i know they MEAN sin cosin and tangent, but i need some help with the actuall uses and meanings and definitions and stuff for it, in context of a triangle.
212938
In order to define the trigonometric functions for the angle A, start with an arbitrary right triangle that contains the angle A:\n\nWe use the following names for the sides of the triangle:\n\nThe hypotenuse is the side opposite the right angle, or defined as the longest side of a right-angled triangle, in this case h. \nThe opposite side is the side opposite to the angle we are interested in, in this case a. \nThe adjacent side is the side that is in contact with the angle we are interested in and the right angle, hence its name. In this case the adjacent side is b. \nAll triangles are taken to exist in the Euclidean plane so that the inside angles of each triangle sum to π radians (or 180°); therefore, for a right triangle the two non-right angles are between zero and π/2 radians. The reader should note that the following definitions, strictly speaking, only define the trigonometric functions for angles in this range. We extend them to the full set of real arguments by requiring that they be periodic functions.\n\n1) The sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. In our case\n\n. \nNote that this ratio does not depend on the particular right triangle chosen, as long as it contains the angle A, since all those triangles are similar.\n\nThe set of zeroes of sine is .\n\n2) The cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse. In our case\n\n. \nThe set of zeroes of cosine is .\n\n3) The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side. In our case\n\n. \nThe set of zeroes of tangent is .\n\nThe remaining three functions are best defined using the above three functions.\n\n4) The cosecant csc(A) is the multiplicative inverse of sin(A), i.e. the ratio of the length of the hypotenuse to the length of the opposite side:\n\n. \n5) The secant sec(A) is the multiplicative inverse of cos(A), i.e. the ratio of the length of the hypotenuse to the length of the adjacent side:\n\n. \n6) The cotangent cot(A) is the multiplicative inverse of tan(A), i.e. the ratio of the length of the adjacent side to the length of the opposite side:\n\n. \n\nMnemonics\nThere are a number of mnemonics for the above definitions, for example SOHCAHTOA (sounds like "soak a toe-a" or "sock-a toe-a" depending upon the use of American English or British English. Can also be read as "soccer tour"). It means:\n\nSOH ... sin = opposite/hypotenuse \nCAH ... cos = adjacent/hypotenuse \nTOA ... tan = opposite/adjacent. \nMany other such words and phrases have been contrived.
363538
we have in a triangle\n\nsin B/b = sin C/c = k (say)\n\nsin B= bk  ...1\nsin C = ck  ...2 \nadditionally\n\nb^2 = c^2+a^2-2ac cos B\nso cos B = (c^2+a^2-b^2)/2ac ...3\n\ncos  C = (a^2+b^2-c^2)/2ab ...4\nfrom1 and 3 by deviding 3 by 1\ncot B = (c^2+a^2-b^2)/2abck\n\nor 2abck = (c^2+a^2-b^2) tan B  ..5\nsimilarly from 2and 4\n2abck =(a^2+b^2-c^2) tan C\n\nfrom above 2 we get the result as both are same
55588
The trigonometric identities have as their primary purpose to simplify calculations.  For example, why evaluate 2sinxcosx when you can evaluate sin2x.\n\nAs for the applications, oblique triangles are used in surveying and mapping.  For example, suppose you want to know your distance from an object on the opposite side of a river.  You can mark off a line of a fixed distance, say 100ft.  Then, go to the opposite ends of the line and use a transit to measure the angle between the object and the line you measured.  You can now determine all three sides with this information.
238702
I'm not giving you the answer, but Cos is adjacent over hypotenuse.  Also, if you wanted the question answered correctly, us answering people might want to know where all of the letters are in relation to the triangle.
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SOH-CAH-TOA.\n\nRight triangle:\nOpposite side = 15\nAdjacent side = 8\nHypotenuse = 17 (sqrt(8² + 15²) = 17)\n\nSo\nsin(angle) = O/H = 15/17\ncos(angle) = A/H = 8/17\ntan(angle) = O/A = 15/8\n\nI agree with you and think the book has it wrong...\n\nAnother confirmation:\ntan(angle) should equal sin(angle) / cos(angle).\ntan(angle) = 15/17 divided by 8/17\ntan(angle) = 15/8\n\nSo cos(angle) should definitely be 8/17... must be a typo.\n\nThe only explanation is that perhaps it was the cotangent?  cot(angle) = 1/tan(angle) = 8/15...
