Find the missing side of a triangle?
927845
Well, if you're in Geometry, you can simply remember that 5, 12, 13 is a Pythagorean triple. (Just like 3, 4, 5)\nIf you don't want to memorize those things, you can simply use the Pythagorean theorem. (a^2 + b^2 = c^2) \nIn this case, a=5, b=12, and c is your hypotenuse (unknown).\nSo... (5)^2 + (12)^2= (x)^2\n25+144=169\n13 x 13= 169\nTherefore, your answer is 13!
876350
Sorry, but your question doesn't make sense.  Maybe you mean the midpoint of one of the sides of a right triangle.
968872
Let the shorter leg be 'x'\nLonger leg = 'x + 1'\nHypotenuse = 'x + 2'\nx^2 + (x + 1)^2 = (x + 2)^2\nx^2 + x^2 + 2x + 1 = x^2 + 4x + 4\n2x^2 + 2x + 1 = x^2 + 4x + 4\nx^2 + 2x + 1 = 4x + 4\nx^2 -2x - 3 = 0\nx^2 + x - 3x - 3 = 0\nx(x + 1) -3(x + 1) = 0\n(x - 3)(x + 1) = 0\nEither x - 3 or x + 1 = 0\nIf x + 1 = 0, x = -1 which cannot be true for the triangle\nIf x - 3 = 0, x = 3\nTherefore, the sides are:\nShorter leg = 3cm\nLonger leg = 3 + 1 = 4cm\nHypotenuse = 3 + 2 = 5cm
89549
Plot the three boundary lines, shade the correct sides; the answers occur at the vertices of the shaded area (a triangle).  Just plug in these vertices to C = 4x + y and see what the highest and lowest outcome is.
109605
Just use Cosinus Theorem: c^2=a^2+b^2-2abCos(gamma)\n1.\nc=sqrt(4+64-2*2*8*sqrt(3)/2))\nc=6.34\n\n2. Is not a right Triangle.. but anyway lets find the angle A:\n22^2=15^2+33^2-2*15*33*Cos(A)\nCos(A)=830/990\nA=ArcCos(83/99)\nA=33°1' 48.54"
302936
Without knowing the measurements of the sides, you can only figure this out if all sides of the polygon are equal in length. And this is how...\n\nSee, first you draw any polygon, except a triangle. So before doing it, you should know that all the angles of a triangle must add up to 180 degrees. (You can do it with a triangle too, but you would be using what you’re hoping to discover.) From the center of the polygon, draw a line to each vertex. This will create a number of triangles, inside the polygon, equal to its number of sides. Knowing that the angles inside each triangle must add up to 180 degrees, you simply multiply 180 by the number of triangles, which is also the number of sides. Then, realizing that the angles in the center of the polygon, which do not contribute to the angles of that polygon, form a complete circle, and therefore must add up to 360 degrees, you subtract 360 from the result. So the angles in any polygon must add up to 180x-360, or 180(x-2), where x is the number of sides. To find the measure of each angle, in a polygon where all sides and angles are equal, you divide that by the number of sides. The equation then works out to (180(x-2))/x.
381599
i think its a triangle
331507
sorry hunny you need at least 2 side lengths in order for us to help you figure it out... plus once you have that the formula for figuring it out is using the pathagorean theorem! a2+b2=c2 (s = squared) say for instance... the 2 sides of your triangle are 7 and 3inches. 3>2+7>2=c>2\nwhich would turn into 9+49=c>2~~~~~~~ then 58=c>2 then to get c by itself(which is needed) you would take the square root of 58! your hypotenuse would equal 7.6 or rounded, 8!\nnow if you use my steps you should be able to do your problem! I'm a 10th grader in geometry and it's the most lovely subject!
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.
937186
WTF IS THE TRIANGLE AND THE U FOR AND HOW DO U MAKE THE TRIANGLE... THERE ISNT A KEY OR ANYTHING...
