Number of possible triangles for A = 60 degrees a = sqrt6 and b = 2?
543158
Yes.  By the law of sines, sin A/a = sin B/b, so B must be 45 degrees.  Thus C = 75 degrees, but maybe my geometry is a bit rusty.
904256
there are 6 vertices and you form 4 triangles from the first 2, then 3, then 2, then 1 for a total of 14.\n\nNow we form other triangles that overlap and it gets really complex...\n\nI count 18 smaller triangles, for a total of 32 so far...\n\nNow middle triangles...\n\n36, for a running total of 68
950897
Draw a triangle, an equilateral triangle, that is : each angle is 60 degrees. Each particle is placed at the vertex.
672664
first you figure out the number of possible outcomes you can have. which would be 6x6 = 36. \nThen you figure out how many combinations you can make to make it greater than 7:\n5 and 3\n5 and 4\n5 and 5\n4 and 4\n4 and 5\n3 and 5\nwhich is 6.\n6 over 36 is one sixth. therefore your answer is 1/6.
808242
Yes. These are intriguing geometry problems. What is YOUR question? Be specific. Post ONE question and be specific about where you are stuck. Right now, this just looks like you want someone to DO your homework for you. But you wouldn't cheat on your homework like that, would you?\n\nHere's a hint -- supplemental angles add up to 180\n\nFor example, if one angle is 60 degrees, its supplement is 120
367069
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
1167592
6√2\n\nRemember it's a square so the sides are all the same length, and the area is s^2\n\nso s^2=36==>s^2-36=(s-6)(s+6)=0 so s is either 6 or -6.  -6 makes no sense here, so it is obviously 6.\n\nNow, since we have a square we can "draw" (imagine) a line at the diagonal, and that makes a right triangle (since the corners of a square are all 90º) with legs 6 and 6.  using the Pythagorean relation, we have d^2=s^2+s^2=6^2+6^2=72.  Thus d^2-72=(d-6√2)(d+6√2)=0.  Thus d is either 6√2 or -6√2, and again is obviously the positive one :)
609578
definitely 63 square inches. The formula to find the area of a triangle is A=(1/2) x base x height. That is 0.5 times base times height. So, if you plug in the numbers, you get\n0.5 x 14 x 9 square inches\n(0.5 x 14) x 9 square inches\n(7) x 9 square inches\n63 square inches\n\nMake sure that whenever you look for the area of the triangle, the problem tells you that the numbers are the height and base.\n\nYes, 24 is correct.
1010584
Just think....\n\nA triangle (3 sides) always totals 180 degrees\nA quadrilateral (4 sides) always totals 360 degrees\nA pentagon (5) will total 540 degrees\n\nSo, the angles, A, will always equal 180 degrees per side, for each side (N) after the first 2.\n\nIn other words,   A = 180 * (N-2)\n\nif A is 32040, then 32040 = 180 * (N-2)\n\nN - 2 = 178\n\nN = 180
942453
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
710344
C - 3 possible answers\n\nIn any given right triangles with the two other angles being a and b,\n\nsin a = cos b AND sin b = cos a,\n\n\nas such, the first and fourth combination will give one answer, the other two will either give square trigonometric values of each angle i.e. sin^2 a and cos^2 a  OR sin^2 b and cos^2 b.\n\nThe only condition that will not satisfy this is when the angle is 45 degrees, which means a = b and contradicting the given statement of a being less than b.\n\nThus the answer is C.
965204
Pythagorean Theorem\n\nc²  = √a ² = b²\n\nc²  = √(20)² + (15)²\n\nc² = √400 + 225\n\nc² = √625\n\nc = 25\n\nThe answer is c = 25\n\nThe length os the hypotenus is 25 cm\n\n- - - - - -s-
736780
In a triangle you have 3 angles that add up to 180°.  So an equilateral triangle would each have angles of 60°.\n\nIn a square you have 4 angles that add up to 360°.  So a square would each have angles of 90°.\n\nThe pattern continues for other regular polygons:\nPentagon - 5 angles add to 540° (360° + 180°) so each angle is 108°.\nHexagon - 6 angles add to 720° (540° + 180°) so each angle is 120°, etc.\n\nInterior angles:\nEquilateral triangle --> 60°\nSquare --> 90°\nPentagon --> 108°\nHexagon --> 120°\nSeptagon --> 900°/7 = 128.57°\netc.\n\nThe answer for a square is 90°.
618490
Just draw a right triangle with short sides of 10 and 24.\n\nThe hypotenuse can be calculated by the Pythagorean theorem as the square root of 24^2 + 10^2.  That works out to 576 + 100 or 676.  The square root of 676 is 26.\n\nThe direction is roughly ENE (East North-East).
1167473
Here is the answer to your question (using the Pythagorean Theorem):  a^2 + b^2 = c^2, a and b are legs of the triangle, c is the hypotenuse.  So c = 26.\n\nso you have:  a^2 + b^2 = c^2\n                       a^2 + b^2 = 26^2\n                       a^2 + b^2 = 676\n                       \nNow, you need two squares that add up to 676.  The only squares that will work are 10 and 24.  So, here is what it looks like:\n\n                     10^2 + 24^2 = 26^2\n                     100 + 576 = 676\n                          676 = 676\nSince you get the same thing on both sides, you have the right answer.  This means that 10 and 24 are the lenglths of the other two sides of the triangle.\n\nHope this answers your question.\n\nChris (Math Teacher)
795933
looks like 240 degrees or  300 degrees. cause sine is +ve in 1st and second quadrant and -ve in the third and fourth quandrant and sin 60 = sqrt3/2. so i think above two answers are some possibilities( note there are infinite angles at which this is possible these are just some possibilities)
237985
Cut a 2x2 piece off. Cut the remaining 2x3 piece in half lengthwise.  Put the two resulting 1x3 pieces one atop the other and cut the stack diagonally.  You now have 4 triangles with sides = 1, 3 and sqrt(10).  Arrange them with their right angles touching the corners of the 2x2 square piece, 3-meter sides parallel to and touching the sides of the square piece, and all "pointing" in the same direction, clockwise or counterclockwise.  The result is a square that is sqrt(10) on a side.  Area = 2x5 = sqrt(10)^2 = 10 square meters.  Three cuts were used to make five pieces.\nTo solve this I started with the requirement of the large square being sqrt(10) on a side.  Sqrt(anything) suggests the hypotenuse of a right triangle.  A little mental fiddling produced the sides of 1 and 3 meters.  Luckily a 2x2 piece remained.