How many different triangles are determined by 8 points in a plane, no three of which are collinear.?
110980
8C3.  8!/(3!5!)  Since every three points forms an unique triangle.
841435
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.\nSeptagon - 7 angles add to 900°, so each angle is 128.57°\nOctagon - 8 angles add to 1080°, so each angle is 135°\nNonagon - 9 angles add to 1260°, so each angle is 140°\n\nThe answer for a nonagon is 140°.\n\nThe general formula for the interior angles an n-sided figure is:\n180 * (n-2) / n\n\nSo for n = 9:\n180 * 7 / 9\n= 20 * 7\n= 140°\n\nAn alternate way to figure this is to calculate the *exterior* angles.  The formula for the exterior angle is 360° / n.  And since the interior and exterior angles are supplementary you have:\n180° - 360°/n\n180° - 360°/9 = 180° - 40° = 140°\n\nEither way the answer for the *interior* angle of a regular nonagon is 140°
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!
502159
?\n\n(x-coordinate, y-coordinate)\n(0,0)\n(1,3)\n(2,6)\n(3,9)\n\nCos y-coordinate is 3 times of x-coordinate?
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
703879
This sounds like plane geometry where you have an n sided polygon and want to know how many diagonals you can draw.  From each vertex, you can draw a diagonal to every other vertex except to yourself and the two next to you (those two line make sides of the polygon not diagonals).  So for n sides (which means n vertices), you can draw n-3 diagonals at each vertex.  This gives (n-3)*n.  But this just counts the starting end of each line. Whenever you draw one of these lines, it goes to another vertex.  If we drew (n-3)*n lines, we would traces over each diagonal twice, once from each end so there are only half as many diagonals as that.\n\nThe answer then is: (n-3)*n/2
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
563121
a right angled isosceles triangle\n\nplease give me 10 points now
905903
There are three medians in a triangle, not just one.\n\na) is your answer.
908743
A regular 2D polygon with 13 vertices is called a triskaidecagon.  There are no regular 3D polyhedron with 13 vertices.  There exists a polyhedron with 13 vertices and 11 faces, an augmented hexaongal prism, but mathematically studied polyhedra with 13 vertices are very rare.
302936
Without knowing the measurements of the sides, you can only figure this out if all sides of the polygon are equal in length. And this is how...\n\nSee, first you draw any polygon, except a triangle. So before doing it, you should know that all the angles of a triangle must add up to 180 degrees. (You can do it with a triangle too, but you would be using what you’re hoping to discover.) From the center of the polygon, draw a line to each vertex. This will create a number of triangles, inside the polygon, equal to its number of sides. Knowing that the angles inside each triangle must add up to 180 degrees, you simply multiply 180 by the number of triangles, which is also the number of sides. Then, realizing that the angles in the center of the polygon, which do not contribute to the angles of that polygon, form a complete circle, and therefore must add up to 360 degrees, you subtract 360 from the result. So the angles in any polygon must add up to 180x-360, or 180(x-2), where x is the number of sides. To find the measure of each angle, in a polygon where all sides and angles are equal, you divide that by the number of sides. The equation then works out to (180(x-2))/x.
699841
Three are three angles in a triangle which all add up to exactly 180 degrees.
651972
A nonagon has 9 sides and 9 vertices.
1022630
None of those are "planes".\n\nA plane is drawn as a square simply because of the limitations of a sheet of paper.\n\nA plane is infinite in two directions.  It has no start or end point on the xy axis.  A sheet of paper of course does have start end end points.  Usually 8x11\n\nNone of rectangels, squares, or triangles are planes.  They are all subsets of them.  And darned small ones at that.\n\n-Dio
598966
1- line segment\n2- angle\n3 -triangle\n4 - square, rectangle, rhombus, diamond, parallelogram, or quadrilateral\n5-pentagon\n6-hexagon\n7-heptagon
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
736929
I think I'd start with 9 isosceles triangles.\n\nEach angle at the apex is 360 divided by 9 degrees\n\n=40degrees.\n\nThe other two angles are equal, so subtract 40 degs from 180 degs and halve what's left.\n\nEach base angle of the triangle = 70 degs.\n\nYou already have 40degs and the side opposite it (0.80m)\n\nSo you can use a over sinA = b over sinB to find the other two (equal) sides of the isosceles triangle.\n\nNow drop a perpendicular from the apex of the isosceles triangle to the (middle of) its base, forming two right angled triangles.\n\nEach of these triangles is half of the isosceles triangle.\n(remember the base of the right angled triangle is half the 0.80m)\n\nNow you can use \n\nsin = opp/hyp,  cos = adj/hyp,  tan = opp/adj to find the length of the perpendicular height.\n\nNearly there!\n\nArea of a triangle = half the base times perpendicular height \n\nWhen you've worked this out for one of the isosceles triangles, you just have to remember back in line one, that you are finding the area of 9 identical triangles, so you multiply the area of one by 9 to get the area of the nonagon.\n\nThis uses lots of fairly sinple steps, which you should be able to remember in an exam.  PS don't try to remember the order of the steps we used here, just have confidence that when you start, the solution will unfold for you.  If you wanted to practice, try it in a week with a different polygon!