How do I find the Sine/Cosine/Tangent of 105 degrees?
538675
For Sine 105- Perpendicular/Hypotenuse\n\nFor Cosine 105- Base/Hypotenuse\n\nFor Tangent 105- Perpendicular/Base\n\n\n* Perpindicular, base and hypotenuse are the three sides of the triangle.
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
333253
You can solve this using x, y, and r (the lengths of the 3 sides of the triangle made by the angle).  The trigonometric functions are originally defined as:\n\nsin = y/r\ncos = x/r\ntan = y/x\nsec = r/x\ncsc = r/y\ncot = x/y\n\nYou need to find x, y, and r for angles a and b.  Do this using the given sin and cot values and the equation\n\nx^2 + y^2 = r^2\n\nSo for a:\n\nsin a = 3/5\ny=3, r=5\n\nx^2 + 9 = 25\nx^2 = 16\nx=-4\n(this is negative because the angle is in the 2nd quadrant)\n\nFor b:\ncot b = -12/5\nx=-12, y=5\n(x is the negative one because this is in the 2nd quadrant)\n\n144 + 25 = r^2\n169 = r^2\nr=13\n\nNow you need to turn tan (a-b) into a form where you can plug in your x, y, and r values, and then plug and solve.  Remember that tan = y/x.\n\ntan (a-b) = (tan a + tan b) / (1 + tan a * tan b)\n=(-3/4 + -5/12) / (1 + (-3/4)(-5/12)\n=(-14/12)/(1 + 15/48)\n=(-7/6)/(63/48)\n=-8/9\n\nTherefore, your final answer is -8/9.\n\nHope this helps!
886349
SOHCAHTOA \nSOH =sine=Opposite /Hypotenuse\nCAH cos=adjacent/hypotenuse\nTOA tan=opposite/adjacent\nIn finding the adjacent,we use CAH\ncos19=adjacent/13.8\nadjacent=cos19x13.8\nadjacent=13.048
730760
θ = α + β  where α = tan^-1 (1/3) and β = tan^-1 (1/2)\nie tanα = 1/3 and tanβ = 1/2\n\ntanθ = tan(α + β)\n= (tanα + tanβ)/(1 - tanαtanβ)\n= (1/3 + 1/2)/(1 - 1/3 * 1/2)\n= (5/6)/(5/6)\n= 1\nSo θ = ....., π/4, 5π/4, .....\nSo cosθ = ±1/√2 = ±√2/2
459338
Go to http://mathworld.wolfram.com/Cosine.html
831941
You're supposed to print a table, right? The question is poorly written, so it's not clear, but I'll assume that's the goal.\n\nWell, it sounds like you can write your loop, so shouldn't you just print out the sine, cosine, and tangent of the angle in the body of the loop?  It's not clear what the stumbling block is.  You realize that there are built in functions that compute sine, cosine, etc.?
861187
0.4*(100kg*g+F*sin(x)) = F*cos(x). ignoring any rotational component
1150095
its sooo easy  \nsinA-sinB = 2 sin[(A-B)/2] * cos[(A-B)/2]\ncosA - cosB = 2 cos[(A-B)/2] * cos[(A-B)/2]\nthus LHS = \n(2 sin[(A-B)/2] * cos[(A-B)/2]) / (2 cos[(A-B)/2] * cos[(A-B)/2])\n= sin[(A-B)/2] / cos[(A-B)/2]\n= tan[(A-B)/2 \n= tan (A/2-B/2)\n=RHS \n\nhence proved
419071
You can cheat and use a calculator to get the values, but I don't think this is what your teacher is looking for. \n\nFirst, use the relationship sin^2(x)+cos^2(x)=1 to find |sin(a)| and |cos(b)|. Pay close attention to the ranges of a and b to get the sign right-- sin(a) and cos(b) should both be negative. Doing this you should find:\n\nsin(a) = -12/13 \ncos(b) = (-2 sqrt(2))/3\n\nHere are two useful identities:\n\ncos(a-b) = cos(a)cos(b) + sin(a)sin(b)\ncos(2b) = 1 - 2sin^2(b)\n\nSince you now know cos(a),cos(b),sin(a) and sin(b), you can find that:\n\ncos(a-b)=(12-10sqrt(2))/39\ncos(2b)=7/9
826563
multiply the tangent of the given angle by the known side as long as it's not the sloping side (i.e. the hypotenuse). You will need a scientific calculator to do this. I am assuming of course that you are dealing with a right triangle. This will give you the other side of the triangle. Example:\n\nlets say you have side A of a right triangle and an angle of 30 degrees and A = 10. Take your scientific calculator press 30 and then the tan button depending on the type of caculator you have. Now multiply that result by 10 and you will get side B.\n\nSide C is found by A^2 + B^2 = C^2. take the square root for your answer.\n\nNow if the given angle is opposite the given side. Just calculate the angle of the given side by adding the given angle to 90 degrees and subtract the result from 180, and then multiply the tangent of the new new angle by the given side.\n\nTheoretically you can check the answer by multiplying the new side by the tangent of the new angle. You should in theory get the given side. But my trig is so rusty I may have forgotten more than  you will ever know. good luck.
1053261
70cm hieght of the topstep\n4 degree angle off the ground\n\na) 70/(sin 4*) = 1003 cm or 10.03 m\n\nb) arcsin times 70/900 = 4.46* \n\nyes it will meet wheel chair code
349408
The answer is B) 1.122\n\nFor the different trigo. functions:\n\nLet:\na = side opposite angle n\nb = side adjacent angle n\nc = hypotenuse of the right triangle\n\nTherefore:\n\ncos n = adj./hyp. = b/c\nsin n = opp./hyp. = a/c\ntan n = opp./adj. = a/b\nsec n = hyp./adj. = c/b\ncsc n = hyp./opp. = c/a\ncot n = adj./opp. = b/a\n\nYou will notice the following:\n\n1) the secant is the reciprocal of cosine.\n2) the cosecant is the reciprocal of sine.\n3) the cotangent is the reciprocal of tangent.\n\nTherefore: \n\ncsc n = 1/sin n = 1/0.891 = 1.122\n\nFor the remaining trigo. functions, solve for n first:\n\nn = arcsin 0.891 = 62.99917663˚ or 62˚59'57.04"\n\nTherefore:\n\n1) cos 62˚59'57.04" = 0.454\n2) tan 62˚59'57.04" = 1.963\n3) sec 62˚59'57.04" = 1/cos 62˚59'57.04" = 2.203\n4) cot 62˚59'57.04" = 1/tan 62˚59'57.04" = 0.510\n\nI hope that the solution was clear enough...
1096282
cos^2(A) / ( 1 - sinA )\n=( 1 - sin^2(A) ) / ( 1 - sinA )\n=( 1 + sinA )( 1 - sinA ) / ( 1 - sinA )\n=1 + sinA
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
882948
Just make the following substitution:\nr=sqrt(x^2+y^2)\ntheta=arctan(y/x)\nand then simplify.  Bring everything over to one side and you get your answer.\n\nIf tan(theta)=y/x, then sec(theta)=sqrt(x^2+y^2)/x.\n\nThe problem is a little ugly, but it is doable and then at the end everything ends up canceling nicely.
982428
LHS\n= (sin^2B-tan^2B)/(1-sec^2B)\n= (sin^2B-tan^2B)/(-tan^2B)\n= (sin^2B/-tan^2B) + (-tan^2B/-tan^2B)\n[splitting the numerator up]\n= (-cos^2B) + 1\n[cosx= sinx/tanx]\n= 1 - cos^2B\n= sin^2B\n= RHS