ok you might want to draw this out, we are doing inscribed angles. you have a circle and a triangle is in it. the pints of the triangle all connect an angle A , B, C (left to right) inside the triangle angle A = 2x AB on the outside of the circle equals 6y , BC(outide of the circle or arc) = 6y and AC(on the arc) equals 4x. find the values of x and y. im really confused, thanks lot for ur help.
213764
Your question is difficult to understand.  From what I make of it, here is my answer (and I could well be wrong)\n\nI assume x and y are angle measures (in degrees)\n\nBC=2(angle A)\n6y=2(2x)\n6y=4x  (eq. 1 to be referenced below)\n\nAB+BC+AC=360 degrees\n6y+6y+4x=360\nUsing (eq. 1)\n6y+6y+6y=360\ny=20\nx=6y/4=30
188932
Let A be the largest angle, B be the middle angle, and C be the smallest angle.  From the question:\n\n1) A = 2C\n2) A + C = 2B\n3) and since it's a triangle, A + B + C = 180 degrees.\n\nYou can add B to both sides of equation 2, and get.\n2-rev) A + B + C = 3B\nCompare to equation 3, and you discover\n3B = 180.\nSo B = 60.  Substitute B into equation 2.\n\n2) A + C = 2B = 120\nAnd you know from equation 1 that A = 2C, so substitute\n2C + C = 120\n3C = 120\nC = 40\n\nSubstitute B and C into equation 3, and\n3) A + B + C = A + 60 + 40 = 180\nA must be 80.
199251
All you have to use is the fact that for any triangle, the measures of the three interior angles add to 180 degrees.  So if you have Triangles ABC and DEF, then you know:\nangleABC + angleBAC + angleBCA = 180 degrees, and\nangleDEF + angleEDF + angleEFD = 180 degrees.\n\nIf two angles are congruent, then they have the same measure in degrees.  So, if angleABC is congruent to angleDEF, and angleBAC is congruent to angleEDF, then:\nangleABC = angleDEF, and\nangleBAC = angleEDF.\n\nUsing substitution:\nangleABC + angleBAC + angleBCA = angleDEF + angleEDF + angleEFD\nangleDEF + angleBAC + angleBCA = angleDEF + angleEDF + angleEFD\nangleDEF + angleEDF + angleBCA = angleDEF + angleEDF + angleEFD.\n\nthen simplify by subtracting (angleDEF + angleEDF) from both sides:\n\nangleBCA = angleEFD.\n\ntherefore, the third angles are congruent.
124576
first draw a right triangle and label the angles and sides. now angle B cut in half is 15 degrees and angle A is 90 degrees.  lets assign AM a value of x.  so the tangent function is equal to the opposite leg over ajacent leg=> tan(15degrees)=x/(2*sqrt(3)).  take the tan of 15 and then multiply that by 2*sqrt of 3 and that will be your answer for the length of AM which is approx=.9282  to find MC you must first find the length of AC with the same tangent function using 30degrees as your angle so lets assign AC a value of y. so tan(30degrees)=y/(2*sqrt(3)) which is = 2 so MC=AC-AM= 1.0718
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.
251348
After 2.5 hrs, Boat A has gone 50 km, and Boat B 80 km. Let the distances of A and B be two sides of a triangle, a 70 degree angle connecting them.\n\nThus we have a triangle with two known sides and one known angle.\n  \nLabel it A, B and C starting from the bottom left angle. Label each angle's opposite side its respective lowercase (a,b, and c). Fill in for b and c, and angle A. We don't know if this is a right triangle or not, so we're going to need to use the law of cosines:\n\na^2 = b^2 + c^2 - 2*b*c*cos(a)\nSubstitute for each known variable:\na^2 = (50)^2 + (80)^2 - 2(50)(80)cos(70 degrees)\na^2 = 2500 + 6400 - 8000 * cos (70 degrees)\na^2 = 8900 - 8000 * .34202\na^2 = 6163.84\na = 78.5101\n\nThus, using the law of cosines, the distance (a) from the two boats is approximately 78.5101 km. It is not exact as there is no exact value for cos (70 degrees).
466051
let the radius of the cylinder be x,  drawing a cylinder in a cone on the paper, we see 2 triangles - the cone & the triangle with it's base as the top flat surface of the cylinder. follow me?\n\nif u do, then let's assume the height of the second triangle to be y. y can be found in terms of x as:\n(since both triangles are similar)\n\nx:4 :: y:12  \ni.e. x/4 = y/12\n=> y=3x\nthus ht of cylinder = ht of cone - y\n=12-3x\n\nvolume of cylinder = pi*r2*h  (r-> radius, h-> height of cylinder)\nv= pi * x2 * (12-3x) = pi * (12x2 - 3x3)\n  (the second term being 12 *x*x - 3*x*x*x ....sorry, i don't have a math equation display software here)\n\npartialy differentiate above equation wth respect to  x, we then get\ndv/dx = pi * (24*x - 9*x*x)\n\nequating dv/dx to zero, (to find a value of x that makes the volume value either maxium or minimum. u can verify it's max by again differentiating dv/dx with respect to x, equating to zero & finding if the value solved for is positive or negative)\n\nanyways, we get\n\neliminating  pi,\n24*x - 9*x*x=0\n\n=> x=24/9 = 8/3= 2.67\n\nnow, simply substitute this value of x in the volume equatiion above to get\n v= pi * (12*2.67*2.67 - 3*2.67*2.67*2.67)\n= 89.372 or so.\n\n\nuh...i guess i only confused u up, hope it helped u atleast a bit.
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!
