How to make a 4 pointed star with 2 triangles?
142784
an isoceles triangle pointing up\nand another pointing down right on top of it
328129
just draw a triangle has right angle which dimensions 1,sqrt(3),and hypothenus 2. if you want it, i can send to you
322891
The obvious way is by trisecting each of the triangle's 60 degree angles and projecting the lines to the circle's circumference.\n\nBut trisecting an angle is impossible with simple geometric tools.  \n\nSee this for an interesting discussion:  http://www.jimloy.com/geometry/trisect.htm\n\n\n\nn.b.  a compass is a magnetic navigational instrument.\n\nThe drawing instrument used for scribing arcs and circles is a pair of compasses.
927347
Draw lines from the top of the vertical lines in the H to the opposite lower corners. There you go.\n\nIn response to the comment below:\nBTW\nI guess the middle line of the H would have to be not directly centered between the lines.  It has to be a little highter or lower so the 2 lines do not intersect on the center line.  Then you come out with 5 individual triangles and not triangles within triangles.
1023518
Use the Law of Sines.\n\nsin A/a = sin B/b\n(sin 36)/24 = sin B/47\nsin B = (47 sin 36)/24\nsin B = 1.151079452\nAt this point you have a problem--it is impossible for the sine of an angle to have a value greater than 1.\n\nSo either you copied the problem wrong, or you are being asked to recognize that a triangle cannot be formed with these dimensions.
861879
Just as some of the other answerers have said,\nfrom 4cos^2(x) = 3, you get cos^2(x) = 3 / 4 and\nso cos(x) = ± sqrt(3) / 2.\n\nI'm imagining that this is where your trouble starts.\nOnce you have the basics, it's useful to remember\nnumbers such as sqrt(3) / 2, sqrt(2) / 2 and 1 / 2,\nwhen it comes to working out angles, as they come\nin handy lots of times.\n\nDraw a right-angled triangle with legs equal to 1 unit,\nthen by Pythagoras, the hypotenuse = sqrt(2) units.\nThis triangle is half of a unit square, so the 2 equal\nangles in the triangle must each be 45º.\nKnowing that sin = opp / hyp and cos = adj / hyp,\nwe have : sin 45º = 1 / sqrt(2) = sqrt(2) / 2, and\nthis is also true for cos 45º. Also, tan 45º = 1.\n\nNow draw an equilateral triangle with each side equal\nto 2 units. Then halve it by constructing a line from\none vertex to the mid-point of the opposite side.\nThis presents us with 2 right-angled triangles.\nUsing Pythagoras again on either of the 2 triangles\nformed, we find that the hypotenuse = 2 units, while\none leg = 1 unit and the other leg = sqrt(3) units.\nThe original equilateral triangle has all angles equal\nto 60º, so each half-triangle will have the angles 90º,\n60º and 30º.\nNow we have sin 30º = cos 60º = 1 / 2 and\nsin 60º = cos 30º = sqrt(3) / 2.\nAlso, tan 30º = sqrt(3) / 3 and tan 60º = sqrt(3).\n\nThese commonly used angles have relatively simple\nratios. Other angles have more complex exact values.\n\nBack to your question,  we have cos(x) = ± sqrt(3) / 2,\nor what is the same, ± cos(x) = sqrt(3) / 2.\nAs we have seen, the principal value of x is 30º. The\nprincipal value is the least positive angle for which the\nequation is true.\n\nTo find more angles, there are a couple of relations you\nprobably know. One is that cos(A) = cos(-A). This says\nthat cos(30º) = cos(-30º) = cos(360º - 30º) = cos(330º).\nHere, 30º is in Quadrant I and 330º is in Quadrant 4.\n\nThe other relation is : -cos(A) = cos(180º ± A).\nThus, -cos(30º) = cos(180º ± 30º) = cos(150º) or cos(210º).\nAs we've seen, it doesn't matter if cos(30º) is positive or\nnegative. In either case, the other 2 angles are 150º, which\nis in Quadrant 2, and 210º, which is in Quadrant 3.\n\nThus, all the angles x, are 30º, 150º, 210º and 330º.\n\nIn terms of π, knowing that π radians =180º, the\nangles in radians are π/6, 5π/6, 7π/6 and 11π/6.\n\nThese are the only angles between 0º and 360º,\nor, between 0 radians and 2π radians.
