Two angles of a triangle measure 52° and 34°. What is the size of the third angle?
812747
a + b + c = 180\n52 + 34 + c = 180\nc = 94
899919
16.72 inches squared.\n\n\nDivide an equilateral triangle into two equal right triangles (meaning, draw the height).  The right triangles will have angles of 30, 60, 90, right (an equilateral triangle has 3 60 degree angles)?\n\nThe ratio of the sides in such a right triangle (the smallest leg being a) is a:a(sq(3)):2a, where 2a is the hypotenuse and a(sq(3)) is the other leg.\n\nThe "other leg" in this case is the height.  So a, half of the length of the side of the whole equilateral triangle, is 5.382/sq(3).\n\nMultiply this by 2 to get the whole length of the side.  We have our base and height.\n\n\nThe area of an equilateral triangle is 1/2bh.  b=2(5.382)/sq(3)\nh=5.382\n\nPlug those in and you get 16.72.
726903
since the sum of two  angles is equal to the one angle it  imlies the third angle is 90 degree since \nbecause:\nLet the angles be a,b,c now a+b=c therefore a+b+c=180 now a+b=c therefore 2c=180 this implies c=90 degrees. therefore it is a right triangle hence the measure of the third side according to pythagoras therom is square root of the sum of the squares of both sides
579191
Area of a triangle:\n\nA = 1/2 (B)(A)\nwhere B = base which could be any of the three sides\nand A = altitude (or height), which is the perpendicular distance to the base B from the opposite vertex.
776798
Every side of this triangle has a pair of ends and all the three pairs are cyclic. ie, if AB,BC & CA are the three sides of the given triangle , then the side AB has the ends (4,5),(1,1), BC has the ends  (1,1)and (-2,5) & CA has the pair of ends (4,5),(-2,5).\nBy simple inspection of these pairs we can find that the points C and A are at the same height from the abcissa hence parallel to it. By this we can conclude easily that the length of CA=6 units.\nFurther, to be an isocelous triangle in this case the point B should be on the mid line of CA, and it is normal to CA. The mid point of CA is three units away from its ends C&A. As CA is parallel to absissa we can clearly conclude that any point between (1,0)&(1,5) will be the vertex of an isosceles triangle. Here the point B is (1,1) on the said line. Thus proved that the given points form an isosceles triangle.\nWe alredy know that CA = 6\nAnd the other sides are equal to sqrt(4-1)^2 +(5-1)^2 = 5 units\nNow, we can find the area of this triangle by Heron's Formula \nie sqrt[8(8-6)(8-5)^2] =12 sq units
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.
890630
+---------65km--------------->\n|\n|\n|\n35km\n|\n|\n\nNotice that this is a right (90 degree) triangle.Formula for a triangle with sides a, b, c is:\na*a+b*b=c*c  where c is the hypotenuse\n\n35*35+65*65=c*c\n\nc= Square root(35*35+65*65)  \n=73.8
551883
The answer is C\nwell, one side's length of a triangle should be shorter than the other 2's plus (2 sides plus should be longer than one side)\nthen: 3x+4+8x+2>6x-1\n      8x+2+6x-1>3x-4\n      6x-1+3x+4>8x+2\n--> x>-7/5\n    x>3/11\n    x>-1\nThus, the answer should be x>3/11\n:)
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.
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!
833343
Let the triangle be ABC\n\nAB = BC\n\nso angle C = angle A (opposite angle of equal sides)\n\nAB = CA\nso angle C = angle B\n\nso all angles are same\nso equiangular
1099275
Oh c'mon! Smiddy hasn't had geometry for 20 years! but this is basic. Pythagorean theorem states that on a triangle A-squared plus B-squared = C-squared where C is the hypotenuse. The hypotenuse of a triangle is the diagonal bit if it's two halves of a square and we know by definition that on a square A and B are equal. Therefore 2X squared = 8-squared or 2x^2 = 64\n\nx^2 = 32\nx = sq.rt 32\n\nthe area of the original square is x times x so you're back at 32 square centimeters
1114001
c = 108 degrees\nyou can find this since A + B + c = 180\nA=B+C+72\nThensubstitute to get A + A + A + 72 = 180\n3A + 72 = 180\nA = 36 so then C= 108
1030265
a=c*sin(30)=.5c since it is apposite the 30 degree angle\nb=sqrt(c^2-a^2)= .866c\nor b=c*cos(30)=.866c\n\na=b=c*sin(45)=.707c=
315798
I think what you're trying to explain is that all points A,B,C are on the cirumference of a circle, with the line AC being a diameter of the circle.\n\nThis forms a right angled triangle with the right angle at point B. That's just a rule of circle geometry when 1 side of a triangle in a circle is a diameter. By angle sum of a triangle, the angle at C must be 62 degrees.\n\nDraw in the line 0B, forming an iscoceles triangle OBC, since OB and OC are equal radii. In this triangle, angles at B and C are both 62 degrees, so angle at centre is 56 degrees. This is the angle that subtends the arc you want to find.\n\nIf you want the arc length you need the radius and use the formula\nl = 2 * pi * r * 56 / 360
628423
The law of cosines lets you solve for sides and angles for any triangle.  It goes like this:\n\na^2 = b^2 + c^2 - 2abcos(A)\n\nNow in our triangle, the capital letters are angles and lowercase ones are sides.  Angle A is between b and c, angle B is between A and C, and angle C is between a and b (remember that the angle and the side opposite it share the same letter).\n\nSo in your case, we have\na^2 = b^2 + c^2 - 2bccos(A)\na^2 = 2^2 + 3^2 - 2(2)(3)cos(35)\na^2 = 4 + 9 - 12cos(35)\na^2 = 3.17\na ~ 1.78\n\nNow that you have all the sides, you can tweak the law of cosines to find the other angles:\nb^2 = a^2 + c^2 - 2accos(B)\n2^2 = 1.78^2 + 3^2 - 2(1.78)(3)cos(B)\n4 = 3.17 + 9 - 10.68cos(B)\n10.68cos(B) = 8.17\ncos(B) = 0.7650\nB ~ 40 degrees\n\nc^2 = a^2 + b^2 - 2accos(C)\n3^2 = 1.78^2 + 2^2 - 2(1.78)(2)cos(C)\n9 = 3.17 + 4 - 7.12cos(C)\n7.12cos(C) = -1.83\ncos(C) = -0.2570\nC ~ 105 degrees\n\nYou could also have used the property of angles f a triangle adding to 180 to find C, FYI.
748474
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.