186357
From your question, we know the 3 points of the triangle are:\n(0,0), (0,y) and (x, y).\n\nThis points us to the fact that the right angle is at point (0,y).\n\nHence, area A of triangle is given by the product of base and height divided by 2. Therefore, we have:\n\n Area A of triangle = 1/2 [base.height]\n                    = 1/2 y.x\n\nWe are told y = x2, therefore:\n Area A of triangle = 1/2 y.x\n                    = 1/2 x3 (read as half x-cube)
17241
Jan 17 Note: If this is hard to understand, after sleeping on it, I may have an easier construction.\nIf you need further explanation, add another note to your question.\n\n-----------\n\nI think I have a solution.\nI'm making the assumptions that the Babylonians didn't have an acurate value for pi and used the value three. If I do that, every thing falls out.\n\nDraw a cirlce. Mark off 2 arcs. Make one the top portion of a circle say from 10 o'clock to 2 o'clock, the other the matching arc from 4 o'clock to 8 o'clock. Note: These arcs are just to get a representation of the problem and are not an acurate length.\n\nNext draw the two chords for the arcs, that is, the straight line from 10 to 2 and the other from 8 to 4.\nNow draw in the diameter of the circle from 12 o'clock to 6 o'clock.\nThe two chords cut off pieces on the ends of the diameter. Label these as length 2 (2 is the perpendicular) "Double 2 and get 4" That's the length of the two end pieces of the diameter.\n"Take 4 from 20 and get 16" This is where using 3 as a value for pi comes in. If pi 3, then a circle of circumference 60 will have a diameter of 20, so let the length of the diameter be 20. Taking 4 from 20 gives you the length of the diameter that is between the 2 chords, 16.\n\n"Square 20 get 400, square 16 get 256, take 256 from 400 get 144. Find the square root of 144. 12, the square root, is the chord."\n\nWe're obviously doing some Pythagorean theorem here so where is the triangle? \nDraw the diameter of the circle from 8 o'clock to 2 o'clock, that's the hypoteneuse as has length 20.\nThe distance between the two cords is 16. That's also the distance from 8 o'clock to 10 o'clock. Draw that line and label it 16. (Remember the drawing isn't accurate just enough to show us how things are layed out).\nThat leave the 3rd side of the triangle as the chord between 10 o'clock and 2 o'clock. \n\n20^2 - 16^2 = 12^2  so 12 is the length of the chord.\n\nHow does that sound?\n\nIn modern English, the statement of the problem would be something like this:\n\nLet AB be an arc of a circle of circumference 60. Let X be the chord connecting A and B.\nLet Y be the perpendicular bisector of the chord. If the distance from X to the circle along Y is 2, find the length of X. (You may assume that pi = 3 for the purposes of solving this problem.)
495612
which is angle 1 and which is angle 2?where is the diagram/description?\ninstead of saying angle 1 and angle 2 can you please name the angles?it will be very helpful
547300
sorry, i don't have a calculator w/ me right now but i'l try\n1. use the law of cosines\na squared = b squared + c squared -2bc*cosA\n2. in a 30-60-90 triangle the sides are x- the side across from the 30; x*square root 3- the side across from the 60; 2x-the side across from the 90\n3. i would try the law of sines\na/sinA = b/sinB = c/sinC\nremember that you know two angles, the 90 and the 61\nso the 61 is the degree the ladder and the ground forms.  set up the problem like this: x/sin61 = 18/sin90\nit should work\n\nto find secant, type into the calculator 1/cos(36) and check what mode you're in (degree or radians)\ni'm pretty sure cot60 = 1/the square root of 3\nsec30=2/the square root of three
577643
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.
350677
C - 3 possible answers\n\nIn any given right triangles with the two other angles being a and b,\n\nsin a = cos b AND sin b = cos a,\n\n\nas such, the first and fourth combination will give one answer, the other two will either give square trigonometric values of each angle i.e. sin^2 a and cos^2 a  OR sin^2 b and cos^2 b.\n\nThe only condition that will not satisfy this is when the angle is 45 degrees, which means a = b and contradicting the given statement of a being less than b.\n\nThus the answer is C.
301676
Area = 1/2 ac Sin B\n\nwhere a,b,c are the sides and A B C are the opposite angles. \nFrom the 10 sided figure you get an internal angle of 360/10 - 36 degrees so the other two angles are 180-36 all divided by 2 = 144/2 = 72 degrees.\n\nand the outside edge is 240/10 = 24\n\nSo using ac = 24 and Sin 36.. plug into the formula above.. this gives you the area of one triangle, then multiply by 10
43338
Just use Cosinus Theorem: c^2=a^2+b^2-2abCos(gamma)\n1.\nc=sqrt(4+64-2*2*8*sqrt(3)/2))\nc=6.34\n\n2. Is not a right Triangle.. but anyway lets find the angle A:\n22^2=15^2+33^2-2*15*33*Cos(A)\nCos(A)=830/990\nA=ArcCos(83/99)\nA=33°1' 48.54"
424744
Hi, \n\nThe answer lies in the pythagorean theorem. \n\nFor any right (one of the angles is 90 degrees) triangle with sides a,b and c where c is the side opposite the 90 degree angle:\n\na^2 + b^2 = c^2\n\nso for your first excercise:\n\na = 4\nb = 4 Sqrt(3)\nc = ?\n\na^2 + b^2 = c^2\n4^2 + {4 Sqrt(3)}^2 = c^2\n16 + 4^2 * 3 = c^2\n16 + 16 * 3 = c^2\n64 = c^2 \nc = +- 8 (since we're dealing with a triangle, -8 doesn't make sense, so just ignore it.)\n\nc = 8. \n\nFollow that same pattern, and you should be all set. \n\nHope that helps, \nMatt