b^2 = c^2\n\n\nApplying the Pythagorean Theorem twice you get a three dimensional version \n\na=x\nb=y\nx^2 + y^2 = c^2\nc^2 + z^2 = h^2 (using c as the side of a new triangle)\nx^2 + y^2 + z^2 = h^2\n\nwhich is the length of any line from (0,0,0) to (x,y,z)\n\nthe distance formula just finds the change in x, y, and z and puts it into this formula\nx = x1-x2\ny = y1-y2\nz = z1-z2\n\nThe midpoint formula is really unrelated to the other two.\nAll it does is average each of the coordinates of the other two points.  ( (x1+x2)/2 , (y1+y2)/2 , (z1+z2)/2 )\nThe only thing it has in common is the use of the same type of coordinate system
411756
Start by drawing a picture of this.  Draw a right triangle.  I put the right angle at the bottom left.  Then draw an arbitrary line from the upper left vertex down to the base.  This separtes the large right triangle into a smaller right triangle and what I will call the "outer triangle."  This gives you the 1000 Jim moved; it's to the right of where the line intersects the base.  I called this leg a.  I drew in the angles I was given, and labeled all of the line segments.  The hypotenuse of the large right triangle is b, the hypotenuse of the smaller right triangle (also the only unlabeled leg of the outer triangle) is c, and the vertical leg of the right triangle is d.  d is the height of the escarpment.  I named the angles after the sides they face.  So the 10 degree angle is C, the 32.5 degree angle is D, and angle A is opposite leg a, and is an acute angle in the outer triangle.  Since I was given angle D, I could find its supplement, B, which is an obtuse angle in the outer triangle, opposite leg b.  B = 180 - D = 180 - 32.5 = 147.5.  Since the sum of angles in a triangle is 180, and I know C and B, I can also find A = 180 - B - C = 180 - 147.5 - 10 = 22.5.\n\nNow that I know every angle in the outer triangle and the length of one leg, a, I can use the law of sines.  The law of sines says that a / sin(A) = b / sin(B).  Rearranging, b = a*sin(B) / sin(A) = 1000*sin(147.5) / sin(22.5).  Now look at the larger right triangle.  You know the length of the hypotenuse, b, and you know the angle C opposite the unknown d (I called it C because it is also opposite leg c in the outer triangle.).  By trigonometry, d = b*sin(C) = b*sin(10), where we use b from above.
121903
Bob, I used to enjoy your show before you started doing all those really expensive houses.  Then, of course, you left.\n\nOh, well.\n\nDon't you know, Bob, that isosceles triangles don't have "right sides?"  An isosceles triangle has two "legs" of equal length.  If it is a right isosceles triangle, the side opposite the right angle is called the hypotenuse.\n\nSay hi to Norm for me!
477825
Listen to the first guy.\nThe distance on the y axis is 1 ((-5)-(-4)) and on the x axis it is 3 ((-3)-0).\nThis gives you a right angles triangle with sides of 3 & 1. \nUnder Pythagoras theoem the sum of the squares is equal to the square of the hypoteneuse so\n1^2 + 3^2 = 10.\nTherefore the distance is the square root of 10.
278634
Memorize that for any triangle having sides of length a, b, and c and angle C opposite side c,\n\nc^2 = a^2 + b^2 - 2abcos C .\n\nNote that for C = 90 degrees where cosine of 90 degrees is zero, the rule reduces to the pythagorean theorem.
402278
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.
391197
if you have a right triangle with sides of 3,4, and 5, you can see what this means very easily\n\nif you sqaure the 3 and the 4, you get 9 and 16\nthen you add these two together to get 25\nthen you take the square root of 25 and you get 5\n\nso\n\nx^2 + y^2 = r^2\n\nit only applies for right triangles
540608
Draw a right angled triangle.  the vertical is 35m and the sloped line, the hypotenuse, is 50m.  We have to find the angle between the string and the ground.\n\n35 is the opposite, 50 is the hyp, so it is a sine problem.  Sin x = opp/hyp = 35/50.\n\nusing the inverse sine function on the calculator, we find that the angle is 44.4 degrees\n\nhope this helps
477852
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
577643
Use the Law of Sines.\n\nsin A/a = sin B/b\n(sin 36)/24 = sin B/47\nsin B = (47 sin 36)/24\nsin B = 1.151079452\nAt this point you have a problem--it is impossible for the sine of an angle to have a value greater than 1.\n\nSo either you copied the problem wrong, or you are being asked to recognize that a triangle cannot be formed with these dimensions.
137770
its just triangle mathe. remember soh -cah -toa... sine =opposite side/hypotenuse,  cosine=adjacent side/hypotenuse, tangent= opposite side/adjacent side....cosecant(hypotenuse/opposite side) is the opposite of sine, secant(hypotenuse/adjacent side) is the opposite of cosine and cotangent(adjacent side/opposite side) is the opposite of tangent
444617
Draw a picture of the triangle formed to see that the side opposite the angle is 24 and the side adjacent is k, with 25 being the hypotenuse.  Sine = opp. leg/hyp. which is 24/25\nCosine = adj. leg. / hyp. = k/25\n\nIf you need to, you can find k using the Pyhhagorean Theorem\n(k squared = 25 squared - 24 squared), keeping in mind it can be positive or negative.
390162
For any given triangle ABC with sides a, b and c.\n\nCos C = (a^2 + b^2 - c^2)/2ab  This is the cosine rule.\n\nSin C =sqrt(1 - Cos^2 C). You may need to adjust the sign of the sign to match the triangle.
513971
Use the fact that the sum of the angles of a triangle add up to 180 degrees. Let A, B, C be the angles of the triangle. The m<A + m<B + m<C =180. If you know m<A and m<B, you can find m<C.\n\nWhen I write m<A, I mean the measure of angle A, not m is less than A.