Trigonometry.. i already know the answer but do not know how to get it..?
822564
It is best that you draw a triangle.\n\nLet x be the distance between fire and tower A.\nLet y be the distance between fire and tower B.\n\n(sin 34)/x=(sin(180-34-14))/30\n(sin 14)/y=(sin(180-34-14))/30\n\nx=22.574, y=9.766\n\n0.5*x*y*(sin 132)=0.5*d*30 (area of triangle)\nplug in the numbers and d=5.46km\n\nThere might be simpler methods, but this is the first thing to comes to my mind.
756840
cosx=tanx\ncosx=sinx/cosx\ncos^2(x)=sinx\n1-sin^2(x)=sinx    {pythagoras}\nsin^2(x)+sinx-1=0\nuse quadratic formula\nsinx=(-1+or-sqrt(5))/2\n=0.6180339888\nor -1.6180339888\nsine has to lie between +/-1\nsinx= 0.6180339888\nx=arcsin(0.6180339888)\n  =38.17270763\n+ve value for sine lies in 1st and 2nd\nquadrants \ntherefore,x=38.17270763 degrees or\n141.8272924 degrees\n\nthese two values were checked in the \noriginal equation and were found \nto be correct\n\ni hope that this helps
1096282
cos^2(A) / ( 1 - sinA )\n=( 1 - sin^2(A) ) / ( 1 - sinA )\n=( 1 + sinA )( 1 - sinA ) / ( 1 - sinA )\n=1 + sinA
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.
542056
Start by factoring:\n\n(cos^2(x))(sin^2(x))-(cos^2(x)) = (cos^2(x))(sin^2(x) - 1)\n= -(cos^2(x))(1 - sin^2(x))\n= -(cos^2(x))(cos^2(x))\n= -cos^4(x)
20199
The basic idea behind trigonometry is that once you know the angles of a triangle, you'll know the ratios of the sides of the triangle.  And so if you know one side, you can use trig to find out the two other sides.\n\nSine, cosine, and tangent deal with right triangles.  They tell you the ratios of different pairs of sides in the right triangle.\n\nHere's a brief intro:\nhttp://opera.answers.com/trigonometry
718931
The only possible answer for your fixed number would be zero.  So you just have to do some inverse trig functions to get the value for x.  \n\nSolved:\nLet x=1.0038848219 [radians]\nTan(1.0038848219)=1.570796327=pi/2\nCos(pi/2)=0\nSin(0)=0
748469
The first thing I would do is rewrite both a and s in terms of h.\n\nThen you use the following rules to find the derivative.\n\n1.  Use the quotient rule to find the derivative of  sin(a)/s^2.  If you don't know the quotient rule -- look it up and memorize it.\n\n2.  The derivative of the sin is the cosin -- use the chain rule to get cos(a)*da/dh\n\n3.  The derivative of the bottom is  2*s * ds/dh (again using the chain rule).\n\nOnce you get the derivative, solve for zero.\n\nGood luck.
835859
First figure out whether to use sin or cos, I decided on cos because we look for the max and min.\n\nNext find the amplitude which is 6. a = amplitude/2 so a*cos(x).\n\nThird find the midpoint of the curve which is (max+min)/2 = m. a*cos(x) + m.\n\nThe fourth step is to get the max or min at the x cord. that you want. With cos you subtract max from x. a*cos(x-max) + m.\n\nFinally you find your wave length, which is simply (max-min)/pi = wl for cos. a*cos((x-max)/wl) + m\n\nso you go from (-2,1) to (10,7)\n\na = 3\nm = 4\nmax = 10\nwl = 12/pi to give us\n\n3*cos ((x-10)/(12/pi)) + 4
86738
The angles all add up to 180 degrees.\n\nIf you know the lengths of all the sides you can use the "Law of Tangents" to find all the angles.  This is advanced geometry and Trigonometry, so rather than type, I am including a link to a pretty good explanation.\n\nhttp://answers.google.com/answers/threadview?id=38131
949562
Use csx =1/sin and  1-cos^2=sin^2  and cot=cos/sin\n Ready...Go\ncscx - (cosx)(cotx) =  1/sin - cos^2/sin = (1-cos^2)/sin=sin^2/sin =sin\n       ole, ole , infree
1080980
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.
623818
We can find angle B by subtracting the others from 180:\n\nB = 180 - 32 - 44\n= 104\n\nThe length of the sides can be found using the law of sines:\n\n(sin A)/a = (sin B)/b = (sin C)/c\n\nSo,\n\n(sin B)/b = (sin C)/c\n(sin 104)/b = (sin 44)/20\nb = 20(sin 104)/(sin 44)\n~ 27.94\n\n(sin A)/a = (sin C)/c\n(sin 32)/a = (sin 44)/20\na = 20(sin 32)/(sin 44)\n~ 15.26
191873
First of all, the best way to learn trigonometry is to learn it from the beginning, which means take your first textbook which has trigonometry and study from the beginning.\nhttp://mathworld.wolfram.com/Trigonometry.html\nhttp://www.du.edu/~jcalvert/math/trig.htm#Trig\nhttp://www.syvum.com/math/trigonometry.html
925769
sin(θ) ≈ θ\nSo cos (θ) ≈ 1  because cos (0) = 1\nMore accurately cos (θ) ≈ 1 - θ²/2
1154463
cos(x)*cos(x) = cos^2(x)= 1-sin^2(x)\ntherefore the problem is now down to 1-sin(x)-Sin^2(x)\nthen integrate each one separately. \nfor inter of 1 is just x\nintegral of  -sin(x) is cos(x)\nand for sin^2 use the half angle identity \nsin^2(x) = .5*(1-cos(2X))\nand therefore -sin^2(x) = -.5*(1-cos(2X))\nintergral of which is -.5x+.25sin(2x)\n\nand the final answer add them and you'll get \n.5x + cos(x) + .25sin(2x)\nor \n.5x + cos(x) + .5sin(x)*cos(x)\nbecause sin(2x) = 2sin(x)*cos(x)
104581
It takes practice!  And its like a puzzle.  Just use the main trig identities as reference and a good math tutor.  Unfortunately, this place isn't the best place to ask that kind of question because we are limited by text explainations and trig identities need a lot more than that!