Is the sine and cosine of a 60 degree angle A in a right triangle always the same?
369074
No.\n\nHere are the "sames'".\n\nSin(60 deg) = Cos(30 deg) = 0.866\nSin(30 deg) = Cos(60 deg) = 0.5\nSin(45 deg) = Cos(45 deg) = 0.7071\n\nIn general, Sin(x deg) = Cos([90 - x] deg)\nFor all real x.\n\nRegards
969737
sine 30° 15'\n\nCos ( 90 - A)\n\n89 60\n30 15\n- - - - -\n\n59 45\n\nThe answer is cos 59° 45'\n\n- - - - - - - -s-
463801
The cosine laws apply to triangles which are NOT right angled. I think you have maybe read that wrong:\n\nhttp://www.themathpage.com/aTrig/law-of-sines.htm\n\ngives you a full guide to using the cosine laws to solve problems like these.
628550
same angle degree, same lenght of a side.
785502
This is a right isosceles triangle.  It's right because the Pythagorean Theorem works and it's isosceles because 2 sides are equal.
595637
if you dont have one already, make a nice big copy of the entire circle with everything labeled.  repetition and review is really just the best way to remember anything.  when i try to think of the unit circle values for 30-60-90 triangles, i just think of it in terms of the long side and the short side.  the short side is always 1/2 and the long side is always (sq. root of 3)/2.  if i need to remember the values for, say, a 120 degree angle, i always draw a quick little sketch- just enough to show which are the short and long sides.  the horizonal is short, so the cosine is 1/2.  the vertical is long, so the sine is (root 3)/2.  this method works for all angles that are multiples of 30 degrees.  45, 135, 225, and 315 degrees are easy because you only have to remember one value for all the sine and cosines: (root 2)/2\n\njust dont forget to look at your sketch to see which of the values is negative! (and if the picture method doesnt work for you, theres a trick to knowing those too . . .\n\nhere are the quadrant numbers: (imagine the axis in between)\n\n2 1\n3 4\n\nuse this acronym to remember what trig functions are positive in each quadrant:\n\nALL---------> ALL of them are positive\nStudents----> Sine is positive\nTake--------> Tangent is positive\nCalculus----> Cosine is positive
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
509207
Yes. You can use sine or cosine to determine the angle.\n\nSine x = Opposite / Hypotonuse\nSin x=5/13\nSin x=.38\nx=22.6*\n\nCosine x = Adjacent / Hypotonuse\nCos x=12/13\nCos x=.92\nx=22.6*
1023518
Use the Law of Sines.\n\nsin A/a = sin B/b\n(sin 36)/24 = sin B/47\nsin B = (47 sin 36)/24\nsin B = 1.151079452\nAt this point you have a problem--it is impossible for the sine of an angle to have a value greater than 1.\n\nSo either you copied the problem wrong, or you are being asked to recognize that a triangle cannot be formed with these dimensions.
722996
Drop a perpendicular to side of length 10 from the vertex where the side of length 8 ends.\n\nA right angle triangle is formed,\n\nthe perpendicular is 8*sin 60 = 4sqrt(3)\n\nArea of ||gm is 10*4*sqrt(3) = 69.282 approx.
865033
okay. so. there's a special set of rules for a 30-60-90 triangle.\n\nthe sides are proportional to each other by the ratio \n\nx : v/( 3)x : 2x\n\nwhere the v/ is the square root.\n\nthat means that the greatest angle corresponds to the greatest length.\n\nbecause the longest side is 7.5 meters, that means that is 2x.\n\ndivide by two for x, and that is the 30* length. (because its the smallest angle, therefore is opposite the smallest length)
714757
(sinX + 1) = 0---> sinX = -1--->x = 3pi/2 + (2pi)n, radians, or x = 270 + 360n degrees, where n is any whole number including 0. \n\nThe first step was simple algebra.  In the second step, I just remembered that the sine is how far up on the unit circle that angle points, so -1 means it points straight down, which is 3/4 of a circle.  Adding a whole circle puts it right back where it started from.\n\ncosX + 1/2 = 0---> cosX = -1/2 ---> x = 2pi/3 or 4pi/3 + (2pi)n radians, 120 or 240 + 360n degrees.  \n\nAgain, the first step was simple algebra.  As for the second, my trig teacher made us memorize the sines/cosines of 30, 45, and 60 degrees, so I just know that the cosine of 60 degrees is 1/2.  Go around to the other side of the unit circle and the cosine of 120 degrees is -1/2.  Go down to the bottom of the unit circle, and the cosine of 240 degrees is also -1/2.
534053
It really depends upon your problem.\n\nIf you are asked to find the sine, cosine, and tangent then for any right triangle:\n\nsine = opposite side/hypotenuse  = y/r\n\ncosine = adjacent side/hypotenuse = x/r\n\ntangent = opposite side/adjacent side = y/x\n\nwhere opposite side is the side opposite the angle in consideration.\n\nwhere adjacent is the side closest to the angle.\n\nwhere hypotenuse is the longest segment in the triangle.
