Really confused on how to solve this using double angles\n\nHave a whole bunch of others and the concept is still kinda fuzzy to me
208067
sin2x = 2sinxcosx\nsin2xsinx = cosx\n(2sinxcosx)sinx = cosx\n2(sinx)^2cosx - cosx = 0\n2(1-(cosx)^2)cosx - cosx = 0  Factor out cosx\ncosx[2(1-(cosx)^2)-1] = 0  Distribute the 2\ncosx[2 - 2(cosx)^2 - 1] = 0  Combine 2 and -1\ncosx(1-2(cosx)^2) = 0\nso cosx = 0 or 1-2(cosx)^2 = 0\ncosx = 0 when x = 90 degrees and 270 degrees\n1-2(cosx)^2 = 0 when\n2(cosx)^2 = 1  Divide both sides by 2\n(cosx)^2 = 1/2 Take the square root of each side\ncosx = sqrt(1/2) or -sqrt(1/2)\ncosx = sqrt(1/2) when x = 45 or 315 degrees\ncosx = -sqrt(1/2) when x = 135 or 225 degrees.\nThus, the answers are\n45, 90, 135, 225, 270, and 315 degrees.
463762
1. So, you have two sides and included angle.  Find the 3rd side and either use Pythagorean theorem to calculate the altitude or use the tangent of 1/2 the angle and 1/2 the base.\n\n2a.  You have two sides and the included angle\n2b.  You have two angles and the included side
75852
I have no idea what possessed me to answer this question so thoroughly. I think it was the sheer rudeness of the previous "answerer" Emma. The problem is not "hard" but you can't assume anything you're not given (i.e. you can't just assume the other segment is parallel) so you have to proceed carefully. \n\nNone of the previous answers is sufficient to truly prove what you asked. For instance, you cannot just say "as constructed, they are right angles." It's true, but just saying so doesn't make it true. It's not a convincing argument, it's a proof.\n\nIn the sources, find two graphics. The first is the parallelogram and the second is a *rigorous* proof of the answer.\n\nThanks for the question. It was anything but trivial if you're actually constructing a true proof. (And please forgive me if I am off by a letter here or there...it's 1 a.m. my time and I tried to stampede through the proof before you got more flames instead of helpful solutions.)
397826
Let the length of side "b" be x ft.\nTherefore side a=2x and side c=2x+3\nAccording to the problem,\nx+2x+2x+3=93\n=>5x+3=93\n=>5x=93-3=90\n=>x=90/5=18\nTherefore.side a=2X18=36 and side c=36+3=39\nAnswer a=36ft. b=18 ft and c=39 ft
612432
You have that r=2sqrt(2). Now factor this out of your original number.\n\nSo you have\nz=r[1/sqrt(2) - 1/sqrt(2) i]\n\nNow you have to find an angle with cos(theta)=1/sqrt(2) and sin(theta)=-1/sqrt(2).\n\nWhat you did wrong: 1) it is cos and sin of some angle. You have to find this angle.\n\n2) You didn't factor out the r correctly.
259401
The angles of the triangle must add to 180, and since it is isosceles, then the two base angles are the same, so the other base angle is 62 and the last angle is 180 - 2(62) = 56.  If a is the length of the base and b is the lengths of the other two sides, then.\n\na/sin56 = b/sin62\nso a = b*sin56/sin62.\n\nThen use the pythagorean theorem.  If we take a line and cut the triangle in half,  then the two halves are right triangle with hypotenuse b and one leg a.  If we let the other leg be c, then that turns out to be the height of the triangle.  So\n\nc = sqrt(b^2-a^2)\n\nThen the area is ac/2 (one-half base times height).\nnow a + 2b = 42.\nso b(2+sin56/sin62) = 42\nso b = 42/(2+sin56/sin62).\n\nThen solve for a.
497877
So give the first one x+6, the second one x, and the third one 3(x+x+6)-4\nnow the second step, x+(x+6)+(6x+14)=180\nnow, 8x+20=180\nthen, 8x=180-20\n         8x=160\n         x=160/8\n         x=20.\ntherefore the first angle =26, second, 20 and the third 134
645627
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
172941
you may be dealing with "right" triangles, but you don't identify the angle that is 90, so your question is bout impossible to answer. this type of question usually involves  \na squared + b squared = c squared (a^2 + b^2 = c^2) where a and b are the 'legs' of the triangle in question. (the two 'legs' of a right triangle make the 90 degree corner) :)
351973
First add the fractions on the left handside getting: b+a/ab=1/c\nthen cross multiply:\ncb+ac/ab=ab/ab then cancel out the a's and b's getting\n2c=1. divide by 2 getting c=1/2. hope that helps!!!\n\nsecond question is \nsubract R from both sides getting: I-R=nE.\nThen divide both side by E getting I-R/E=n and your done!!!
