Trig ?   Sin angle is .7071    0pp = 5       H = 7.07    solve right angle for theta   How?
156129
The sine of an angle is equal to the opposite leg over the Hypotenuse,\n\nsin(theta) = opp/ hyp\n\nSo to solve for the angle, theta, take the arc sin of opp/hyp.\narcsin(opp/hyp) = arcsin(.7071) = theta = 45 degrees.\n\narcsin essentially undos what sin does to the angle to allow you to reverse the process.\nArcsin is more commonly seen on calculators as sin^-1
838146
x  =  3pi/2  is one possible solution to the equation.  3pi/2 (or 270° if you're using degrees) is on the boundary between the third and fourth quadrants, so its sine has the same value as sin pi/2 (90°) but the opposite sign; i.e. sin 3pi/2  =  -1 .\n\nNow, (-1)³  -  4(-1)²  -2(-1)   =   -1  -  4(+1)  +2   =   -3\nSo  3pi/2 is a solution.\n\nHowever, it's not the only solution.  You could do a complete revolution into the 7th/8th quadrant (x  =  7pi/2) and you would get the same value for sin x.\nIn general, x  =  (4n + 3)pi/2  will satisfy the equation, where  n  is zero or any whole number, positive or negative.
767880
y = r sin theta, x = r cos theta (by drawing a triangle or otherwise.)\nSubstitute them in:\n2r cos theta + 3r sin theta = 6.\nOr r(2 cos theta + 3 sin theta) = 6.
98672
use the tangent ratio ( trig function ) or inverse tangent of (6/ (43/8)) OR iNV TANGENT (1.116) = 48.14 degrees and to get the other angle, just subtract it from 90 degrees or 41.86 degrees. hope it helps how I got it. go over the trig function or cohsahtoa rule
710047
Since theta is between 90 and 180 degrees, it is in the II quadrant. Tan = opp/adj. The y value (opp) is +3, and the x value (adj) is -4.  By the Pythagorean theorem, hyp^2 = opp^2 + adj^2. So the hypotenuse  is the sqrt of 3^2 +4^2 and is equal to 5. Cos is add/hyp, and is -4/5. So the answer is -4/5.
857524
sec(72.3pi)=1.701301618\n\nfirst change the mode of ur calculator to radians and then press the sequence of keys.\n\nand sec =cos-1 not sin-1\nso if u want sec , find cos first and then its inverse.\ncosec=sin-1\nsec=cos-1 \ncot=tan-1
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.
924113
sin(pi/7)*sin(2pi/7)*sin(3pi/7)\nsin (pi/7) = sin (0.45 radians) =.4350\nsin (2pi/7)=  sin(.88 radians) =.771\nsin(3pi/7)=sin(1.35 radians) = .976\nThus the abswer is .435*.771*.976=.327
925769
sin(θ) ≈ θ\nSo cos (θ) ≈ 1  because cos (0) = 1\nMore accurately cos (θ) ≈ 1 - θ²/2
145134
I will use Q for theta here:\ncosQ = 2 - 2cosQ\nNow we add 2 cosQ to both sides to get rid of it on the right.\nSo we get:\n3cosQ = 2\nDividing both sides by 3 gives:\ncosQ = 2/3\nAnd solving for Q, we take the inverse cosine of both sides:\nQ = cos(-1)(2/3)\nHope this is helpful :-)
608164
I hope that this will help: \nsin(70) = 0.7739,sin(110) = -0.0442,sin(47) = 0.1236,sin(133) = 0.8689, sin(69) = -0.1148,\nsin(111) = -0.8645,cos(70) = 0.6333, cos(110) = -0.9990,cos(47) = -0.9923,cos(133) = 0.4948,\ncos(69) = 0.9933, cos(111) = -0.5025
928486
sin 90 = 1, right?\n\n40/sin A = 70, so\n\nsin A = 40/70 = 4/7\n\nA = arcsin(4/7)
966822
Draw a right angled triangle.  the vertical is 35m and the sloped line, the hypotenuse, is 50m.  We have to find the angle between the string and the ground.\n\n35 is the opposite, 50 is the hyp, so it is a sine problem.  Sin x = opp/hyp = 35/50.\n\nusing the inverse sine function on the calculator, we find that the angle is 44.4 degrees\n\nhope this helps
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.
830888
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.
769161
Its a pain to do this with X and Y components, but there is an easier way....\n\nRange Formula - used when start and end heights are the same\nQ = Theta\nd = ((Vo^2)(sin 2Q))/g\n\n301.5 = ((Vo^2)(sin 2(25)))/9.81\n\nVo = 62.13 m/s\n\nNow that u have initial velocity, find the y component\n\nVo Sin Q\n\n62.13 sin 25 = 26.26 m/s\n\nThen No Time Formula\n\nVf^2 = Vo^2 + 2gd\n\n0 = (26.26^2) + 2(-9.81)d\n\nd = 35.15 m\n\nHope this helps
27656
I used the acronym sohcahtoa to remeber that sin=opp/hyp cos=adjacent/hyp and tan=opp/adj, so you have a angle you want to measure the degree of that angle, using the lengths of the sides of the triangle opposite, adjacent and hypotenuse and then using your calculater you divide the lengths and hit the button sin-1, cos-1 or tan-1 key, also cot=1/tan, hope that helps