Here i go again...(u might want to draw this out) so  you have a circle. A is on the outside and C is on the outside, both are between the top and bottom of the circle. Then B is on the top, on the side where C is at. There a diagonal line from A and B in the inside of the circle. Then from A to C there is a line. in the middle of that line( A to C) it's point O- the middle of the circle. Angle A is 28 degrees. What is the measure of the arc BC? \n\nIf you don't understand something please tell me.\nThanks A lot! in advanced.
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
119907
Your question is a little bit generic. You can draw a lot of "angles and triangles" with a compass.\nThe simplest procedure that comes to my mind is this:\n1) Trace an horizontal segment from a point A to another point B.\n2) Now point your compass in A, with radius equal to AB and trace an arc over the segment.\n3) Then point in B with the same radius and trace another arc. The two arcs will intersect in a third point C.\n4) If now you join A with C and C with B, you have a triangle with all segments and angles equal.\nBe more specific if you need more help.
149708
It's the same as the pi you know and love (3.14......).  This is to do with units of measuring angles--specifically, radians and degrees.  Just as we can use different units for measuring length, such as inches and centimetres, there are a number of units for angles.  \n\nPi radians and 180 degrees are the same angle, equivalent to half of a complete rotation.  A radian is about 57 degrees.\n\nHow is a radian defined?  It is the angle such that, if you have a sector of a circle with angle 1 radian at the centre of the circle, then the length of the arc is equal to the radius.
244909
I know how to do it, but it's going to cost me a lot of work, and sorry I don't have the time!\n\nI can give you some tips to start off though:\n\nDraw an x,y diagram where the center of the circle is the origin. Draw AB as a horizontal baseline which is 21 below the origin.\n\nThen you will have to create functions for the circle and then for the lines that have the tangents with the circle that together form the triangle. Particularly the formulas for the triangle lines will be hard.\n\nYou will have to calculate at which other points than P the triangle lines touch the circle. For that you will need the derivative of the function for the circle.\n\nEnglish is not my native language, so I hoped I used the correct English terminology. Good luck!\n\nP.S.:  aytu_y, what you say about equilateral triangles is correct, but it doesn't work in this case because here point P is not in the middle! (PA = 23 while PB = 27)
439095
The crux of this problem is for you to realize that the given angles are "complementary pairs", which add up to 90 degrees.\n\nYou are expected to know that the "co-function" of angle A gives the same value as the "function" of angle A's complement.\n\nSpecifically:\n\n cos(A) = sin(90-A)\n\nA graphical way of looking at this statement is that "cosine and sine are the same function, but offset by 90 degrees!"  We call that "out of phase by pi/2 or 90 degrees."\n\nThe same works for tangent and its co-function:\n\n tan(A) = cot(90-A)\n\nSo the answer to the second problem is 0.\n\nIf one of the fuctions in the first problem had been sin, the quotient would have been 1.  However, that's not the case, and I leave the upper problem as an exercise for another Answerer.\n\n Hope this helps!
487769
There is a theorem (I think you remember it better). I just put it with my own words. Let points A, B, C & D belong to the same circle. Then the angle ABC=(Arc ADC)/2\nThus the Sum of 7 Arcs under 7 Angles will be = to 2pi radians, that means that the Sum of those Angles = pi radians, the main demand being they belong to the circle.
529836
OK, a triangle has a total of 180 degrees so angle C is 180 - (50 +70) = 180 - 120 = 60\n\nUsing the formula sin A/a = sin B/b = sin C / c\n\nsin 50/14 = sin 70 / b\nb sin 50 = 14 sin 70\nb = 14 sin 70 / sin 50 \nb = about 17.173542356679484521363125314633\n\nsin 50 / 14 = sin 60 / c\nc sin 50 = 14 sin 60\nc = 14 sin 60 / sin 50 \nc = about 15.827222247859905497585022514782
440390
The angles formed by crossing lines are supplementary (add to 180º).\n\nSo it is a simple matter of subtracting:\n180º - 145º = 35º\n\nThe acute angles will each be 35º.