929308
take a knife and cut 4 times thru the circle and u end up with 8 pie wedges...then line the wedges up-----<<<<<<<<------like that and cut thru on the dashed line, the fifth cut, giving u 16 wedges...not very mathematical but it works
557771
Draw a straight line.\nDraw a good cuve (mirror), using a compass or a protractor(for later mark a point as centre and place the protractor on this point coinciding with the axis.)\nThe curve should be of equal length on each side of principle axis.\nNow mark the pole, radius and focal point (twice the radius).\nIn case u use protractor, take focal lenght as twice the distance between the marked pont and pole.\n\nRemember how the rays reflect/refract, and accordingly draw two incident and two reflected/refracted rays.\nGive the standard sign convention...I hope plane mirror and prism are easier.
454825
We can do this 2 different ways.\n\nFirst: Use the Pythagorean Theorem. Since all angles are right angles, the diagonal (measure from corner to corner) is the hypotenuse of the right triangle formed.\n\nBy the Pythagorean Theorem, 10^2 + 10^2 =x^2, where x=diagonal\n\nx^2=200\nx=sqrt(200)\nx=10 sqrt 2\n\n2nd way: use 45-45-90 Triangles\n\nWhen you draw any diagonal in a square, you get a 45-45-90 degree right triangle. This is because the 2 sides are the same, so the 2 base angles must be also. The top is 90, so the others must be 45 and 45 to equal each other.\n\nThe sides to a 45-45-90 right triangle are always in the ratio 1:1: sqrt(2) where sqrt(2) is the hypotenuse.\n\nYou know that the sides are 10 feet. Therefore the hypotenuse is 10 sqrt2 feet
176268
If you are constructing an angle that is equal to another angle, here are the steps.\n\n\n\nOn the original angle, label the vertex v1.\nPut the point of the compass on v1.\n\nSwing an arc that crosses both sides of the angle. Label the points where the arc crosses the sides of the angle a1 and b1.\n\nDraw a straight line where you want to construct your new angle.\n\nMark a point on the line to be the vertex of the angle you are constructing. Label it v2\n\nWithout changing the compass setting put the point of the compass on v2.\n\nSwing an arc that crosses the line and is wider than the angle you want to draw. Label the point where the arc crosses the line a2.\n\nPut the point of the compass at a1.\n\nAdjust the width of the compass so that the pencil is at b1.\n\nWithout changing the compass setting put the point of the compass at point a2.\n\nSwing an arc that crosses the first arc you drew. Label the intersection b2.\n\nDraw a straight line from the b2 to v2.\n\nDone.
908730
Sure, how about a triangle with a base of 3cm and a height of 4cm?\n\nA = 1/2 b * h\n6 = 1/2 b * h\n12 = b * h\n\nSo any numbers where b * h = 12 will work...\n1 cm x 12 cm\n2 cm x 6 cm\n3 cm x 4 cm\n0.5 cm x 24 cm\n1.2 cm x 10 cm\n2.4 cm x 5 cm\netc.\n\nAnd you can have a right triangle, or an acute triangle or an obtuse triangle... doesn't matter, as long as the base and height multiply to give 12...
