start with an equal triangle and draw a circle around it\nthe points of the triangle on the circle's line.
136709
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.
351705
Draw the chord. Observe that , unless it goes thru the center of the circle, that it splits the circle into a large part and a smaller part. Now construct a perpendicular bisector on the chord. Note\nthat it's length to the edge of the circle on the larger part is greater than the radius else it would be on the smaller part. \nAlso the distance to the edge in the smaller part is less than\nthe radius. So somewhere on the perpendicular bisector the\ndistance to the edge is equal to the radius. At any\npoint on the bisector draw a pair of lines to the end points of the chord. They make a pair of congruent triangles by S.A.S. therefore the lines are of equal length. At some point the lengths must be equal to the length of the radius. The only place that can be is at the center.
19643
1.  Graph the inequalities like they were equations.  These lines are the boundaries of an area.\n2.  ">" is the area above your boundary and "<" is the area below your boundary\n3.  Using different colors or cross-hatches, shade in the areas defined by each inequality in your system.\n4.  The intersection of these areas (i.e. the part of the graph that is contained by every one of the areas) is the solution. It can be just one point, an area in a plane, or the empty set.
376353
It's straight ahead geometry.\nDraw a circle with center O.  Draw a diameter horizontally through the circle. Label the left intersect as A and the right intersect as B.  Now draw a vector from O to a point P on the circle.  What tdo you know about triangles AOP and BOP in terms of the angle the vector makes with the diameter?  Use law of sines, law of cosines, trig, etc. to show that angle APO and angle BPO add up to 90 degrees and your done.\n\n\nDoug
211121
Trigonometry is hard to do in you head, so start out by making a drawing. You have two buildings A (tall) and B (short) - Draw them far apart, not like this:\n\nA         B\n|         \n|         \n|    \n|         |\n|         |\n|         |\n|         |\n\nNow draw one line from the top of A to the top of B - this is the angle of depression of the top and it is 46 degrees. That is, if you drew a straight, horizontal line across the top of building A, the angle between that line and the line joining the top of A to the top of B is 46 degress.\n\nSimilarly, draw a line from the top of A to the *bottom* of B - this is the angle of depression of the bottom and it is 77 degrees. That is, the angle between the horizontal line across the top of A, and the line between top of A and bottom of B is 77 degrees.\n\nNow can you find out the angle between the line from A-top to B-top, and the line from A-top to B-bottom? (It's very important for you to draw this!) Let's put this value aside for a minute. \n\nLooking at your drawing, can you see that you know the value for *part* of the angle between the horizontal line across A's top and length of building A? What is that part you know? Now, see how the line across A's top and length of A make a right angle? So how many degrees in a right angle? So if you know the total of that angle, and part of that angle, how do you get the other part? **Notice that this part we're trying to get is actually the angle between the line A-top to *B*-bottom, and the line A-top to *A*-bottom. \n\nYou should also notice that now we have two triangles:\nTriangle 1: A-top to B-bottom to A-bottom. Solve this triangle for line A-bottom to B-bottom for distance between the two buildings.\nTriangle 2: A-top to B-top to B-bottom. Solve this triangle for line B-top to B-bottom for height of building. \n\nDon't forget, we have two important angles that will help solve these two triangles!
379111
The 14 foot line is called a secant.  Bisect it, and draw a radius through the point of bisection.  You now have two identical right triangles, with hypotenuses of 12 feet, and known sides of 7 feet.  Use trigonometry to calculate the angle at the center of the circle of either triangle, and then a simple proportion (involving the circle's total circumference) will give you the arc length you want.
78279
Draw all this out so you can see it - it won't be to scale, but don't let that fool you.....\n\n\ndiameter = 50cm (radius = 25cm)\nchord = 20cm from diameter (thus it's 20cm from the center)\n\nOK, you'll need to assume that the line segment forming the diameter touches the chord at point B on the circle - i.e. we'll say the diameter is line segment AB and the chord is line segment DB. \n\nC is the center of the circle, so we have radius CB = 25cm. \n\nAlso, we'll have line segment CE drawn perpendicular from the center (C) to point E on the chord, forming a right angle and thus *bisecting* chord DB.\n\nYou have the hypotenuse of triangle ECB (radius CB) as 25cm, and side EC as 20cm (distance from the center to the chord - always perpendicular).\n\nLine segment EB will be the second side of a right triangle, so your formula:\n\nEC^2 + EB^2 = CB^2\n20^2 + EB^2 = 25^2\nEB^2 = 25^2 - 20^2\nEB^2 = 625 - 400 = 225\nEB = 15cm\n\nSince EB = half the length of the chord DB, DB = 2*15cm = 30cm.
