Find the sum of the measures of the five acute angles of a star...?
821150
first draw a circle, and then put the star in it w/ the points touching the edge. that divides the circumference of the circle into 5 equal parts. so to find the degree of each one of those section divide 360 by 5 = 72. now look at the angle that is made from the lines from the ends of the individuals sections to the point of the star it goes to, which is the inscribed angle. now that angle which is one of the acute angles is half of its intercepted arc. so that makes 72 divided by 2= 36. so since you wanted to know the measure of all the acute angles and not just one multiply 36 times 5 = 180
816840
Use the sine law.\n A/sin(a) = B/sin(b) = C/sin(c)\nwhere angles a, b, c are opposite sides A, B, C respectively.\n\n   In your example  5/sin(prq)=4/sin(40)\nsin(prq)=5(sin(40))/4=\n(5/4)(.6428)=.803\nsin^(-1)(.803)=53.4degrees  and the sine is positive in the\n2nd quadrant as well as the first so angle qrp may also be\n180-53.4 = 126.6\n\nSo if it's 53.4 then angle pqr is 180-(53.4+40)=86.6\n\nAgain use the sine law to solve for pr.\npr/sin(86.6)=4/sin(40)\npr=4(sin(86.6))/sin(40)\npr=4(.998)/.6428\npr=6.21\n\nAnd if it's 126.6  angle pqr is 180-(126.6+40)=13.4\npr/sin(13.4)=4/sin(40)\npr=4(.232)/.6428\npr=1.44
483665
<c=<a, <a+<d=180 (laws of parrallelograms....?!) \ntherefore 180=(5x+30)+(x)\n180=6x+30\n150=6x\nx=25\ntherefore<c=5x+30\n<c=5(25)+30\n<c==125+30\n<c=155 assuming <c is not less than a right angle (look at you picture in the text book)
209291
8/5 + 3 1/9= 8/5+ 28/9 =  (40+5x28) /45 = (40+140)/45 = 180/45= 4\nP.S. I suppose that the star is in fact nine.
62729
Use the law of cosines.\n\n7^2=3^2 + 5^2 - 2(3)(5)cos(C)\n49=9 + 25 - 30cos(C)\n49=34 - 30cos(C)\n15= -30cos(C)\ncos(C)= -15/30\ncos(C)= -1/2\nC= 120 degrees.\nThe angle opposite the side of seven is 120 degrees.\n\nTo find the next angle, use the law of sines.\n\n7/sin(120) = 3/sin(A)\n3sin(120)= 7sin(A)\n(3sqrt3)/2= 7sin(A)\n(3sqrt3)/14= sin(A)\nA= 21.7867893\nThere are 21.7867893 degrees in the angle opposite the side of 3. \n\nTo find the final angle, add the two known ones and subtract their sum from 180.\n\n180 - (120 + 21.7867893) which equals 38.2132107 degrees.\n\nThere are 38.2132107 degrees in the angle opposite the side of 5.
27065
I've been studying this stuff recently, and I'm pretty sure this is how it goes:\n\nK is now (-2,1) and D is now (-3,-2).\n\nFor future reference on determining rotations, simply turn the graph counter-clockwise one time for 90 degrees, upside down for 180 degrees, and for 270 degrees, just turn the graph clockwise once to see what the new cooridnates are.
450833
join the 2 pointz frm da arc vit da radius.measure out da value of da angle forned here.then substitute da values 4 this formula:  2*pie*radius/da angle u measured\n\nhappy solvin!
73727
Angle of deviation, d = i1 + i2 - A\nA= r1 + r2\nIn minimum deviation, µ = {sin(A+D)/2}/(sinA/2)
199013
add 2 other angle in degree and substract from 180\n ie. c=180 -(a+b)\nwhere a ,b are known angles \nand c is unknown angle
884011
The sum of the measures of the EXTERIOR angles is always 360.  So subtract 162 from 180 to get an exterior angle (18) then divide 360 by 18 to see how many sides.
236377
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)
158498
sin(5)=0.08715574.....
733210
You basically have 2 options. Use a protractor and measure them (The internal angles total 180 degrees).\n\nThe other option is using the trig functions (Sin, cos and tan), however you must have some info on the triangle eg. length of sides to apply any of these.
1159205
pair up the  opposite sines, for example sin(10) and sin(350) and add them:\nsin(10) + sin(350) = 0\nsin(20) + sin(340) = 0\n...\nuntil you only have sin(180) and sin(360) which both equal 0\n\nPS. I assumed you were using degrees (sexagesimal). Won't work in radians (RAD) or metric (decimal) angles.
934709
Plot 3 points, connect them with lines. you get a triangle\n\nList most acute (sharp) angle 1st, most obtuse (blunt) angle last, and remaining one in the middle.\n\nIt looks like C will be the most obtuse
390709
Two angles whose sum is 90 degrees.
1036046
1) 180 degrees\n2) 34678900\n3) 10^10\n4) 4.026 * 10^6\n5) 702600