780884
As with many problems, it helps to draw a diagram.  I made a triangle, and then put in each of the altitudes, since they are my only known values.  With these in place, we now have 6 principal right triangles, of which we know the value one leg (the altitude).  In each of these triangles, the hypoteneuse is a complete side of the original triangle, and the angle opposite the known leg is also of the original triangle.  Personally, I chose to work with cosines and tnagents, but you could just as easily have used sines (though I think it might add a step or two).  Remember that the cosine of an angle in a right triangle is merely the ratio of the side opposite the angle to the hypoteneuse.  Don't worry that we don't know the value of any of these angles; I went through the entire problem without ever needing it.\n\nSo, since each angle, in its entirety, is in two of our right triangles, they give us two ratios to compare.  That is:\n\nIf the sides of the original triangle are x,y,z\ncos a1 = 20/x = 15/y\ncos a2 = 20/z = 12/y\ncos a3 = 12/x = 15/z\n\nWith a little cross multiplication and rearrangement of terms, we find that:\n12z = 15x = 20y.  \n\n(Your x, y, z may be in a different order depending on how you labeled your triangle, but if you manipulate your triangle a bit, you should be able to get it to match.  This will keep your coefficients identical to mine.  But even if you don't match it up, in the end you will get the same answer.  That's the beauty of math.)\n\nThe tangents are a little trickier, because they involve a fraction of the original side.  This involves the addition of a new variable on each, such that one portion of the line is "a", and the remainder is x - a; and then b and y - b; c and c - z.  The tangents then follow:\n\ntan a1 = 20/(y - b) = 15/a\ntan a2 = 12/(z - c) = 20/b\ntan a3 = 15/(x - a) = 12/c\n\nRearranging each gives us the three equations:\n20a = 15(y - b)\n12b = 20(z - c)\n15c = 12(x - a)\n\nFrom there, you can solve for either a, b, or c.  It's your choice.  I opted to solve for a.  Substituting the third equation into the second, and then that new equation into the first, we get:\n\n20a = 15(y - (20/12)(z - (12/15)(x - a)))\nwhich reduces to\n8a = 4x + 3y - 5z\n\nNow, side a is in a right triangle with hypoteneuse y and second leg 15, meaning that:\na^2 + 15^2 = y^2   or\na = sqrt(y^2 - 225)\n\nUsing this equation, plus the ratios we received from the cosine functions, we solve for y:\n\n8 * sqrt(y^2 - 225) = (16/3)y + 3y - (25/3)\n\nConveniently, the right half is equal to zero, so dividing by 8 and squaring:\n\ny^2 - 225 = 0\ny^2 = 225\ny = 15\n\nEnter this into our ratio:\n\n15x = 20y\n15x = 20*15\nx = 20\n\nand\n\n12z = 20y\n12z = 20*15\nz = 25\n\nSo finally we have the lengths of all three sides, and adding them together, the perimeter is\n15 + 20 + 25 = 60\n\nThis turns out to be a right triangle, but that is merely happenstance.  This process can be used to solve for any triangle with three known altitudes.\n\nHope this helps.
1136647
I think so, actually...\n\n1. Draw 2 perp. lines.\n2. Draw 2 hypotenuses to each base such that the triangles made are 45-45-90 triangles (extend the bases as well, if you need to).\n3.  You have just made an isoceles triangle since the base angles are congruent.\n4.  The line that is perp. to the base is now the perpendicular bisector because the triangle is isoceles.\n\nViola!
519523
try cutting it up into squares and triangles
729410
You can  find two different values for x that are easy to find 1/7 of, such as 7 and 14; see what y would be at those points [ -4, -3]; then plot the points in either order and connect the dots.  It is smart to also plot a third point to reduce the chance of error, because if one of the three is wrong, they won't line up.
1078788
For any line length X, it can be divided into 7 equal parts like so:\nStart with your segment --we'll call it AB-- and at point A, draw a ray going towards the top of your paper.  Then, with your compass, mark off seven equidistant points, starting at A (put your compass on A, make a mark on the ray; put your compass on that mark, make another on the ray, etc.) Then, draw a segment from your 7th point to point B.  Then just draw 6 more parallel lines (constructing congruent arcs and such) and your segment AB is divided.
740750
easy \n\nU make a triangle between the centers of the 3 circls \nmeaning it would have three lines = 2*R = 2m\nso the hight of the triangle would be root of ( R*R - 0.5R * 0.5R)\nso it would be the root of (3) which is( 1.73)\nthe hight would be 1 + 1.73+ 1 = 3.73 m