780884
As with many problems, it helps to draw a diagram.  I made a triangle, and then put in each of the altitudes, since they are my only known values.  With these in place, we now have 6 principal right triangles, of which we know the value one leg (the altitude).  In each of these triangles, the hypoteneuse is a complete side of the original triangle, and the angle opposite the known leg is also of the original triangle.  Personally, I chose to work with cosines and tnagents, but you could just as easily have used sines (though I think it might add a step or two).  Remember that the cosine of an angle in a right triangle is merely the ratio of the side opposite the angle to the hypoteneuse.  Don't worry that we don't know the value of any of these angles; I went through the entire problem without ever needing it.\n\nSo, since each angle, in its entirety, is in two of our right triangles, they give us two ratios to compare.  That is:\n\nIf the sides of the original triangle are x,y,z\ncos a1 = 20/x = 15/y\ncos a2 = 20/z = 12/y\ncos a3 = 12/x = 15/z\n\nWith a little cross multiplication and rearrangement of terms, we find that:\n12z = 15x = 20y.  \n\n(Your x, y, z may be in a different order depending on how you labeled your triangle, but if you manipulate your triangle a bit, you should be able to get it to match.  This will keep your coefficients identical to mine.  But even if you don't match it up, in the end you will get the same answer.  That's the beauty of math.)\n\nThe tangents are a little trickier, because they involve a fraction of the original side.  This involves the addition of a new variable on each, such that one portion of the line is "a", and the remainder is x - a; and then b and y - b; c and c - z.  The tangents then follow:\n\ntan a1 = 20/(y - b) = 15/a\ntan a2 = 12/(z - c) = 20/b\ntan a3 = 15/(x - a) = 12/c\n\nRearranging each gives us the three equations:\n20a = 15(y - b)\n12b = 20(z - c)\n15c = 12(x - a)\n\nFrom there, you can solve for either a, b, or c.  It's your choice.  I opted to solve for a.  Substituting the third equation into the second, and then that new equation into the first, we get:\n\n20a = 15(y - (20/12)(z - (12/15)(x - a)))\nwhich reduces to\n8a = 4x + 3y - 5z\n\nNow, side a is in a right triangle with hypoteneuse y and second leg 15, meaning that:\na^2 + 15^2 = y^2   or\na = sqrt(y^2 - 225)\n\nUsing this equation, plus the ratios we received from the cosine functions, we solve for y:\n\n8 * sqrt(y^2 - 225) = (16/3)y + 3y - (25/3)\n\nConveniently, the right half is equal to zero, so dividing by 8 and squaring:\n\ny^2 - 225 = 0\ny^2 = 225\ny = 15\n\nEnter this into our ratio:\n\n15x = 20y\n15x = 20*15\nx = 20\n\nand\n\n12z = 20y\n12z = 20*15\nz = 25\n\nSo finally we have the lengths of all three sides, and adding them together, the perimeter is\n15 + 20 + 25 = 60\n\nThis turns out to be a right triangle, but that is merely happenstance.  This process can be used to solve for any triangle with three known altitudes.\n\nHope this helps.
997563
You're looking for the angle that has a sine of 1/2.\nDrawing a triangle representing this is very helpful.  You will see that it is a 30-60-90 triangle.  \nIf you can find it in degrees, then you can convert it to radians by multiplying it by pi/180.\nsine(30) = 1/2\n30 degrees = 30(pi/180) = pi/6
1015200
Let a, b, and c be the lengths of the legs of a triangle opposite angles A, B, and C. Then the law of sines states that\na/(sinA)=b/(sinB)=c/(sinC)=2R,\n\nwhere R is the radius of the circumcircle.\n\nIn other words,\n\nIn any triangle, the ratios of the sides to the sines of the opposite angles are equal.
433960
ok. well if y = sin^-1 (x/a), then that means that \nsin y = (x/a). \nso if you do implicit differentiation, you get \ncos y * dy/dx = 1/a.\n\nso thatz not hard. now if you draw a right angle triangle, and label 1 of the angles y. so sin of y is x/a, but it is also equal to opposite/hypotenuse. so then that meanz the opposite is equal to x and the hypotenuse is equal to a. now if you use pythagoras theorem, the other side (the adjacent side) is sqrt(hyp^2 - opp^2). because \nadj^2 + opp^2 = hyp^2\nso adj = sqrt(hyp^2 - opp^2)\nso adj = sqrt(a^2 - x^2).\n\nso now you need cos of y, which is adj/hyp\nso cos y = sqrt(a^2-x^2) / a.\n\nnow back to the first equation\ncos y * dy/dx = 1/a\nsqrt (a^2 - x^2)/a  * dy/dx = 1/a\nsqrt (a^2-x^2) * dy/dx = 1\nso dy/dx = 1/sqrt(a^2-x^2)\n\nyou can do this with any inverse trig derivative stuff. i hope you understood that. thatz the way i remember it but there might be other easier ways (like just memorize the relations), but there you go. hope that helped. let me know if you need more help