489638
This is a neat problem, and I am sure that there are many ways to solve it. I will be interested in seeing if the different ways to solve it end up producing answers that are at all similar.\n\nHere is what I did. I'll start with part (b), since I found it to be easier.\n\nIf x > 7, then the largest angle is the angle opposite the side having length x. This is the included angle of the sides having length 4 and 7.\n\nUse the law of cosines to relate the lengths of the two shorter sides and their included angle to the side having length x.\n\nc^2 = a^2 + b^2 - 2abcos(theta)\nx^2 = 4^2 + 7^2 - 2(4)(7)cos(theta)\n\nx^2 = 16 + 49 - 56cos(theta)\nx^2 - 65 = -56cos(theta)\ncos(theta) = (65 - x^2)/56\n\nOr, theta = arccos[(65 - x^2)/56]\n\nIn this case, x must be greater than 7 and less than 11.\n\nIf x is less than 7, then the largest angle is the angle opposite the side having length 7. This is the included angle of the sides having lengths 4 and x.\n\nMy first idea didn't work, so let's try something that could be even uglier :D\n\nCall the included angle of the sides of length 4 and 7 alpha.\n\nUsing the law of cosines, \n\nc^2 = a^2 + b^2 - 2abcos(alpha)\nx^2 = 4^2 + 7^2 - 2(4)(7)cos(alpha)\n\nx^2 = 16 + 49 - 56cos(alpha)\nx^2 - 65 = -56cos(alpha)\ncos(alpha) = (65 - x^2)/56\n\nLet's derive sin(alpha) from cos(alpha).\n\nSince sin^2(alpha) + cos^2(alpha) = 1,\n\nsin(alpha) = sqrt[1 - cos^2(alpha)]\n=sqrt{1 - [(65 - x^2)/56]^2}\n\nAt this point, you're probably wondering what this has to do with theta.\n\nNow that we know sin(alpha), we can use the law of sines on two of the sides of the triangle, and the angles opposite them.\n\na/sin(A) = b/sin(B) = c/sin(C)\n\nIn this case, theta is still the angle opposite the side having length 7. alpha is the angle opposite the side having length x.\n\nSo:\n\nx/sin(alpha) = 7/sin(theta)\n\nor\n\nsin(theta) = 7sin(alpha)/x\n\nWe know sin(alpha):\n\nsin(theta) = 7*sqrt{1 - [(65 - x^2)/56]^2}/x\n\nSo theta equals arcsin [7*sqrt{1 - [(65 - x^2)/56]^2}/x]\n\nThat's not very pretty.\n\nIn this case, x must be greater than 3 and less than 7.
160059
x = 75 degrees\n\nSince all the angles must add to 360 degrees\n\n4x + 15 + 45 = 360\n\n4x = 300\n\nx = 75
589622
It's too cumbersome to rewrite it all here, but check out http://www.euclideanspace.com/maths/algebra/vectors/angleBetween/index.htm\n\nThat's 2D, but the formula is the same for 3D.
33882
How about this? Straight multiplying gives you the following:\n\n18b^3/b^2\n\nb^3 = bxbxb\nb^2 = bxb\n\nYou have 3 bs on top and 2 on bottom. Whenever you have the same number on top and bottom, it cancels out. Since there are 2 bs on bottom, 2 of the ones on the top cancel out as well. This leaves you with 18b.\n\nDid that help at all?
375445
F(Y)=117.6N becasue W=mg\n\nm=12kg\ng=9.8m/s/s\n12*9.8=117.6N\n\nbut i disagree with the second part of f(X)=55N. though i could be wrong i guess\n\nstart with some trig.\n\nsin^-1(2.4/2.8)=58.99 deg.\n\nyou know your Y component is 117.6(opposite side in a trangle from the angle just found) and you are looking for your X component(adjacent side to your angle) and that tan=opp/adj\n\nso:\n117.6/tan(58.99)=70.68N\n\ni'm not saying your teacher is wrong, but if he is just getting the numbers out of the answer book then the book could be wrong. i see this all the time even in my college books.
10918
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
321442
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck