then, what is sine, cosine and tangent in a triangle??
605025
sin(48) = 4/c\nc = 4/sin(48)\nside a can then be figured out using the pythagorean theorem.
743164
Remember the definitons of tangent, cotangent, etc. as they relate to right triangles.  Draw a triangle for both α and β.  Use the pythagorean theorem to finr the hypotneuses.  You will find the sides of the α triangle to be 3, 4, 5.  so:\n\nsin α = 4/5\ncos α = 3/5\n\nDo the same for β and the triangle is 5, 12, 13 with:\n\nsin β = 12/13\ncos β = 5/13\n\nFinaly, use the identity:\n\n sin(α-β) = sin α  cos β - sin β cos α \n\nSubstituting:\n\nsin(α-β) = (4/5)(5/13) - (12/13)(3/5) = (20 - 36)/65 = - 16/65
447477
In a triangle the sum of the angles is 180°, but the sum of the tangents is not anything in particular.  The tangent function is 0 for 0°, but it increases without bound as the angle approaches 90°.  So if the triangle contains one angle very very close to 90° and the other 2 are the rest of the 180°, the sum of tangents can be as large as you like.  On the other hand, if two of the angles are very very close to 0 and the third very close to 180, the sum of the tangents can be as close to 0 as you like.
209404
trigonometry is the study of angles in one or several triangles. it is a little like geometry, in the sense that it deals with shapes, but it also has characteristics similar to algebra and basic number functions. it's not too difficult once you learn how to work a scientific calculator.
830426
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!
96766
1) tan A = x/y\n2) sin A = x/r\n3) cos A = y/r\n4) sin A/cos A = (x/r)/(y/r)\n5) sin A/cos A = xr/yr\n6) sin A/cos A = x/y\n7) sin A/cos A = tan A\n8) tan A = sin A/cos A\nReasons\n1-3)DEFINITION OF CIRCULAR FUNCTIONS\n4)MPE (STATEMENTS 2 AND 3)\n5)SIMPLIFYING COMPLEX FRACTIONS\n6)CANCELLATION\n7)SUBSTITUTION (STATEMENT 1)\n8)SYMMETRIC PROPERTY\n\nQED
722736
the angles are just unknown if i get you correctly.\nthere are sin, cos, and tan proportions you can work out to find the rest. i think you only need three of the measurements to work out the rest of the triangle, like 2 sides and an angle or 2 angles and a side.  not positive.
200427
tangent = opposite/adjacent (toa)\nsine = opposite/hypotenuse (soh)\ncosine = adjacent/hypotenuse (cah)\n\nTom's old aunt sat on her coffin and howled.\n\nI learned that 10 years ago and I still remember it for some reason. Then cotangent is just the flip of tangent, secant is the flip of cosine, and cosecant is the flip of sine.\n\nAll, sine, tangent, cosine (all students take calculus) tells you which of the trig functions are positive in the various quadrants I-IV, in that order.\n\nThese are the only things in trig that you need to sit down and memorize (more or less). Almost else can be justified using this information, specifically the trig identities. The best way to remember these guys is to derive them yourself. If you try to memorize them all, you're bound to forget them since they don't have convenient mnemonics, but if you derive them yourself and understand why they hold, you tend to remember them.\n\nSpecifically you can prove easily that tan = sin/cos, and that sin^2 + cos^2 = 1. Once you've proven these, you can prove a ton of other identities.
779945
1. cosθ=A*B / (|A|*|B|)=(1*5+0*2+3*6) / 5√ 26\n            =81.171\n\n2. assume the angle of vectors of a and b is "m"  \n   so cos m= a*b/(|a|*|b|)=0   because cos90=0\n   m=90...so it is a right triangle...\n   |a| and |b| are representing the right-angled sides of the triangle.\n   |a|=√17; |b|=√26\n   S=|a|*|b| / 2 =10.5\n\n3. P(1,3,5) to cylindrical coordinates.......\n                         r=√1+9 =√10....\n                         θ= tan^(-1)3=71.6\n                         z= 5 \n       In terms  of x , y , and z\n                         x= r.cosθ =√10*71.6 =226\n                         y= r.sinθ = 0.95*√10= 3\n                         z=5\n                        so its (226,3,5)\n            ditto, T 's doesn't have cylindrical coordinates \n                     S 's cylindrical coordinates is (0.6, -0.8, -10)\n  ......................(that's all i know..)
303035
If the triangle is a right triangle, you can find the other side by Py's Th and then by the formula,(A=1/2bh).\nIf it is not, you should know the angle they make each other or an adjacent angle. Only if, you can find the area by sine laws or cosine laws.
488283
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.
937438
180 degrees = π radian    ok!\n\nThe sum of the three angles of a triangle = 180 degree\nor, x + y + z = 180 deg.\nor, x + y = (180 - z)deg.\nor, x + y = 180(1 - z/180)deg.\nor, x + y = 180(1 - z/π) radian (pvd)
161223
law of cosines : a^2 = b^2 + c^2 -2bc * cos(A)\nwhere a, b,c are side lengths and A is the angle oppisite side a.
736970
Here are some real tough ones.(Looked tough to me)\n\nProve that cot(90 - theta)sin(90 - theta) = sin theta\n\nFind Cos(3x)\n\nFind Sin(3x)
626033
I wanted to give a partial answer that wasn't presented elsewhere:\n\ne.g. Sine 10 deg =0.174, while cosine 80 deg= 0.174. how ironic is that?!!\n\nIt's not ironic at all.  If you have a right triangle, you know one angle is 90°, and the sum of all angles must be 180°, so the other two angles must add to 90.  You give sin 10°, so the other angle must be 80°.\nThe sin of any angle must be equal to the cos of its complementary angle (x and 90°-x are complementary), because of the definitions.  The side opposite the x angle is adjacent to the the 90°-x angle.\nThus sin 10° = cos 80°.
610782
A circle has 360 degrees. Now imagine that from the icosagon you can make 20 equally-sized triangles. The angle made by each triangle is 360/20=18 degrees. The other two angles of each triangle are both 81 degrees.\n\nThe basic formula for the area of a triangle is 1/2 base x height. How do we find these? Well, imagine that from each of the 20 triangles we make two additional right triangles. In doing this we know several things:\n\n-The hypotenuse is equal to 6.5 cm (the radius).\n-The angles of the smaller triangle are now 9 degrees, 81 degrees, and 90 degrees.\n-The sine of 9 degrees is equal to the opposite side divided by the hypotenuse. If we multiply this value by 6.6 we get the length of the opposite side ("1/2 base").\n-The sine of 81 degrees multiplied by 6.5 gives us the "height".\n-Once you compute each value (1/2 base and height) you multiply them by each other and you get the area of the triangle.\n-Multiply the area of that triangle by 20, since you have 20 of them. Voila.
956162
It depends on which leg you know the length of.  If it's the "adjacent"leg--the one that, along with the hypotenuse, forms the angle--then remember that the cosine of an angle is defined as the length of the adjacent angle divided by the length of the hypotenuse.\n\nIf it's the "opposite" leg--the one that isn't part of the angle--then you use the fact that the sine is defined as the length of the opposite leg divided by the length of the hypotenuse.\n\nEither way, you use the inverse trig functions on your calculator to find the angle.