How do I calculate a simple angle from longitude and latitude?
491873
Hope these are of use...
1043313
It's too cumbersome to rewrite it all here, but check out http://www.euclideanspace.com/maths/algebra/vectors/angleBetween/index.htm\n\nThat's 2D, but the formula is the same for 3D.
23041
You can measure the distance from the angle to the pole and use trigonometry.  But it may be easier to just move nearer or farther from the pole until the angle is 45 degrees; then the distance to the pole will equal the height of the pole (and you won't need a trig table).
91781
Do you mean lattitude and longitude?\n\ntry http://earth.google.com
40284
(1) Subtitute the damn values for what you have.\n\n(2) Sin theta=opp/hyps,\n    Tan   "  =opp/adj,\n    Cos   "  =adj/hyps.\n\nNow usually a blind man follows his stick now you might know where you are at the moment - get it???
546529
1. Make sure the angle of interest is in radians.  Radians are the "natural" (non)-units of angles, because they relate the angle measurement to the length of arc the angle cuts off on a unit circle.  So 2*Pi radians are equivalent to 360 degrees.  Basically, if you have that sucker in degrees, multiply it by 2 * Pi / 360.  Unit conversion is done. \n \n2. Make sure the angle is between -Pi and Pi.  Because sine is a periodic function, with period 2*Pi, just subtract or add 2*Pi enough times to get the angle between those two points.\n\nnow for the meat of the problem:\n\nsin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - ... + \n+ (-1)^(n-1)*x^(2n-1)/(2n-1)! + ...\n\nExample: sin(15)\n\nStep 1: 15 degrees = 15*(2*Pi/360) radians = Pi/12 radians \n \nStep 2: I need Pi/12 in decimal (let's see, Pi=3.14159 (memorize that much, it's pretty easy... "Pi when calculating sine, 3 point 1 4 1 5 9!")) \nPi/12 approx = .261799 (we only had 6 sig figs, so we can't add any more than that) \n \nNow to the series: \n         first term:  x=.261799 \n         1st estimate: .261799 \n         second term: -x^3/3! = -.0029905  \n         2nd estimate: .258809 \n         third term: x^5/5! = .0000102     \n         3rd estimate: .258819 \n         fourth term: -x^7/7! = -.167*10^-7  \n         stop here - past precision\n\nThe answer is positive \nsin(15) = 0.258819\nsin(18) = 0.309017
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.
66035
Well suppose you want to find antipodal positions on the earth. The earth is only approximately a sphere (it is actually closer to an oblate spheroid), so the mathematical definition of an antipodal point on the earth does not quite hold. \n\nIf one location on earth is at 23 degrees North latitude, and 10 degrees West longitude, the antipodal point will be at 23 degrees South latitude, and 190 degrees West longitude (but 190 degrees West longitude will be more often called 170 degrees East longitude). \n\nThat is, just add 180 degrees to the longitude, and switch the North latitude to South latitude (or vice versa).
506648
use a protractor - thats' the semi circle type thing...\n\nmeasure the angle on a straight line which is 180 degrees in total, the obtuse (larger) angle and the acute (smaller) angle will equal 180 degress so calc the obtuse one and subtract this by 180 leaving the acute angle.
154864
If you have a TI graphing calculator...your calculator is probably similar to mine.  For my TI, I simply hit the MODE button and arrow down to ANGLE, and hit the right arrow to switch it from RADIAN to DEGREE.
928764
You would just do 180-56 to equal 124 degrees\n\nsupplementary angles add up together to equal 180 degres or a straight line
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
504335
Draw two lines from any tow angles to two opposite base
1044084
That's easy enough to calculate with some basic trigonometry.  If you draw the angle (a) you are looking for, you see that you can bisect it to form two right triangles whose smaller angle is (a/2) and where one leg equals 5 light years and the other leg is half the diameter of the star.  The diameter of the sun is about 1.39x10^6 km, and a light year is about 9.46×10^12 km, so using these values, we find that the angle a is:\n\ntan(a/2) = 1.39x10^6km / 5*(9.46x10^12km)\n\nSo...\n\na = 2*tan-1 [ 1.39x10^6km / 5*(9.46x10^12km) ]\n\n= 5.877x10^(-8) radians\n\n...or about 0.000003367 degrees.
1054686
Job one, convert everything to trig functions of a single angle, namely x.\n\nsin2x=2sinxcosx\ncos2x=cos^2(x)-sin^2(x)=2cos^2(x)-1=1-2sin^2(x)\n\nThen substitute in and simplify:\n2sinxcosx+2(2cos^2(x)-1)=1+sinx-4cosx\n2sinxcosx+4cos^2(x)-2-1-sinx+4cosx=0\n4cos^2(x)+2sinxcosx-sinx+4cosx-3=0\n\nFactor this into:\n(2cosx-1)(2cosx+sinx+3)=0\n\nset each parantheses equal to zero\n2cosx-1=0   \ncosx=1/2\nx=arccos(1/2)\nx={pi/3, 5pi/3} + 2pi*n  where n=0,1,2....\n\nAnd\n\n2cosx+sinx+3=0\n\nSolved numerically gives imaginary solutions of\n{-i*ln((-2/5)-(i/5),-i*ln(-2-i)}\n\nSolved....
1183908
There is no such equation. \n\nThe post office keeps a database with the latitude and longitude of each post office -- each zip code. You can look up the latitudes and longitudes of the zip codes and use them to compute the distance. For a spherical earth:\n\nLet Lat1 and Lon1 be the latitude and longitude of the first post office. Let Lat2 and Lon2 be the latitude and longitude of the second post office. Then compute the angles:\n\ndLon = | Lon1 - Lon2 |   (make sure to account for any difference in the east- or west- of Greenwich longitudes)\n\nD = 2 arcsin[ sqroot{ sin^2((Lat2 - Lat1)/ 2)+ cos(Lat1) cos(Lat2) sin^2(dLon/2) } ]\n\nIf you computed D in radians, convert D from an angle to a distance (in miles) using the equation: \n\nMiles = D * 12450 / pi\n\nIf you computed D in degrees, use the equation:\n\nMiles = D * 69.167
763822
Divide the right angle into three equal parts, or three 30-degree angles.\n\nSounds like homework to me (which, hey, I'm cool with!), as if you owned this parcel this is NOT the way you'd want to divide it for real -- you wouldn't want three neighbors all bunched up like that at the confluence of the former right angle.\n\n.