What are real world applications of trigonometry?
82068
surveying does use it alot and so does desgining and architecture. Engineering, construction and factories and industries all use it. There are many jobs out there were trig can help. I hated it but it shocked me how much I have used it now.
36505
Great question.\n\nI think a maths history project would be wonderful.\n\nSpecifically, I would like to see a paper which explains how the trigonometric functions were calculated in history.\n\nFor example, what angles did the Greeks know trig functions for?\n\nWhat resources did Isaac Newton have for trig functions?\n\nHow were these tables calculated before electronic computers?\n\nThe combination of history and maths would be very interesting.\n\nCheers!
824701
Well maths is the foundation of most ares of science, especially physics, and physics relates to many jobs in fields like engineering etc. Trigonometry in maths at first seems to just relate to triangles, but apart from finding unknown sides and angles, it has many applications. the relationship between the sides of a triangle, in terms of sine, cosine and tangent apply to many areas of physics. You may have heard of trig graphs, and in physics, you can see how many things like waves, simple harmonic motion and AC current all relate to trigonometrical concepts, that is, looking at objects in motion in terms of a reference circle
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^_^
908924
Because of (one of) the definition(s) of sin and cos.\nWhich is: take a unit circle, and draw in any radius at some angle theta. Then sin theta = the y value, and cos theta = the x value. Those are obviously less than 1.\n\nOr you could just take a typical right angled triangle - the length of the opposite/adjacent sides are always less than the length of the hypotenuse.
249103
well it depends on the angle between quantities. \nat 0 degree, simply add the given magnitude\nat 90 degree, use the pythagorean theorem\nat 180 degree, simply subtract the magnitude\nat any angle, other than 0, 90 and 180, use the laws of sine and cosine
680870
sounds like you need to use some basic trig functions.\n\nthe button "tan" on the calculator stands for tangent. the tangent of an angle of a right triangle (other than the right angle) is = to the ratio of the side opposite the angle to the side adjacent to the angle.\n\nthe button "sin" on the calculator stands for sine.  the sine of the angle of a right triangle is = to the ratio of the side opposite the angle to the hypotenuse.\n\nthe button "cos" on the calculator stands for cosine. the cosine of an angle is = to the ratio of the side adjacent to the angle to the hypotenuse.\n\nbasically, you decide which function to use that best fits the information given.  you enter the given angle in the calculater and press the appropriate function.  then you set that value (which you round off to the nearest ten-thousandth) equal to the appropriate ratio and then you solve.
231249
You may never have to know the sine of an angle in real life, but the point of your math classes is to teach you how to think logically. What the different subjects show you is their applications in the real world. If you become an English teacher, you won't need to know the area under a curve.\n\nBut the thing with math is that the numbers provide a numerical visual for the concepts. When you get the right answer, you can see how the concept works via that example (sometimes called a "problem" or a "question"). And that's it. It's somewhat akin to writing a paper - you make a thesis statement, and then follow by offering proof/support of your statement. You're showing why your statement works by using words. Math often uses numbers for the same purpose.
449531
Knowing trig makes learning calculus easier.
126358
Before they had GPS, altimeters, and satellites to measure altitudes, they estimated the height of a mountain using trigonometry.  If the distance to the mountain can be estimated or determined, the height of the mountain can be calculated from the angle between looking straight and looking up at the top of the mountain.  \n\nThe link in sources gives an example.
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
567245
simple x=r*cos(theta), y=r*sin(theta), r^2=x^2+y^2\nsubstitude, not hard
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???
895625
This may be unexpected, but a good site with the formulas you need.\n\nhttp://en.wikipedia.org/wiki/Trig
214507
http://www.awesomelibrary.org/Classroom/Mathematics/Middle-High_School_Math/Trigonometry.html\nhttp://www.analyzemath.com/Trigonometry.html
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.
785854
my school asked us to do just one: the daily applications of geometry (with special emphasis on Similar Triangle, Trig and Pyth theorem)