213860
They are a set of Matricies that map one set of {x,y,z} to another set of {x',y',z'} through a rotation angle.  In oder to use them you need to know your starting set of coordiantes and the three angles through which you want to rotate.\n\nOne such set of angles that forms a perfectly good rotation basis are the Euler angles.\n\nIf you know the rotations, you can actually write the whole transformation as one matrix (usefull if you have to apply it over and over again.
87432
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..
553694
sine opposite/hypotenuse\ncosine adjacent/hypotenuse\ntangent opposite/adjacent
222717
I think that what you really want here is the tangent ratio. You know the slope of the line, right (hint, it is right there in the equation)? You know that the tangent ratio is a over b (where a and b are the non-hypotenuse sides of the triangle), right? You should be able to use that to figure out the tangent of the angle made by the line. Now all you have to do is figure out the angle that has that number as a tangent. That would be the "tan -1" (inverse tangent) function on your calculator.\n\nIf you are working from trig tables rather than a calculator, find the number in the second column of the tangent table that is closest to your tangent number then look in the first column for the angle.\n\nIf you are required to use sin and cosine remember your Pythagorean theorem and use it to find the hypotenuse.\n\nASIDE TO Drakestone: Your solution only finds the sine of the angle, not the angle itself. You need to apply "sin -1" to your answer to get the final answer.\n\n\nASIDE TO eternal...: You are wrong too. graph y=5x+3 and you will see that it is way too steep to be an 11 degree angle. You mixed up your opposite and adjacent values in calculating your tangent.
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Adjacent means  "next to"\n\nIn trigonometry (which you are calling maths), the term is used with respect to angles. When you consider the right angle in your figure above, "a" and "b" are the adjacent sides of the right angle (the 90 degree angle) and the side "c" is the hypotenuse and also the "opposite" side.
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If the circles are all tangent to each other, then you can draw lines from center to center to center.  Thus, you'd have a triangle with sides 8, 12, and 10 (each side is formed by adding two radii).  So you can find the angles using the law of cosines:\n\nc^2 = a^2 + b^2 - 2abcos(c)\n8^2 = 12^2 + 10^2 - 2(12)(10)cos(c)\n64 = 144 + 100 - 240cos(c)\n240cos(c) = 180\ncos(c) = 3/4\nc ~ 41 deg\n\nAnd then just change up numbers to find the other angles:\n\nb^2 = a^2 + c^2 - 2accos(b)\n10^2 = 12^2 + 8^2 - 2(12)(8)cos(b)\n100 = 144 + 64 - 192cos(b)\n192cos(b) = 108\ncos(b) = 9/16\nb ~ 56 deg\n\na^2 = b^2 + c^2 - 2bccos(a)\n12^2 = 10^2 + 8^2 - 2(10)(8)cos(a)\n144 = 100 + 64 - 160cos(a)\n160cos(a) = 20\ncos(a) = 1/8\na ~ 83 deg\n\nAnd this checks out since 41 + 56 + 83 = 180.
