How do people use trigonometry in life?
145573
Drafting, surveying, triangulation (using trigonometric theroms and formulas to determin distance, heighth, or location). Generally you'll fing the following types of people using it: architects, engineers, surveyors, cartographers, sailors, any and all branches of the militairy, navigaters, physicists, and of course geometry and trigonometry teachers.
191873
First of all, the best way to learn trigonometry is to learn it from the beginning, which means take your first textbook which has trigonometry and study from the beginning.\nhttp://mathworld.wolfram.com/Trigonometry.html\nhttp://www.du.edu/~jcalvert/math/trig.htm#Trig\nhttp://www.syvum.com/math/trigonometry.html
406591
Switch your calculator from degrees (DEG) to radians (RAD). That should do the trick. \nHere are some quick trig rules you need to know:\nIn radians just remember that pi is 180 degrees. Pi/3 is 60 degrees, pi/2 is 90, pi/4 is 45 degrees and so on. \nIn a unit circle, the hypotenuse is always 1.\nSin is always the far side away from the angle (the vertical side). We call that the opposite side.\nCos is always the adjacent side (or horizontal side).\nIn a right triangle with angles 60 and 30, the shortest side is always 1/2 the hypotenuse, so in a unit triangle, it is just 1/2. You can use the pathagorean theorem to find that the longer side is the square root of (3/4), or (sqrt(3))/2.\nIn a right triangle with angles 45 and 45, the opposite and adjacent sides are equal, and we can use the pathagorean theorem to find that the sides are the square root of (1/2), or 1/(sqrt(2)).\n\n1. Sin 1/3 pi. This is the same as sin pi/3 which is a 60 degree angle. If you draw the triangle on a unit circle you would see that the opposite side is the longer side of a 90/60/30 triangle. We have seen above that the length is (sqrt(3))/2.\n\n2. Cos pi. Easy. pi is 180 degrees, so draw the unit circle. You see that it is just a striaght horizontal line going the other way so the answer is -1.\n\n3. cos pi/4 is a 45 degree angle. We know from above that both sides have to be 1/(sqrt(2)).\n\n4. sin pi/4. Same as #3.\n\n27. What is 315 degrees? Well draw your unit circle. Go, 90, then to 180, then to 270, then go 45 more degrees and you will end up at 315 degrees. Draw your triangle. We know that the hypotenuse is 1, so the coordinates will simply be, (cos 315, sin 315). Cos 315 is just a 90/45/45 triangle in the bottom right quadrant so the adjacent side is positive. The opposite is negative, so you end up with coordinates (1/(sqrt(2)), -1/(sqrt(2))). Do the same to solve the last question youeself.
831941
You're supposed to print a table, right? The question is poorly written, so it's not clear, but I'll assume that's the goal.\n\nWell, it sounds like you can write your loop, so shouldn't you just print out the sine, cosine, and tangent of the angle in the body of the loop?  It's not clear what the stumbling block is.  You realize that there are built in functions that compute sine, cosine, etc.?
215627
well from what i learn we have a formulae for that right??!! then the formulae to find theta is.... to find cosine the formulae is adjecent over hypothenus in shortform its CAH or C=A/H. and to find sine is opposite over hypothenus or SOH or S=O/H.. and to fine tangent is opposite over adjecent or TOA or T=O/A.. so to remember it you just remmber it in the short form.. as if you have a chinesse friend name SOH CAH TOA.. soh for sine n cah for cosine n toa for tangent.. i hope thats what you want for the answer.. if not.. im sorry.. sooo sorryy..
105824
Trigonometry.  The good old days when I still loved math... ahhh!
892355
Trig is just one of those courses prior to specialization that most science/math students have to take.  Case in point, during my first yr mech engineering studies, I had to take geology.  As for aviation, trig is defifitely applicable.  It will used to calculate al types of parameters such as forces related to lift, drag and trajectory.
890202
in trigonomertic form\n\nz = r cos t + ir sin t = 1 + i\n\nso r cos t = 1\n r sin t = 1\n\nsquare and add r^2 = 2\nso r = sqrt(2)\n\nnow cos t = 1/sqrt(2) and sint = 1/sqrt(2)\n\nso t = pi/4 (as cos pi/4 = 1/sqrt(2) and sin is also positive so 1st quadrant)\n\nso t = pi/4 and r = sqrt(2)
472052
ok so i assume you know the basic rules of trigonomerty for a triangle.\ni.e sinR = opposite/hypotenuse and cosR= adjacent/hypotenuse.\n\ndraw the triangle on a piece of paper.\nYou will see that sin R = ST/SR\n\n(where ST is the edge running from S to T and similarly SR is the edge running from Sto R)\n\nalso cos R= RT/SR\n\nso sinR^2 + cosR^2 = (ST)^2/(SR^2) + (RT)^2/(SR^2)\n\n= [(ST)^2 + (RT)^2]/(SR^2)\n\nNow we use pythagoras' theorem. I assume u know this? The sum of the squares of the two shorter sides of the triangle equals the square of the longer side (the hypotenuse)\n\nso (ST^2) + (RT)^2 = SR^2\n\nSO WE NOW HAVE\nsinR^2 + cosR^2 = SR^2/SR^2 = 1\n\nthere you go\n\n\nCANT E-MAIL YOU BACK WESTILE BECAUSE IT SAYS YOUR E-MAIL IS NOT VERIFIED
1117196
