If you are given values for two sides of an oblique triangle and an angle opposite one of them, how would you?
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you need to use the cosine rule.\n\nA^2 = B^2 + C^2 - 2BC*cos(a)\n\nwhere Capitals are the lengths of sides, and lower case is an angle opposite the side of the corresponding letter.
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The angles all add up to 180 degrees.\n\nIf you know the lengths of all the sides you can use the "Law of Tangents" to find all the angles.  This is advanced geometry and Trigonometry, so rather than type, I am including a link to a pretty good explanation.\n\nhttp://answers.google.com/answers/threadview?id=38131
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Can you give angle measures ?  Cant do much without those or some relationship between the other sides.
406038
The oblique answer is correct. As for the slanted 3-D rectangle, I believe you are referring to a parallelepiped: http://mathworld.wolfram.com/Parallelepiped.html .
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Draw a triangle. Side a is 35m. Side b is 5m, and Angle C (the angle between sides a and b) is 155 degrees (180-25).\n\nsides a and b are the runners path. The remaining side, c, is the total displacement. By the law of cosines, the following is true of a triangle:\n\nc^2 = a^2 + b^2 - 2ab cos C.\n\nc^2 = 35^2 + 5^2 - 350 cos (155)\n\nAnd solving for c will get apprx  39.59 meters.\n\nYou can then plug this back into the law of cosines to get the angle B:\n\nb^2 = a^2 + c^2 -2ac cos B\n\n5^2 = 35 ^ 2 + 39.59^2 - 2 * 39.59 * 35 cos B\n\nSolving for B will get you about 3.06 degrees.
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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
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Wallstream is correct -- start with a right triangle.\n\nLabel the interior angles a, b and c, with c being the 90-degree angle.\n\nNow notice that a and b are complementary angles that add up to 90-degrees.\n\nLet x = sina.  Since the angles are complementary, we also have that x = cosb.\n\nSo we can rewrite as follows:\n\nIf x = sina, then a = arcsinx\nIf x = cosb, then b = arccosx\n\nBut a+b = 90-degrees (aka pi/2).\n\nSo arcsinx + arccosx = pi/2\n\nHope that helped!
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no equal side = scalene triangles\n2 equal sides = isosceles triangles\n3 equal sides = equilateral triangles\n3 equal angles = equiangular triangles\n\ni hope it will help you
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if a is the angle between them then, the formula is given as;\n    if 360/a is even then formula is\n n=(360/a)-1\n if 360/a is odd then,\n n=360/a (if they are placed unsymmetrically)\n n=(360/a)-1 (if they are placed symmetrically)
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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
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Given: \n1) a right triangle\n2) hypotenuse = 2\n3) the other two angles are 70 and 20 degrees\n\nFind:\nLengths of the remaining sides\n\nNote: \nWe have to be more specific here.\n\nLet:\na = side opposite the 70 degree angle (side adj. 20 degrees)\nb = side opposite the 20 degree angle (side adj. 70 degrees)\nc = hypotenuse = 2\n\nSolution:\n\n1) For side a:\n\nsin 70 = opp./hyp. = a/c\nTherefore, a = csin70 = 2sin70 = 1.88 ANSWER\nOR\ncos 20 = adj./hyp. = a/c\nTherefore, a = ccos20 = 2cos20 = 1.88 ANSWER\n\n2) For side b:\n\ncos 70 = adj./hyp. = b/c\nTherefore, b = ccos70 = 2cos70 = 0.68 ANSWER\nOR\nsin 20 = opp./hyp. = b/c\nTherefore, b = csin20 = 2sin20 = 0.68 ANSWER\nOR\nPythagorean Theorem: a^2 + b^2 = c^2\nb^2 = c^2 - a^2\nb^2 = 2^2 - 1.88^2\nb^2 = 0.4656\nTherefore:\nb = 0.68 ANSWER
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If you are given the arc length of s,\n\nThe angle may be found by\n\ns/(2*PI*r) * 360 degrees.
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Since there are four quadrant that make up the 360degrees. Draw the two axis x and y. Assume the top right part as quadrant 1, top left part as quadrant 2, bottom left as 3 and bottom right as 4. Now the calculation of angle starts from quadrant 1 in anticlockwise direction to make an angle of 360degrees. Now use "ASTC" rule i.e. assign 'A' to quadrant 1, 'S' to quadrant 2, 'T' to 3 and 'C to 4. \n\n'A' states that in this quadrant all i.e. sin, cos and tan have positive values.\n\n'S' states that in this quadrant sin have positive value and cos and tan have negative values.\n\n'T' states that in this quadrant tan have positive value and cos and sin have negative values.\n\n'C' states that in this quadrant cos have positive value and sin and tan have negative values.\n\nthat's it simple enough.
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From symmetry, you have all the angles\n69 for two, and 180-69= 111 for the other two.\nUse the law of cosines: for triangle abc with opposing angles A,B,C we have the length of the 3rd side c for given a,b from:\n\nc = sqrt(a^2 +b^2 -2abcos(C))\n\nIn your case: a=4, b=6 and C = 69 so\n\nBD = sqrt(36+16-48(cos(69)) = 5.9\n\nFor C = 111 we have:\n\nAC = sqrt(36+16-48(cos(111))= 8.32
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It is possible only when the given triangle is isosceles or equilateral. It's also possible for right-angled triangles but only when the given angle isn't 90 degrees.
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1. All angles are equal\n2. All angles are less than 180° \n3. Size of each angle = (180 - 360/n)° where n = number of sides
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We can find angle B by subtracting the others from 180:\n\nB = 180 - 32 - 44\n= 104\n\nThe length of the sides can be found using the law of sines:\n\n(sin A)/a = (sin B)/b = (sin C)/c\n\nSo,\n\n(sin B)/b = (sin C)/c\n(sin 104)/b = (sin 44)/20\nb = 20(sin 104)/(sin 44)\n~ 27.94\n\n(sin A)/a = (sin C)/c\n(sin 32)/a = (sin 44)/20\na = 20(sin 32)/(sin 44)\n~ 15.26