help on a trigonometry word pr0blem !!! yikes !!! i`m stuck.?
602899
I always find it easier if you draw it first, which this stupid bos wont let me do.\n\nYou know one angle (theta) and one side of the triangle on parts a and b.  Use Tan Theta = x / 30 to find how far the truck has traveled in one second, and convert to MPH.\n\nFor C, you have the distance traveled in MPH and one side (30 ft).  Convert 55 MPH to ft / sec and find the angle Theta.  Tan Theta = 55 MPH / 30 ft\n\nYou are going to have to figure the answers out for yourself.  Good luck
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.
507417
r/SinR = s/SinS\nSinS = 24*(Sin(38)/20) = 0.73879....\nAngle S = Sin^-1 (0.73879.... = 47.628 degrees\nAngle T = (180-(38+47.628))94.372\nUse same procedure to find side t.
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
1102373
i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
703752
simply subtract 7 from both sides. This leave you with\n\nx<_3
892068
Well, you got me stumped!\n\n-------------------------------------\n\nI started working it out and got pretty far.\n\nThe above given information might be incorrect.  Maybe?  I don't know.\n\nI tried it several times.  I even double and triple checked my work!\n\n---------------------------------------------\n\nThis is as far as I got and then I got stumped:\n(D-E) (B+C) = (10J + AFGH) / A\n\n------------------------------------------------------\n\nGreat question!\n\nI look forward to the continuing discussion.\n\nI love the way you worded your question.  Right on!!!\n\nKeep those challenge questions coming!  ;o}
1159205
pair up the  opposite sines, for example sin(10) and sin(350) and add them:\nsin(10) + sin(350) = 0\nsin(20) + sin(340) = 0\n...\nuntil you only have sin(180) and sin(360) which both equal 0\n\nPS. I assumed you were using degrees (sexagesimal). Won't work in radians (RAD) or metric (decimal) angles.
895625
This may be unexpected, but a good site with the formulas you need.\n\nhttp://en.wikipedia.org/wiki/Trig
1046178
I think...\n\n# means theta\n\nsin(90-#)*cos(90-#)=tan#/sec^2#\ncos#sin#                =(sin#/cos#)/(1/cos^2#)\ncos#sin#                =(sin#cos^2#)/cos#\ncos#sin#                =cos#sin#\n\nUsing: sin(90-#)=cos#\n           cos(90-#)=sin#\n           tan#= (sin#/cos#)\n           sec^2#=(1/cos^2#)\n\nThat work?
40284
(1) Subtitute the damn values for what you have.\n\n(2) Sin theta=opp/hyps,\n    Tan   "  =opp/adj,\n    Cos   "  =adj/hyps.\n\nNow usually a blind man follows his stick now you might know where you are at the moment - get it???
757793
1. Using exact values, verify that Cos 60°= 1-2 sin² 30°\nRHS=1-sin^2 30*\n=1-(1/2)^2\n=1-1/4\n=3/4\nLHS=[(rt3)/2]^2\n=3/4\nLHS=RHS\n\n2. Using exact values, find the value of:\n\na. 12 sin 30° -6 tan 45° + sec45°/52 \n12(1/2)-6(1)+(rt2)/52\n\nb. sin 2A + tan 3A/2 – cos A + sec (A+15°), When A= 30°\n=sin60*+tan45*-cos30*+sec45*\n=rt3/2+1-rt3/2+rt2\n=1+rt2\n\n3. Write the other functions of x if cos x= 3/5..(show work)\nthe right triangle will bea 3,4,5 triangle using the pythagorean triplet\nsinx=4/5\ntanx=4/4\ncotx=3/4\nsecx=5/3\ncosecx=5/4 \n.\n4. In a right triangle sin A =3/4 and c =17. Find a and b. Be sure to draw a right triangle and letter it properly.\nc=17\nletthe oppositeside be 3x and the hypotenuse be 4x\nadjacent side=rt(16x^2-9x^2)\n=(rt7)x\n4x=17\nx=17/4\nrt7 x=17/4\nsquaring\n7x^2=289/16\nx^2=(289/16)*(1/7)\nx=rt[9289/16)(1/7)\n5. The vertex angle of an isosceles triangle is 57° 24’ and each of its equal sides is 375.5 \nft. long.Find the altitude of the triangle.\naltitude/side=sin of base angle\nbase angle=(180-57*24')/2\naltitude=375.5 times sin[(180*-57*24')/2]\n6. Solve the following right triangles:\na. a=117 ft.--B=16.35ft.\ncapital letters are used only for angles\nso B=16.35' is wrong\nb. B=8°29, --a=32.8 ft. \nare you taking <c as the right angle\nif so <a=(90*-8*29')\nuse a/sinA=b/sinB=c/sinC\n\n7. A man standing on top of a building sees an automobile on a street below him.The angle of depression is 36 degrees 22. If eye is 550 ft. above the level of the automobile ,how far is the automobile from him?\nlet the horizontal distance be x\nx/550=cot36*22'\nx=550cot 36*22'\n=>550tan(90-36*22')\n\n8. When the suns angle of elevation is 30 degrees. The shadow of a post is 6 ft.longer than when the angle is 45 degrees . Find the height of the post.(show work)\nlet the height of the post be x\nthe length of the shadow at 30*=xtan30*\nthe length of the shadow at 45*=x tan 45*\ngiven the difference x(tan45*-tan30*)=6\nsolve\n\n9. In a right triangle, given a= 32.8 ft. and A=28° 30’ 15’’,solve for B and C\ntaking <C=90* by convention\n<B=90*-28*30'15" \nnow use a/sinA=b/sinB=c/sinC
675026
Just remember SOH CAH TOA\n\nSin: opposite/hypotenuse\n\nCos: adjacent/hypotenuse\n\nTan: opposite/adjacent
716883
Trying for wow #2---\n\nR = .15 m\nω = 2π/T = 13.963 rad/sec\nA little vector diagram shows that tanθ = rω²m/mg = rω²/g\nbut r = Rsinθ, so tanθ = sinθ/cosθ = Rsinθω²/g  from which\n\n1/cosθ = secθ = Rω²/g = .15*13.963²/9.8 = 2.984 → θ = 70.42°\n\nYou get to do b)\nGood luck..
22859
Subract the two terms on the left giving:\n-2t < 16\n\nAdd 2t to both sides giving\n0 < 16 + 2t\n\nSubtract 16 from both sides\n\n-16 < 2t\n\nDivide both sides by 2\n\n-8 < t
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-
791163
2r*2r*(p/4)\n=4r^2(p/4)\n=pr^2