560385
The first point is not at exactly 210 degrees.  It is a little over 212 degrees.\n\nThe second point (3,3) is obviously at 45 degrees, not 30 degrees.\n\nThe distance between (-8, -5) and (3, 3) is\nsqrt((-8 - 3)^2 + (-5 - 3)^2) = sqrt((-11)^2 + (-8)^2) = sqrt(121 + 64)\n = sqrt(185)\n\nThe original angle from (3,3) to (-8, -5) is\natan2(-8 - 3, -5 - 3) = atan2(-11, -8) = 180 degrees + arctan(8/11)\ncounterclockwise.\n\nThe clockwise rotation subtracts thirty degrees from that, so in the final coordinate system the original (-8, -5) is sqrt(185) magnitude in the 150 degrees + arctan(8/11) direction from the new origin at the original (3,3).\n\nLuckily, sin( arctan(8/11) ) = 8 / sqrt(8^2 + 11^2) = 8 / sqrt(185)\nand cos( arctan(8/11) ) = 11 / sqrt(185).\n\nWe have the "new" x is\n\nsqrt(185) cos(150deg. + arctan(8/11)) =\n = sqrt(185) ( cos( 150 deg) cos ( arctan(8/11) ) - sin( 150 deg ) sin( arctan(8/11) ) )\n = 11 cos(150 deg) - 8 sin(150 deg)\n = 11 (-sqrt(3)/2) - 8 (1/2)\n = -(1/2)(8 + 11 sqrt(3)) or approx. - 13.526\n\nand the "new" y is:\n\nsqrt(185) sin(150deg. + arctan(8/11)) =\n = sqrt(185) ( sin( 150 deg ) cos ( arctan(8/11) ) + cos( 150 deg ) sin( arctan(8/11) ) )\n = 11 sin( 150 deg ) + 8 cos( 150 deg )\n = 11 (1/2) + 8 (- sqrt(3) / 2)\n = (1/2)(11 - 8 sqrt(3)) or approx. -1.428\n\nwhich makes sense, as the original slightly over 212 degrees after the rotation is slightly over 212 - 30 = 182 degrees, so both quantities are negative with the magnitude of x large and the magnitude of y small.\n\nCheck: "new x"^2 + "new y"^2 =\n = ( -(1/2)(8 + 11 sqrt(3)) )^2 + ( (1/2)(11 - 8sqrt(3)) )^2\n = (1/4)( 8^2 + 2*8*11 sqrt(3) + 3*11^2 + 11^2 - 2*11*8 sqrt(3) + 3*8^2)\n = (1/4)( 4 (8^2 + 11^2) )\n = 185 as it should be.\n\nCheck: atan2("new x", "new y") = 180 degrees + arctan("new y" / "new x")\n\n"new y" / "new x" =\n = (11 - 8sqrt(3)) / (- 8 - 11sqrt(3))\n = (11 - 8sqrt(3)) (- 8 + 11sqrt(3)) / ( (-8)^2 - (11sqrt(3))^2 )\n = (-88 + 121sqrt(3) + 64sqrt(3) - 88*3) / (64 - 121*3)\n = (- 352 + 185sqrt(3)) / -299 or approx. 0.105587\nwith arctan 6.02737\nand expected value arctan(8/11) - 30 or approx. 6.02737\nso this checks out, too.\n\nChanged coordinates:\n\n( -(1/2)(8 + 11 sqrt(3)), (1/2)(11 - 8 sqrt(3)) ) or approx ( - 13.526, -1.428 )
393329
180°\n\nYesterday it was asked about a 5-pointed star.\n\nWell let us look at an n-pointed star\n\n1. Draw the circumscribing circle\n\n2. Join the vertices and create a regular n sides polygon inside the circumscribing circle\n\n3. Join each vertex to the centre\n\n4. Each point of the star is half the ange at the centree subtended on the same side.\n\n5. Each angle in the centre is equal as they are all subtended by equal chords as the sides of a regular polygon are equal.\n\n6. The sum of the angles at the centre of the circle = 360°\n\n7 So the sum of the angles at the circumference \n= sum of the angles of the star \n= ½ of 360°\n= 180°\n\nBONUS\n8 So each angle = 180°/n\n\nIf n = 5 each angle = 36°\nIf n = 7 each angle = (25 5/7)°
619857
A = 45 degrees or (pi / 4) radians\n\n1/sin^2A-1/cos^2A-1/tan^2A-1/cot^2A-1/sec^2A-1/csc^2A = -3\nIf i understand the way you wrote the problem above, it's 1/ each term, subtract each other right?