If you have a triangle and you know both degrees and need to find the 3rd degree what do i do?
606300
well, think about it- how many degrees are there in ALL triangles? \n\nif you know 2 angles, you only need one more. Let's say 2 angles were 44 and 72. The two angles combined are 44+72. If you know how many degrees there are overall, you would subtract how many you have from what you know needs to be there!
943161
The faster way to solve this questions is using graphical method. Draw the respective triangles (choosing appropriate scale) and you will get answer.
120928
ok, let the triangle be ABC with sides AB, BC, AC.\ndraw a typical line, get too point on in with the distance of BC, name them B and C. then draw an arc with the center of B and the length of AC and another one eith the center of C and the length of AB. if they had a common point, it would be A and your triangle is there, but if not, there is no triangle with those three sides.
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.
323866
26+64 = 90, and a triangle has 180 degrees, so the other angle must be 90.  A triangle with a 90 degree angle is a right-triangle.
454825
We can do this 2 different ways.\n\nFirst: Use the Pythagorean Theorem. Since all angles are right angles, the diagonal (measure from corner to corner) is the hypotenuse of the right triangle formed.\n\nBy the Pythagorean Theorem, 10^2 + 10^2 =x^2, where x=diagonal\n\nx^2=200\nx=sqrt(200)\nx=10 sqrt 2\n\n2nd way: use 45-45-90 Triangles\n\nWhen you draw any diagonal in a square, you get a 45-45-90 degree right triangle. This is because the 2 sides are the same, so the 2 base angles must be also. The top is 90, so the others must be 45 and 45 to equal each other.\n\nThe sides to a 45-45-90 right triangle are always in the ratio 1:1: sqrt(2) where sqrt(2) is the hypotenuse.\n\nYou know that the sides are 10 feet. Therefore the hypotenuse is 10 sqrt2 feet
728290
The method is entirely dependent on what you know about the triangle in the first place.  If you know the 3 sides, you can use Steve's law for the altitude; ex: (alt to side c) = sqrt(4a^2b^2-(a^2+b^2-c^2)^2) / 2c\n\nIn general, the altitudes of a triangle intersect at a point called the orthocenter, whose location is described in detail at\nhttp://www.cut-the-knot.org/triangle/altitudes.shtml
775751
You need something more, unless you are talking about a right triangle (since you'd then know an angle).
955686
The orthocentre of a triangle is the point where all the altitudes meet\n\nSo if you have ΔABC one altitude is the line through A perpendicular to BC, another is the line through B perpendicular to AC and the last is the line through C perpendicular to AB\n\nUse these facts with the coordinates of A, B, and C and the slopes of AB, AC and BC (Find where two altitudes meet and if you are correct that point will also lie on the 3rd altitude)\n\neg Find the orthocentre of the ΔABC where A ≡(1, -3), B ≡(-1, -5), C ≡(3, 1)\n\nSlope AB = Δy/Δx\n= 2/(2) \n= 1\n\nSlope of perpendicular = -1 (-1 * 1 = -1)\nAltitude to AB goes through C\nSo equation of altitude of form y = -x + b\n(3, 1) lies on altitude\nTherefore 1 = - 3 + b\nb = 4\nThus altitude through C is y = -x + 4\nie x + y = 4 ............... (1)\n\nSlope AC = Δy/Δx\n= 4/2\n= 2\n\nSlope of perpendicular = -1/2 (2 * -1/2 = -1)\nAltitude to AC goes through B\nSo equation of altitude of form y = -1/2 x + b\n(-1, -5) lies on altitude\nTherefore -5 = -1/2 * -1 + b\nb = -11/2\nThus altitude through C is y = -1/2 x - 11/2\nie x + 2y = -11 ............... (2)\n\nSlope BC = Δy/Δx\n= 6/(4) \n= 3/2\n\nSlope of perpendicular = -2/3 (3/2 * -2/3 = -1)\nAltitude to BC goes through A\nSo equation of altitude of form y = -2/3 x + b\n(1, -3) lies on altitude\nTherefore -3 = -2/3 * 1 + b\nb = -7/3\nThus altitude through C is y = -2/3 x - 7/3\nie 2x + 3y = -7 ............... (3)\n\nie we have:\n x + y = 4 ................. (1)\n x + 2y = -11 .................. (2)\n2x + 3y = -7 ............... (3)\n\nEquation (2) - equation (1)\ny = -15\nFrom equation (1) x = 19\n\nCheck in equation (2) \nx + 2y = 19 + -30 = -11 YESSSSSSSSS!!!!!!!!!!!!!\n\nCheck in equation (3) to ensure that this point lies on this altiude also\n2x + 3y = 38 + -45 = -7 YESSSSSSSSS!!!!!!!!!!!!!\n\nSo the orthocentre of ΔABC is (19, -15)
777719
Mark you first two corners 20 m apart.\nThe third marker must be 20 m away from one corner and the other must be 20*sqrt(2) = 28.284 meters from the other corner (The length of the diagonal of a square is sqrt(2) * the side, so that distance will ensure that you have right angles.)\n\nSo tie a length of rope (no stretch), to each of the first two markers.  Cut the ends at 20m and 28.284 m respectively.  Hold the two ends together and find the spot where both ropes are taut.  That's where your third marker goes.\n\nAgain, your last marker should be 28.284m from its opposite.
389433
If you're asking how to do it: Just take the total number of degrees of the angle, divide by two, and there you go, you know the measurements of the 2 equal angles.\nanswered by Fading Away...\n\nAngle bisector Theorem.\n\nIf you are a Geometry student who knows constructions, locate a radi from your vertex and draw an arc. Choose another radii on that new arc that you just drew(That is more than half the arc) from the point where your newly drawn arc intersects the segment. Draw another arc. Repeat process on opposite side. Draw bisectore from vertex through intersection of two arcs.\n\nHope that helps!\nAnswered by Emily\n\n\n\nThis was asked and answered 13 hours ago, click this link.\nhttp://answers.yahoo.com/question/;_ylt=Al6hLACuaPLouhbP7BmwxcIjzKIX?qid=1006041024466
735084
a normal triangle has 3 sides. asking q just for fun?
980809
4 rules\n1.two sides and the included angles of the two triangles equal\n2.2 angles and a side of one triangle equal to 2 angles and a side of theothrtriangle\n3.3 sides of one equal tothe three sides of theother\n4.in 2 right angled triangles if the  hypotenuses and a side of each are equal
963860
where is the right angle in the triangle.  Between the right and bottom, between the bottom and left?  Without knowing this, we can't tell you the answer.\n\nBUT\nin a right triangle\nhypotenuse squared = side 1 squared + side 2 squared
247800
it is basically:\na2 + b2 = c2   -->  (A squared)+(B squared)=(C squared)\n  \n\nif you have a right triangle.you have three sides.the two legs.and the hypotenus(sp?).a and b are the two legs and c is the hypotenus.\n\nin an equation.if you are given a and b.you can square each of them and then add the two values together.the will give you c squared.to find c you then need to take the square root of the value(a squared + b squared).\n\nif you have c and a leg(example:a).you square c.and then subtract a squared from that value(c squared).that will give you b squared.to get b you just get the square root of c squared - a squared.\n\n\nhope that helps.
534053
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.
1030296
for it to be a triangle the angles have to measure to 180, so if you have a 90 degree andlge off one, you could subtract 90 from that, take the remaining 90 and work with that, since you know your need a 1:1 of the legs , then you know you need an equal angle.. angle has to equal 90/2... and thats where i get stumped...\n      gl