It is a circle with a triangle inside. The triangle is ABO, OB=6, O=60*(degrees), and then X is in between A&B..KINDA CONFUSED!SHEEESH.\nArea of the circle?\nArea of sector O-AXB?\nArea of segment AXB?\nCircumference of circle?\n\n\nif you can answer this question i swear you are a friggin genius!!!
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I'm assuming that A and B are on the circle and O is the center of the circle.  Since OB=6, AB=6 as well, since OB and AB are radii.\n\n1.  Area of a circle = pi*r^2\n= pi*(6)^2\n= 36pi\n\n2.  Area of a sector (pie wedge) is easiest if you consider that the pie wedge is a certain fraction of the whole circle.  In this case, your sector is 60 degrees, and the whole circle is 360 degrees.  So the sector is 1/6 of the total area, which we found in part 1.  So the area = 6pi.\n\n3.  The segment is trickier.  If you look at the triangle, the segment is the curved part - almost like the top of the ice cream cone, if you can follow my description.\n\nThe area of a segment is given by\n\nA = r^2[theta - sin(theta)]/2\n\nwhere theta is the angle in radians.  The angle we have is 60 degrees.  To convert to radians, multiply by pi/180.  So theta is pi/3.\n\nNow since we know that the triangle has a 60 degree angle, and that OA and OB are both 6, through some reasoning we can see that this is an equilateral triangle, and AB (the chord) is also 6.\n\nSo the area of the segment then is\n\nA = r^2[theta - sin(theta)]/2\n= (6)^2[pi/3 - sin(pi/3)]/2\n= 3.26\n\n4.  Circumference = 2*pi*r\n= 2*pi*6\n= 12pi\n\nI don't think that X served any purpose except to indicate the region you're supposed to be considering for this problem.
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cool! I'll think about it\n\n---\n\nhere I am.\nUse the two perpenducular segments as a coordinate system with O on their intersection.\n\nSay: the 4-3 segment defines the x axis, and the 2-6 segment defines the y axis.\nYou know four points belonging to the circle: (-4;0),(0;-2),(3;0) and (0,6)\nThe center of the circle has the mean value between -4 and 3 as its x and the mean value between -2 and 6 as its y (property of chords: the center is on the chord's axis)\n\nso the center's coordinates are C(-1/2;2)\n\nNow you can easily calculate the distance between C and anyone of the four points you know to get the radius.\n\nThen finish finding the diameter\n\n----
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Let the two foci be at (a, b) and (c, d) which we assume are distinct -- if they are identical then you have a circle and a much easier problem. Afterwards we can check the result to see if relaxing that restriction would introduce a division by zero.\n\nLet the sum of the distances from the two foci (in the case of a circle, the diameter) be s.\n\nLet theta be the angle from (a, b) following the usual convention that an angle of zero goes from (a, b) to (a + infinity, b) and theta increases counterclockwise.\n\nThe ray going out from (a, b) in the angle theta direction can be parameterized as the curve\n\n(x(t), y(t)) = (a + t cos theta, b + t sin theta) for t >= 0\n\nWe want to find t such that (x(t), y(t)) satisfies that the sum of its distances from the two foci is s:\n\nThe distance from (a, b) is\n\nsqrt((x(t) - a)^2 + (y(t) - b)^2)\n = sqrt((t cos theta)^2 + (t sin theta)^2)\n = sqrt( t^2 (cos^2 theta + sin^2 theta) )\n = sqrt( t^2 )\n = t\n\nImagine that. :-)  (We don't need to say the absolute value of t because we know t >= 0).\n\nSo we want to find t such that\n\nt + sqrt((a + t cos theta - c)^2 + (b + t sin theta - d)^2) = s\n\nsqrt((a + t cos theta - c)^2 + (b + t sin theta - d)^2) = s - t\n\n(a + t cos theta - c)^2 + (b + t sin theta - d)^2 = (s - t)^2\n\nt^2 cos^2 theta + 2 (a - c) t cos theta + (a - c)^2 + t^2 sin^2 theta + 2 (b - d) t sin theta + (b - d)^2 = s^2 - 2 s t + t^2\n\nWell look at that, on the left hand side the quadratic term in t is t^2 (cos^2 theta + sin^2 theta) = t^2 whic is also the quadtratic term in t on the right.  So we can drop those and we get\n\nt (2 (a - c) cos theta + 2 (b - d) sin theta + 2 s) = s^2 - (a - c)^2 - (b - d)^2\n\nIf s is at least the distance between the two foci then the right hand side is >= 0 because it is s^2 minus the square of the distance between the two foci.\n\nSo let T = (s^2 - (a - c)^2 - (b - d)^2) / (2 (a - c) cos theta + 2 (b - d) sin theta + 2 s)\n\n(assuming the denonimator is nonzero) and the point you want is\n\n(x(T), y(T)) = (a + T cos theta, b + T sin theta)
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I knew it as I did the activity.Draw a circle an divide it into 8 equal parts.It should be equal.Then cut it and stick such that the tip of one cone is up and the other is down.See that the edges stick each other close together.Do it until you finish all the 8 parts.\nNow you needed some concentration.You can see that it forms a parallelogram.Now area of parallelogram=base*height and the circumference of circle=Pi*2*radius.The base of the diagram is half the circumference that is Pi*radius,and the height will be the radius.You will also notice that Area of parallelogram=Area of circle.Therefore,Base*height=Pi *Radius *radius.so, now you have got pi*radius square.\nHope you like my answer.