\n1/ ((sinA)^2) - 1/ ((cosA)^2) - 1/ ((tanA)^2) - 1/ ((cotA)^2) - 1/ ((secA)^2) - 1/ ((cscA)^2) = -3\nI'm going to drop the A part, i assume you know that it will still be officially added in the final version\n\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cot^2)] + [-(tan^2)] + [-(cos^2)] +  [-(sin^2)] = -3\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cos^2) / (sin^2)] + [-(sin^2) / (cos^2)] + [-(cos^2) - (sin^2)] = -3\ncombine denominators\n[( 1 - (cos^2) ) / (sin^2)] + [( (-1) - (sin^2) ) / (cos^2)] + (-1) [(cos^2) + (sin^2)] = -3\n[(sin^2) / (sin^2)] + (-1)[( 1 + (sin^2) ) / (cos^2)] + (-1) [ 1 ] = -3\nTrig identity [(cos^2) + (sin^2)] = 1\n[ 1 ] + (-1)[( 1 + (sin^2) ) / (cos^2)] + [ -1 ] = -3\n(-1)[( 1 + (sin^2) ) / (cos^2)] = -3\n[( 1 + (sin^2) ) / (cos^2)] = +3\n1 + (sin^2) = 3 * (cos^2)\nTrig identity 1 = [(sin^2) + (cos^2)] =\n[(sin^2) + (cos^2)] + (sin^2) = 3 * (cos^2)\n2 * (sin^2) = 2 * (cos^2)\nsin^2 = cos^2\nsinA = cosA\nA= 45 degrees of (pi/4) radians
513971
Use the fact that the sum of the angles of a triangle add up to 180 degrees. Let A, B, C be the angles of the triangle. The m<A + m<B + m<C =180. If you know m<A and m<B, you can find m<C.\n\nWhen I write m<A, I mean the measure of angle A, not m is less than A.
375721
1) All you need is the circumference, which is pi * diameter, to get the linear distance traveled per revolution.  Then just divide by 100,000 to get it in terms of km, and multiply it by 60 to get it in terms of "per hour".\n\n2) Are you sure you copied this one correctly? The way I figure it, AD = ½ of AB, BE is ½ of BC and CF is ½ of CA.  This means that the left side can be written "4(½AB + ½BC + ½CA).  You then factor out the ½, and you're left with 2(AB + BC + CA), which will always be less than 3(AB + BC + CA)...\n\n3) Start by graphing a right triangle on an X-Y graph with one of the non-right angles at the origin (0,0). Then figure the circle that is formed by the three points of the triangle. When you are finished, the hypotenuse should be the same as a diameter of the circle.\n\nI hope this helps.
641664
First off......do you know how to do basic trigonometry and use Snell's Law?  If you do, then I'll walk you through it.  Write me at fortitudinousskeptic@yahoo.com\nIf you don't know trig and Snell's law I'm not going to try to explain it.  I'm not that motivated.
492305
1 Degree=2*pi/360 Radian.\n\nsin 30=sin(pi/6)Radian.\n\n(1 Degree=0.01745329 Radian.\n30degree=30*2*pi/360 Radian.\n\n30degree=30*pi/180 Radian.\n\nsin 30=sin(30*pi/180)Radian\nsin 30=sin(pi/6) Radian..)\n\n(360 degrees=2*pi*Radian\n\n1 radian=360/2*pi\n\n1 Radian=57.29577951 degrees.)\n\n(One radian is the angle subtended at the center of a circle by an arc of circumference that is equal in length to the radius of the circle.\n\nIn terms of a circle it can be seen as the ratio of the length of the arc subtended by two radii to the radius of the circle. )
361714
ok: i hope you sketched this out: forget for a minute the solid's diagonal is 26cm. let's use the ratios as actual units. so the one side with a 3*4 rectangle has a 5unit diagonal (from Pythagoras).\nnow-picture the right triangle with the two adjacent sides to the right angle, one side is 5 and one is 12 (that's the solid's diagonal). using OUR units for now, THAT diagonal is 13units (Pythagorean again). luckily, 26 is 2 * 13, so hopefully the measurements are 6,8,24, would such a solid yield a diagonal of 26?
533601
By drawing a sketch diagram, you can work out the system in question can be represented as a right angle triangle, with the hypotenuse equal to 1 (kg) and the opposite side = 0.5 (kg).  the angle you are trying to calculate is between the hypotenuse and adjacent, and is given by the formula \n\nsin a = opp / hyp\n\ntherefore sin a = 0.5 / 1\nsin a = 0.5\n\na = arcsin(0.5)\n\na  = 30 degrees.\n\nAs both weigths are acting in opposition to each other, the angle will be positive, so your answer is option 2, 30 degrees.\n\nHope this helps!