905611
If u r saying that u have three sides of a triangle as a,b,c\nAnd u want to determine angle b/w them, u can use the formula\ncosA=(b^2+c^2-a^2)/2bc where A is the angle formed by the sides b and c.similarly if B is the angle formed by a and c, it can be calculated as\ncosB=(a^2+c^2-b^2)/2ac\nonce u have calculated A and B, subtract their sum from 180, u will get third angle.\nRemember u will have to use inverse table for cos to find A and B
472052
ok so i assume you know the basic rules of trigonomerty for a triangle.\ni.e sinR = opposite/hypotenuse and cosR= adjacent/hypotenuse.\n\ndraw the triangle on a piece of paper.\nYou will see that sin R = ST/SR\n\n(where ST is the edge running from S to T and similarly SR is the edge running from Sto R)\n\nalso cos R= RT/SR\n\nso sinR^2 + cosR^2 = (ST)^2/(SR^2) + (RT)^2/(SR^2)\n\n= [(ST)^2 + (RT)^2]/(SR^2)\n\nNow we use pythagoras' theorem. I assume u know this? The sum of the squares of the two shorter sides of the triangle equals the square of the longer side (the hypotenuse)\n\nso (ST^2) + (RT)^2 = SR^2\n\nSO WE NOW HAVE\nsinR^2 + cosR^2 = SR^2/SR^2 = 1\n\nthere you go\n\n\nCANT E-MAIL YOU BACK WESTILE BECAUSE IT SAYS YOUR E-MAIL IS NOT VERIFIED
548292
Assuming that an area may be dissected into triangles, then the area of each triangle is given by \n\nArea = (s(s-a)(s-b)(s-c))^(1/2)\n\nwhere s = (1/2)(a + b + c)\n\nand a, b, and c are the lengths of sides of each triangle -\n\nThe sought for area is (finally) the sum of all the areas of the triangles.\n\nGiven sufficient patience, any arbitrary area will be found.\n\n(Drawing a diagram while slogging through the exercise and looping around the separate triangles in the same direction can yield a shortcut whereby all of the internal perimeters eliminate each other, leaving only the perimeter of the original figure.)
1176005
no, it won't..\n\nHOWEVER there are some other theorems through which you can calculate the length of a certain side of a triangle given the lengths of the other the sides.\n\nLikewise it is also possible to calculate (through analytical geometry) the area of any triangle given the lengths of all sides.\n\nSo, practically, given the lengths of at least two sides of a triangle, you can calculate the length of the third side and then calculate the total area of the triangle.\n\n(8th grade mathematics)
599733
I believe it would be one-half times the length of each diagonal.\n\nIn other words, if the diagonals are 4 & 6 the area would be:\n\nA = 1/2 * 4 * 6 = 12\n\nIf you don't know the diagonals, but you know the length of the sides, you could construct the diagonals and calculate their dimensions using the Pythagorean theorem:  c^2 = a^2 + b^2, and then solve from there.  Keep in mind that you'll have to do some extra geometric analysis to find the length of each leg of the right triangles that are created by the intersection of the diagonals before you can use the theorem.
324000
ok, we can call the three sides x,y,and z respectively. all angles of a triangle have to add up to 180. so\nx+y+z=180\n\nok. y=2x and z=y+10\n\nsubstitute\nx+2x+2x+10=180 then solve for x\n\n5x+10=180\n5x=170\nx=34 degrees
847510
Several ways. The first that comes to mind is to figure the area with Heron's formula (area = sqrt(s(s - a)(s - b)(s - c)), where s = (a + b + c)/2). The area can also be figured as 1/2 base times altitude, so you know the altitudes of the triangle. The altitudes give you right triangles where you know the hypotenuse (a side of the triangle), and a leg, so you can compute the trig functions of all the angles. Then the angles can be computed as arcsin or arccos of the angle (arcsin(altitude/side))