523401
I can tell you how to get the six points on the circle then you should be able to do the folding and cutting you need.\n\nDraw a circle with a compass and with the radius you just used, mark that length with your compass on the circle and you will get the 6 equal sections.
253965
You have to have a right triangle.\nThen you figure out which is side is opposite or adjacent from the angle you need.
123024
You have to draw the lines and mark the points.\nIf Cp has a 90 degree angle with the line you just draw and crosses it right in the middle, then you have graphic proof that it is the perpendicular (which means it has a 90 degree angle) bisector (which means that it splits another line in two halfs).
289829
If the circles are all tangent to each other, then you can draw lines from center to center to center.  Thus, you'd have a triangle with sides 8, 12, and 10 (each side is formed by adding two radii).  So you can find the angles using the law of cosines:\n\nc^2 = a^2 + b^2 - 2abcos(c)\n8^2 = 12^2 + 10^2 - 2(12)(10)cos(c)\n64 = 144 + 100 - 240cos(c)\n240cos(c) = 180\ncos(c) = 3/4\nc ~ 41 deg\n\nAnd then just change up numbers to find the other angles:\n\nb^2 = a^2 + c^2 - 2accos(b)\n10^2 = 12^2 + 8^2 - 2(12)(8)cos(b)\n100 = 144 + 64 - 192cos(b)\n192cos(b) = 108\ncos(b) = 9/16\nb ~ 56 deg\n\na^2 = b^2 + c^2 - 2bccos(a)\n12^2 = 10^2 + 8^2 - 2(10)(8)cos(a)\n144 = 100 + 64 - 160cos(a)\n160cos(a) = 20\ncos(a) = 1/8\na ~ 83 deg\n\nAnd this checks out since 41 + 56 + 83 = 180.
441271
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
349402
Draw axis of symmetry to the base.  Look at the rightangled triangle on the left side.  It has hypotenuse s, short side s/2.  Use Pythagoras' to show the height is (sqrt(3)/2)*s, then use\narea = (1/2)*base*height
66629
Any circle that is tangent to two adjacent sides of a polygon has to have it's center on the bisector of the angle between them. Construct the bisectors of two of the angles and put the center where the lines cross.\n\nIf you construct a line from the center that is perpendicular to any of the sides, that gives you the radius of the circle.
652711
First draw a circle w/ the centerpoint well defined (e.g. use a compass).  Then use a protractor, with the circle center at your protractor center, to create the angle you want.\n\nIf you want to divide the circle equally for 30 people, then you take 360 degree/30.  Then you need 30 angles of 12 degrees.  Create these w/ your protractor and extend to the circumference of the circle.\n\nKeep in mind, that spreadsheet programs like excel will let you form pie-charts, which are basically this same thing only done much easier.
133050
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
101226
Draw a horizontal line with a circle the size of your desired pentagon that has its center on this line. \nPut your compass' needle where the circle's circumference crosses the horizontal line, and draw a half-circle through the center of your first circle, crossing the circumference of the first circle in two places. Draw a vertical line through the points where the half-circle crosses the first circle. This line will pass through a point we call (a). \nOpen your compass so that you can, when placing the needle in the two intersections between the horizontal line and the first circle, draw a small cross above and below the horizontal line, outside the first circle, with one line of the cross from each point. If you join these crosses you will obtain a line perpendicular to the horizontal line, also passing through the center of the first circle. The point where this line crosses the circumference of the first circle on the top, we call (b). This is the first corner of the pentagon. \nPut the compass' needle in (a) and drawing a circle segment passing through (b) and down through the horizontal line, obtaining a point on this line we call (c). \nPut the needle in (b) and pass a circle segment through (c) and the first circle. These points on the first circle are the second and third corners of the pentagon. \nWithout extending the compass, put its needle in the second and third corners, and draw circle segments passing through the first circle to find the two remaining corners. Extend a line from each corner of the pentagon through the center of the circle made in step 1 of constructing a pentagon to the opposite side of that same circle. \nThe five corners of the pentagon constitute every other corner of the decagon. The remaining five corners of the decagon are those points where the lines of step 2 cross the original circle (but not a pentagon corner).