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.............6\n........______\n....../.|...........|&#92;\n...../..|...........|..&#92; 2\n..../_|_____|__&#92;\n\n\nThe triangle on the right is a 30-60-90, so the shorter side must be 2/1 = 1.  So the length of the bottom part is 6+1+1 or 8.  The height must be 1√3, so the area is 1/2(6+8)√3 = 7√3.\n\nGopal screwed up when he did 2*2cos60.  That's 2*2*1/2 = 2, not 4.
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So AB=S1=OA=OB=OC=R, S2^2=h^2+(S1/2)^2, while h is height of triangle ACB, \nR-h=S1-h is height of AOB.\nThus (AO)^2=(AB/2)^2+(R-h)^2 or R^2-(R/2)^2=(R-h)^2 or R-h=R*sqrt(3/4)\n and h=R*(1-sqrt(3)/2), \nhence S2^2 = R^2*(1-sqrt(3)/2)^2+(R/2)^2, \nS2=R*sqrt((1-sqrt(3)/2)^2+1/4) \n= R*sqrt(1-sqrt(3)+3/4+1/4)\n= R*sqrt(2-sqrt(3)), \nthus next pi=12*S2/(2R) = 6*sqrt(2-sqrt(3)) = 3.105829
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Wallstream is correct -- start with a right triangle.\n\nLabel the interior angles a, b and c, with c being the 90-degree angle.\n\nNow notice that a and b are complementary angles that add up to 90-degrees.\n\nLet x = sina.  Since the angles are complementary, we also have that x = cosb.\n\nSo we can rewrite as follows:\n\nIf x = sina, then a = arcsinx\nIf x = cosb, then b = arccosx\n\nBut a+b = 90-degrees (aka pi/2).\n\nSo arcsinx + arccosx = pi/2\n\nHope that helped!
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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)
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Hi. Don't you hate these? OK you have an unknown angle. Call it a. You have a complement, c, which is 90-a. c=3a-10. Any help?
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WTF IS THE TRIANGLE AND THE U FOR AND HOW DO U MAKE THE TRIANGLE... THERE ISNT A KEY OR ANYTHING...