115364
So you know the ones for sin(2a) and cos(2a). Cool.\n\nNow, tan(2a) is the same as sin(2a)/cos(2a), since tan = sin/cos (definition of tangent).\n\nSub in the formulae for sin(2a) and cos(2a) :\n\n2sin(a)cos(a)\n----------------\ncos(a)^2-sin(a)^2\n\nDivide both numerator and denominator by (cos(a))^2\n\n2tan(a)\n----------\n1-tan(a)^2
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
404829
I think the formula for sin(c-d) is sin c cos d - cos c sin d\n\nso sin (x - pi/3) = sin x cos pi/3 - cosx sin pi/3\n\nwhich is 1/2 sin x - sqrt3 / 2 cos x\n\nMultiplied by the -2 in front, this is -1 sin x + sqrt 3 cos x\n\nSo A = -1 and B = sqrt 3\n\nLook up the original formula, I may have messed up the signs
439446
OK, first let's try to figure out the arctan part.  Arctan(-1) means the angle whose tangent is -1, right?  And the tangent is the sine divided by the cosine.  So you're looking for angles whose sines and cosines are equal, but with opposite signs.  They point out to the right and down, or out to the left and up, at 45-degree angles.  That means the angles are 135 degrees and 315 degrees, or 3pi/4 and 7pi/4 radians.\n\nNow, the question asked you to find the sines of these two angles.  The sine of 3pi/4 is sqrt(2)/2, and the sine of 7pi/4 is -sqrt(2)/2.  So your answer is +/- sqrt(2)/2.
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In quadrant one, sine, cosine, and tangent are positive. In quadrant two, sine is positive, and cosine and tangent are negative. In quadrant three, tangent is positive, and sine and cosine are negative. And in quadrant four, cosine is positive, and sine and tangent are negative.\n\nQuadrant 1 – Sin(+) Cos(+) Tan(+)\nQuadrant 2 – Sin(+) Cos(–) Tan(–)\nQuadrant 3 – Sin(–) Cos(–) Tan(+)\nQuadrant 4 – Sin(–) Cos(+) Tan(–)
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Don't panic ok? Math is something that's dead. It can't kill you so don't worry, we'll get through this =D\n\nLet the sign (>)= angle\n\nLets start off with what you know:\n\n1) Isosceles Triangle means that the bottom 2 angles (in this case >B and >C) is the SAME THING. >B = >C!!!! Remember that.\n\n2) >A is 20 degrees more than 3 times >B. Now how the hell do you write that? Well, I'll show you:\n                                     >A= (3)>B + 20 \nNow look at what I've wrote. I wrote that >A is 3 times the >B plus 20. Yay.\n\n3) All the angles of a Triangle adds up to how many degrees? If you say 180, you are RIGHT! So it makes sense to write:\n                                (>A) + (>B) + (>C) = 180\n                                                 Yay.\n\nNow all you have to do is substitute what we already know. We know that [>A= (3)>B + 20] , and that [>B = >C] \n                                 \n                                  (>A) + (>B) + (>C) = 180\n                                           which becomes\n                            [(3)>B + 20] + [>B] + [>B] = 180\n\nYou see, towards the end I substituted >C with another >B because they are the same damn thing. =D Yay.\n\nNow we add Like-terms.\n\n                            [(3)>B + 20] + [>B] + [>B] = 180\n                                               becomes\n                                         (5)>B + 20 = 180\n\nSee how I got that? I just added the 2 angle Bs in the end with the 3 we started out with.  >B  + >B + (3) >B = (5)>B Yay!\n\nNow we subtract 20 on both sides of the equation. Giving you:\n\n                                          (5)>B = 160\n\nNow lets divide the 5 on both sides so we can get >B alone by itself. Giving you:\n\n                                            >B= 32 degrees!\n                                            which means that \n                                            >C= 32 degrees!\n\nWe stated before that >B is the same as >C so there is your answer:  >C= 32 degrees!!!!!!!!!!!!!!! YEA BABY WE DID IT! WOOOO We solved it YES!!!\n------------------------------------------------------------------------------------\nNow lets check! If we did this correctly, >A + >B +>C should equal 180 right? Damn straight. Lets first find what >A is equal to.\n                                 \n                                         >A= (3)>B + 20 \n                                        Substitute 32 for B...\n                                        >A= (3)(32) + 20\n                                         which means that \n                                          >A=116 !!!!!!!\n\nNow lets plug all the angles in. >A= 116, >B= 32, and >C=32\n\n                               Does 116 + 32 + 32 give you 180?\n\n     OMG IT DOES IT DOES IT DOES!!!! YES!!!! HAHA\n\n                                   Answer to your question:\n                                           >C= 32 degrees.\n\nHope I helped you ALOT hahahaha You're welcome. =D