Solve the triangle, given A=32degrees, C=44degrees, c=20...?
441926
180 - (32 + 44) = 180 - 76 = 104\n\na/sin(A) = b/sin(B) = c/sin(C)\n\n(a/sin(32)) = (20/sin(44))\nasin(44) = 20sin(32)\na = (20sin(32))/(sin(44))\n\nb/(sin(104)) = (20/(sin(44)))\nb = (20sin(104))/(sin(44))\n\na = about 15.25697\nb = about 27.93591\n\nIf you use the formula below\nc^2 = a^2 + b^2 - 2ab cos(C)\nYou will find out that the problem works out.
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.
764538
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.
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!
1104391
The angles in a triangle add up to 180, so the answer is b. 40 degrees.
943973
Tangent 25° 56' 30" = cot (90 - A)\n\n89 59 60\n25 56 30\n- - - - - - -\n\n64 3 30\n\nCotangent 64 degrees 3 minutes 30 seconds\n\nThe answer is. The complemenary angle of tangent 25° 56' 30" is cotangent  64° 3' 30 "\n\n- - - - - - - - - - - -s-
716806
32 + (9/5)50 = 122\n(0C is 32F)
30686
Let the length of the equal ropes be x ft\n    the length of the longer rope is (x+2)ft\n\ntherefore,\n\nx+x+x+(x+2) = 44\n4x = 44-2\n4x = 42\nx = 10.5
861681
subtract 1st equation from the second:\n  8y^2=32\nso y^2=4, and y = +/- 2\n\nplug back into 1st equation:\n4x^2+4=4\n\n4x^2=0\nx=0
1155045
You are dealing with a triangle.  It is not lying flat on the ground, but all three lines are in the same plane.  One side is 350 ft, another is 400 ft, and the angle between them is 50 degrees.  This sounds like a Law of Cosines problem to me.\n\nCalling the unknown side x,\nx^2 = 350^2 + 400^2 - (2)(350)(400)(cos 50)\nLooks like 320.2 ft to me.
555107
MY guess is that your trying to find the Hypotenuse of a triangle.  Thus A^2+B^2=C^2 or in this case H^2+W^2 = C^2, 25 + 196 = C^2, 221 = C^2, C = 14.86.  Missing side is 14.86.
915466
Let a and b the lengths of the sides of the rectangle (with a the side with the semi-circle).\nThe perimeter is 2b+a(pi/2+1)=16, and the area ab+pi/8*a^2. Substituting b in the area (from the perimeter equation), the area is 8a-a^2*(1/2+pi/8). \nSetting the derivative to 0, we have a=32/(4+pi). Then b comes from the perimeter constraint as b=8*(1-(2+pi)/(4+pi)).\nNumerically this gives a=4.48, b=2.24 and an area of 17.92
900288
A=3*sqrt3 *a^2\nhere a =4\nso A= 3*sqrt3*16\n= 48*sqrt3\nor= 16*2.59808\n= 41.56928 sq.m
124621
x-4y=-20\n=> -4y=-x-20\n=> 4y=x+20\n=> y=x/4+5 Ans.\nThanks.
883050
the area of a rectangle is:  a = b*h\n\nSo, the area with the path is 400 (given) = (30-2x)(20-2x)\n400=600-100x+4x^2\n0 = 200-100x+4x^2\n0 = 50 - 25x +x^2\nand then solve from there using quad eq.\n\naddendum:  using the quadratic equation, the answer is about 2.1922.\n\n2 ft is not correct:  the area you are trying to get is 400.  if the path were 2 feet, then the area would be b * h = (30-4)(20-4) [the path goes all the way around so 2 ft from each side].  So 26*16=416.
745313
Not enough information. If the rectangle is also a square, then the length of each side would be 32/4 = 8 cm. Letting x be the length of one of an arbitrary pair of sides and y represent the length of a side from the other pair, you do know of course that x and y must satisfy the equation 2x + 2y = 32 cm --> x + y = 16 cm. Since you have two unknowns but just one equation, there are uncountably infinite many possibilities for (x, y).
423642
Let A be the largest angle, B be the middle angle, and C be the smallest angle.  From the question:\n\n1) A = 2C\n2) A + C = 2B\n3) and since it's a triangle, A + B + C = 180 degrees.\n\nYou can add B to both sides of equation 2, and get.\n2-rev) A + B + C = 3B\nCompare to equation 3, and you discover\n3B = 180.\nSo B = 60.  Substitute B into equation 2.\n\n2) A + C = 2B = 120\nAnd you know from equation 1 that A = 2C, so substitute\n2C + C = 120\n3C = 120\nC = 40\n\nSubstitute B and C into equation 3, and\n3) A + B + C = A + 60 + 40 = 180\nA must be 80.