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We have x=rcos(θ), and y=rsin(θ).  \n\n\nTherefore x=4sin(θ)cos(θ) and y=4sin(θ)sin(θ).  \nAnd x^2+y^2= 16sin(θ)^2(cos(θ)^2+sin(θ)^2) =16sin(θ)=4y.  \nThus x^2+y^2-4y+4-4 =x^2+(y-2)^2-4=0 and x^2+(y-2)^2=4.\n\nAre you sure the second one is given by:\nρsin(θ)=2cos(θ) and not ρsin(φ)=cos(θ)?\n\nif it is ρsin(φ)=2cos(θ), then first notice that φ≠kπ for any integer k.\nThen we have ρ=2cos(θ)/sin(φ)\nNext we have x=ρcos(θ)sin(φ)=2cos(θ)^2, and y=ρsin(θ)sin(φ)=2cos(θ)sin(θ).\n\nThus x^2+y^2= 4cos(θ)^2(cos(θ)^2+sin(θ)^2) =2x,\nand x^2-2x+1-1+y^2= (x-1)^2-1+y^2=0,\nso (x-1)^2+y^2=1
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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
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In polar coordinates area is\n\nA = ½∫r² dθ\n\nThe area of the sector bounded by the curve r = f(θ) and the two radial lines defined by θ1 and θ2 is\n\n½∫[θ = θ1 to θ2] f(θ)² dθ\n\nIn your problem f(θ) = 2 cos 3θ. The curve starts when θ = 0 at (2,0) and continues. It first meets (0,0) when 2 cos 3θ = 0, or θ = π/6 making a region of half a loop.\n\nSo, area of one loop =\n\n2 × ½∫[θ = 0 to π/6] ( 2 cos 3θ )² dθ\n\n= 4∫[θ = 0 to π/6] cos² 3θ dθ\n\n= 4∫[θ = 0 to π/6] ½ ( 1 + cos 6θ ) dθ\n\n= 2 { ∫[θ = 0 to π/6]  dθ + ∫[θ = 0 to π/6] cos 6θ dθ }\n\n= 2 { [θ = 0 to π/6: θ]  + 0 } because the area over the range 0 to π/6 of cos 6θ cancels to 0\n\n= π / 3
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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
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90° < beta < 180° means 2nd quadrant.\nsin beta = 12/13 \nbeta=arcsine(12/13)=67.38° in the 1st quadrant\nor 180°-67.38°=112.62° in the 2nd.\n\ntan beta = 3/4 , sin beta < 0\nsin beta<0 means 3rd or 4th quadrant.\ntan beta>0 means 1st or 3rs quadrant so we are in the 3rd quadrant.\nbeta=arctan 3/4=36.87° in the 1st quadrant. or\n36.87°+180°=216.87° in the 3rd.
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I haven't done one of these in quite a while. \n\nWhat would you differential element be?  How about r*(dtheta)*dr?\n\nthat sort of a ds*dr   where ds depends on the radius.  Try it! \n\nNote that without the sin theta you'd get the int rdthetadr with r 0 to R and theta 0 to 2pi  You'd get the correct area of a circle.\n\nTry the intergral, see if it makes sense.  Then sketch the figure and estimate it's area to compare.  My confidence is increasing as I write this.
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Your question is not well posed, but I'm going to restate it in the way that I'm guessing you're thinking.  \n\nSuppose you have a 20 inch diameter circle.  Take two points on the circle and draw a straight line segment between them that is precisely 8 inches in length (this is termed a "secant", or a "chord").  What is now the arc-length of the segment of circle between those two points?\n\nIf that is your question, then here is how I work the problem:\n\nYou have defined an equilateral triangle, two edges of which are the radii of the circle, 10 inches each, and one which is the chord of 8 inches.  We can now determine the ANGLE between the two 10 inch edges using basic trigonometry.  The easiest way for me to think about this, offhand is to divide the equilateral triangle in half, making two right triangles.  The hypotenuse is 10 inches, the opposite leg is 4 inches, and therefore the sin function of the angle is equal to 4/10.  the angle is therefore (approximately) 23.6 degrees.  Since we "trigged" a right triangle that was half of the original equilateral triangle, we need to double the result, yielding 47.2 degrees.\n\nThe angle between the two radii that subtend the arc who's chord is 8 inches (on a 20 inch circle) is 47.2 degrees.\n\n47.2 degrees represents 47.2/360 of the circle (or 0.13 of the circle).  Therefore the arc length of the segment in question is given by 0.13 x the circumfrence.\n\nThe circumfrence is pi times the diameter or 3.14 x 20 = 62.8 inches, therefore the arc length is given by 0.13 x 62.8 =\n8.2 inches\n\nThe answer here isn't terribly surprising.  The 20 inch diameter circle is quite large compared to the short segment of an 8 inch straight line.  The "gentle curvature" of the circle compared to the line means that the arc length of the circular segment isn't going to be much longer than the straight line.  If  you took it to the extreme and compared the circular segment from a circle who's DIAMETER was about 8 inches, then the arc length would become nearly half the circle, 3.14 x 8 / 2 =12.5 inches.  On the other hand, if the diameter of the circle got really really big (like the diameter of the earth for example) then you'd not be able to tell the difference between the straight 8 inch rod and the curve of the circle at all.  The answer would approach exactly 8 inches.\n\nI hope that answered your question!