i need help with a basic trigonometry question?
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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.
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
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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.
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A = base * height\n\nbase = 8\n\nslant height = 6\nto get the actuall height, we use trig\n\nthink of 6 as the hypotenuse\n\nsin 30 = height / slant height\n\nor height = sin30 * 6\n\nheight = 3\n\ntherefore area = 8 * 3 = 24
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(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???
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Trigonometry is hard to do in you head, so start out by making a drawing. You have two buildings A (tall) and B (short) - Draw them far apart, not like this:\n\nA         B\n|         \n|         \n|    \n|         |\n|         |\n|         |\n|         |\n\nNow draw one line from the top of A to the top of B - this is the angle of depression of the top and it is 46 degrees. That is, if you drew a straight, horizontal line across the top of building A, the angle between that line and the line joining the top of A to the top of B is 46 degress.\n\nSimilarly, draw a line from the top of A to the *bottom* of B - this is the angle of depression of the bottom and it is 77 degrees. That is, the angle between the horizontal line across the top of A, and the line between top of A and bottom of B is 77 degrees.\n\nNow can you find out the angle between the line from A-top to B-top, and the line from A-top to B-bottom? (It's very important for you to draw this!) Let's put this value aside for a minute. \n\nLooking at your drawing, can you see that you know the value for *part* of the angle between the horizontal line across A's top and length of building A? What is that part you know? Now, see how the line across A's top and length of A make a right angle? So how many degrees in a right angle? So if you know the total of that angle, and part of that angle, how do you get the other part? **Notice that this part we're trying to get is actually the angle between the line A-top to *B*-bottom, and the line A-top to *A*-bottom. \n\nYou should also notice that now we have two triangles:\nTriangle 1: A-top to B-bottom to A-bottom. Solve this triangle for line A-bottom to B-bottom for distance between the two buildings.\nTriangle 2: A-top to B-top to B-bottom. Solve this triangle for line B-top to B-bottom for height of building. \n\nDon't forget, we have two important angles that will help solve these two triangles!
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Depends\n\nlim θ→π/2- (ie from below) tanθ = +∞\n\nlim θ→π/2+ (ie from above) tanθ = -∞ \n\nThis is because in the 1st quadrant tanθ >0 and in the second quadrant tanθ <0
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All you need to do is enter these on a scientific or graphing calculator.  Make sure it is in degree mode.  For the second part use the 2nd or shift button to get sin^-1 etc
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It's quite dificult to explain trigonometry this way, so it;s better for you to have an electronic teacher, this teacher is at your service day or night, sun or rain and  you could find it\n\nhttp://aleph0.clarku.edu/~djoyce/java/trig/\n\nhttp://www.sosmath.com/trig/trig.html\n\nhttp://fergusmurray.members.beeb.net/trig.htm\n\nhttp://mathworld.wolfram.com/Trigonometry.html\n\ngood luck
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I don't know any good math websites but I do know how to solve this.  b would be the side adjacent to the 40 degree angle and a is the side opposite it.\n\nTangent = opposite leg/adjacent leg so\ntan 40 = 15/b\n\nb tan 40 = 15\n\nb = 15 / tan 40
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Sin A = -12/13\nCos A = 5/13\n\nFirst, draw a x-axis and y-axix. The angle starts on the positive x-axis going counter-clockwise ends at the line that contains (0,0) and (5, -12). That line is in the fourth quadrant so angle A is bigger than 270 degrees but less than 360. Angle A's sin and cos are the same as the angle that starts on the positive x-axis and goes CLOCKWISE to the that line. Let that angle be angle B.\n\nDraw a line from (5, -12) up vertically to (5, 0) and a line from (0, 0) to (5, -12). You get a triangle with Angle B. It's opposite side is 12 units long, adjacent is 5, and hypotenuse is 13. \n\nTherefore the sin is 12/13 and the cos is 5/13. Now we look at the position of angle B. Because angle B is in the fourth quadrant, its sin must be negative and its cos positive.\n\nWe get sin B= -12/13 and cos B= 5/13. And since angle A's sin and cos are the same as angle B's...\n\n:)
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i took  trig and dont remember much but it should go like this\n(a)\n3x^2+12x=-12\n9x+12x=-12\n21x=-12\nx=-12/21\nx= -0.57\n\n(b)\n7x^2=28\n49x=28\nx=28/49\nx=0.57\n\n\nc im clueless\n\ni was wrong i just fixed it
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Trigonometry.  The good old days when I still loved math... ahhh!
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QUESTION 1:\n\nAsk yourself: WHICH TRIG FUNCTION IS OPPOSITE OVER ADJACENT?  IT IS THE TANGENT FUNCTION, RIGHT?\n\nThe bottom of the tree is 25 feet away from a point that measures 30 degrees from ground level to the top of the tree.\n\nAnswer: tan30degrees = x/25feet...choice 1.\n\nNotice that they are NOT asking you to solve for x.  Get it?  \n\nQUESTION 2:\n\nYou need to write the given equation 3x - 2y = 12 in the y =mx + b\nform.\n\nLet's do that right now.\n\nWe have:\n\n3x - 2y = 12 \n\nI will subtract -3x from both sides of the equation.\n\nWe get:\n\n-2y = -3x + 12...See what I did?  \n\nNext:\n\nDivide BOTH sides of the equation by -2 to find the value of y.\n\nWe get this:\n\ny = (-3x + 12)/-2\n\nWe can reduce the equation to this:\n\ny = 3/2 (x) - 6\n\nSlope is 3/2 and the y-intercept is -6.\n\nFinal answer: choice 1\n\nGuido
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V = √(550² + 80²)\n≈ 556 mph\nDirection tanθ = 550/80 in 3rd quadrant\nθ ≈  81° 43' 27" from West\n= 188° 16' 33" T (clockwise from N)
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im taking Trig. try finding a homework helper web page. or go to the book that ur learning from web page because the majority of the times they have practice help on there.
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x = a*sin(ang)\ny = b*cos(ang)\n\nang goes from 0 to 360 degress, a and b determine how oval it is (if a=b it is a circle)