use cosine law (because you don't know any of the angles) \n\n---- cosine law used when you know the measurement of an angle between 2 known sides or when there're no known angles at all\n\na^2= b^2 + c^2 - 2bc x cosA\n\nb^2= a^2 + c^2 - 2ac x cosB\n\nc^2= a^2 + b^2 - 2ab x cosC\n\nc is side opposite angle C, a the side opp angle A, b the side opp angle B.\n\nyou first have to draw the triangle\n\nafter drawing it, you will see that the side opp angle A is BC, and it is 7 units long (label the sides as a, b, or c if that helps you). what you do afterwards is put the values into the equation.\n\n(7)^2= (12)^2 +(10)^2 - 2(12)(10)cosA\n\nisolate the unknown value\n\n49 - 144 = 100 - 2(12)(10)cosA\n-95-100 = -2(12)(10)cosA\n-195 = -2(12)(10)cosA\n-195/(-2 x 12 x 10) = cosA\ncosA = 0.8125\n\nto find A, just press cos^-1 0.8125 on your calculator.\n\nangle A = cos^-1 0.8125\nangle A = 35.65908.... degrees\n\nrounded to nearest degree, angle A is 36 degrees\n\nhope that helped :)
1180581
When I was first trying to do trigonometry problems like this, I found it helpful to draw a picture.\n\nDraw a horizontal line to represent the ground, and then put a vertical line to represent the radio tower.  We want to know the height of the tower, so label it "x".  The angle between the tower and the ground is 90 degrees, so label it too.\n\nNow draw a dot (or a little stick figure with a protracter if you feel like it) to the side of the tower, on the ground.  Label the line between the tower base and the dot "50", as it's 50 meters from the tower.\n\nNow draw a line between the dot and the top of the tower, representing the person sighting to the top and measuring the angle.  Lable the angle between this sight line and the tower top "67 degrees".\n\nYou now have a drawing of the triangle with two angles and one adjacent side known.  This should be enough to get you started picking the right trig function to apply here.\n\nIf that's not enough to get going, here's another hint: Remember the unit circle with a triangle and the sides labeled with "sin(theta)", "tan(theta)" etc?  Are there any similarities to your tower drawing?  I always found being able to draw that unit circle on my scratch paper at the beginning of a problem set or test helped immensely.\n\nGood luck.
449610
First, you need a protractor. Create a right triangle ABC with C as the right triangle. Let angle B be your angle. Make rough measurements of the three sides of the triangle. now:\nsin B = AC/AB\n\nOnce you know its sine, use these formulae for the other trigonometric functions:\ncos B = sqrt(1 - sin² B)\ntan B = sin B/cos B\n\nSpecial cases are:\nsin 0 = 0\nsin 90 = 1\nsin 30 = 1/2\nsin 60 = sqrt3/2\nsin 45 = sqrt2/2\nFrom these, you can compute all angles.\nremember,\nsin B = sin (B - 360n) where n is an integer\nsin B = cos (90 - B)\ncos B = sin (90 - B)\ncos (-B) = cos B\nsin (-B) = -sin B\n\nexample:\ncalculate sin, cos and tan of -150.\nfirst, sin -150 = -sin 150\n-sin 150 = -cos(90 - 150)\n= -cos (-60)\n= - cos 60 = -sin(90 - 60) = - sin 30\n= -1/2\nsin -150 = -1/2\ncos -150 = cos 150 = sin(90 - 150) \n= sin (-60) = -sin 60\n= -sqrt3/2\ntan -150 = sin -150/cos -150\n= (-1/2)/(-sqrt3/2)\n= sqrt3/3\n\nJust use these sign rules:\nif angle is in Q1 = sin cos and tan are +\nif angle is in Q2 = sin is +,cos is -,tan is -\nif angle is in Q3 = sin is -,cos is -,tan is +\nis angle is in Q4 = sin is -,cos is +,tan is -\n\n^_^
74653
Why ask; there are lots and lots of problems in trigonometry.
618088
It's spelled "pi" and it's used in trigonometry because it's how you measure a circle in radians. Radians is a better way to measure angles - pi is equivalent to 180 degrees, and 2pi is equivalent to 360 degrees.
842369
With these data a lot of angles fulfill that requirement; If  you have alpha first look at cos alpha, after divide by1.33  and look for theta.\n\nDid you not make a mistake. I suppose you mean sin and not cos\n\nSnell law uses sinus
907694
I'm not sure if I get your question right, but I think you're referring to lookup tables (or LUTs).  I think the easiest way to explain the use of these, is through an example from the old days.  Twenty years ago (or so), it took a long time to calculate the sine of an angle.  So if you needed to make a lot of calculations with sines in them, what you did was: calculate the sines of each of the 360 degrees (or more if you needed a better precision) and store the results in an array.  If you then needed the sine of n, all you needed to do is look up the value of the n'th entry in the array --which was a lot faster and therefore efficient than calculating the value.\n\nOf course, this only helped if you needed to calculate a sine more than 360 times, otherwise you wouldn't win back the overhead of pre-calculating all those sines.\n\nHTH.\n\nPS: a conclusion of what I wrote should be that using arrays with[in] loops does NOT *always* improve efficiency.
961572
Try to find a book on analytic geometry (which should also cover trigonometry), and one that has lots of exercises, which will help you retain all that information.  There used to be a book called Geometry & Trigonometry for Calculus, but it's out of print.  As per the algebra, try to find a cheap Dover Classics book through amazon, they are usually only $10-20.  Also, the following site may help you:\nhttp://ltcconline.net/greenl/courses/152a/152a.htm\nhttp://ltcconline.net/greenl/courses/152b/152b.htm\nhttp://ltcconline.net/greenl/courses/154/154.htm\nhttp://tutorial.math.lamar.edu/AllBrowsers/2413/TrigEquations.asp\nhttp://tutorial.math.lamar.edu/AllBrowsers/2413/TrigFcns.asp\n\nGood luck!\n\nSteve