plane trigonometry help?
824139
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
976254
The strategy is as follows:\n\n1.  Find two vectors in the plane.  One can be found using the line.  Another can be found using the point A and any point you like on the line.  \n\n(If (x,y,z) and (u,v,w) are two points in the plane, then the vector (x-u)i + (y-v)j + (z-w)k lies in the plane; or more accurately, this vector is parallel to the plane.)\n\n2. Find the cross product of the two vectors in part 1.  Call this new vector N (for normal).\n\n3. The equation of the plane is m(x-a) + n(y-b) + p(z-c) = 0, where N = mi + nj + pk and where A = (a,b,c).\n\n(Why does step 3 work?  A point B = (x,y,z) lies in the plane if and only if the vector AB is orthogonal to N.  In other words, their dot product is zero.  If you write this out in coordinates, you get exactly the condition that m(x-a) + n(y-b) + p(z-c) = 0.)
791994
First distance: Pythagorean Theorem is a squared + b squared = c squared.  So (2*2) + (3*3) = ???  Then find the square root of ??? \n\nThen the bearing: We first need to find the measure of angle A.  Since we know the length of all 3 sides, any of the SohCahToa formulas work.  Tangent is easiest here.  tan(A) = opposite/adjacent (length of opposite leg divided by length of adjacent leg).  tan(A) = 3/2.  Use the tan-1 key (probably will require shift key).  \n\nWe know the first leg of the trip is south which represents 180 degrees.  So add the found angle to 180 degrees.  (Draw it out, it should make sense.)
1093073
A = 45 degrees or (pi / 4) radians\n\n1/sin^2A-1/cos^2A-1/tan^2A-1/cot^2A-1/sec^2A-1/csc^2A = -3\nIf i understand the way you wrote the problem above, it's 1/ each term, subtract each other right?\n1/ ((sinA)^2) - 1/ ((cosA)^2) - 1/ ((tanA)^2) - 1/ ((cotA)^2) - 1/ ((secA)^2) - 1/ ((cscA)^2) = -3\nI'm going to drop the A part, i assume you know that it will still be officially added in the final version\n\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cot^2)] + [-(tan^2)] + [-(cos^2)] +  [-(sin^2)] = -3\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cos^2) / (sin^2)] + [-(sin^2) / (cos^2)] + [-(cos^2) - (sin^2)] = -3\ncombine denominators\n[( 1 - (cos^2) ) / (sin^2)] + [( (-1) - (sin^2) ) / (cos^2)] + (-1) [(cos^2) + (sin^2)] = -3\n[(sin^2) / (sin^2)] + (-1)[( 1 + (sin^2) ) / (cos^2)] + (-1) [ 1 ] = -3\nTrig identity [(cos^2) + (sin^2)] = 1\n[ 1 ] + (-1)[( 1 + (sin^2) ) / (cos^2)] + [ -1 ] = -3\n(-1)[( 1 + (sin^2) ) / (cos^2)] = -3\n[( 1 + (sin^2) ) / (cos^2)] = +3\n1 + (sin^2) = 3 * (cos^2)\nTrig identity 1 = [(sin^2) + (cos^2)] =\n[(sin^2) + (cos^2)] + (sin^2) = 3 * (cos^2)\n2 * (sin^2) = 2 * (cos^2)\nsin^2 = cos^2\nsinA = cosA\nA= 45 degrees of (pi/4) radians
895625
This may be unexpected, but a good site with the formulas you need.\n\nhttp://en.wikipedia.org/wiki/Trig
932480
sin(A+B)sin(A-B)/(cos²Acos²B)\n= (sinAcosB + cosAsinB)(sinAcosB - cosAsinB)/(cos²Acos²B)\n= (sin²Acos²B - cos²Asin²B)/(cos²Acos²B) (since (x + y)(x - y) = x² - y²)\n= sin²A/cos²A - sin²B/cos²B\n= tan²A - tan²B ← Option c)\n\nPlease note: sin(A + B) = sinAcosB + cosAsinB\nand sin(A - B) = sinAcosB - cosAsinB
733210
You basically have 2 options. Use a protractor and measure them (The internal angles total 180 degrees).\n\nThe other option is using the trig functions (Sin, cos and tan), however you must have some info on the triangle eg. length of sides to apply any of these.
906923
Suppose the component vectors are a and b with magnitudes |a| and |b| and the angle between them θ\n\nThen, on completing the parallelogram of forces, the angle between the end of a and the beginning of b is (180° - θ)\n\nIf the resultant vector is R (which is the required vector), makes an angle of α between a and R, then by the sine rule\n\nsinα/|b| = sin(180° - θ)/|R|\nSo sinα = |b|sinθ/|R| (because sin(180° - θ) = sinθ\nHence α = arcsin{|b|sinθ/|R|}\n\nNote If you need |R| then by the cosine rule:\n\n|R|² = |a|² + |b|² - 2*|a|*|b|*cos(180° - θ)\n= |a|² + |b|² + 2*|a|*|b|*cosθ (since cos(180° - θ) = - cosθ)\n\nThis is for a and b in any orientation on a plane (not necessarily at right angles)\n\nIF θ = 90° then cosθ = 0 and sinθ = 1\nand these equations reduce to\n\n|R|² = |a|² + |b|²\nand \nα = arcsin{|b|/|R|} (which incidentally also equals arctan {|b|/|a|}
601325
I'll get you started by answering the first question.\n\nSay that you have two parallel planes in the form\n\nax+by+cz = d1\nax+by+cz = d2\n\nThe vector normal to these planes is N = (a,b,c) i.e., if you draw pictures of these planes then the vector N defines a line that goes perpendicularly through them both.  This line intersects the planes at the points\n\nP1 = d1/(a^2 + b^2 + c^2)*N\nP2 = d2/(a^2 + b^2 + c^2)*N\n\nThe distance between the two planes is simply the distance between these points:\n\ndist = | P1-P2 | = |d1-d2|/sqrt{a^2 + b^2 + c^2}\n\nFor your equations, I'm going to multiply the second equation by 3:\n\n15x+6y-6z=7\n15x+6y-6z=18\n\nThe distance between these planes is\n\ndist = |18-7|/sqrt{15^2 + 6^2 + 6^2} = √33/9
331507
sorry hunny you need at least 2 side lengths in order for us to help you figure it out... plus once you have that the formula for figuring it out is using the pathagorean theorem! a2+b2=c2 (s = squared) say for instance... the 2 sides of your triangle are 7 and 3inches. 3>2+7>2=c>2\nwhich would turn into 9+49=c>2~~~~~~~ then 58=c>2 then to get c by itself(which is needed) you would take the square root of 58! your hypotenuse would equal 7.6 or rounded, 8!\nnow if you use my steps you should be able to do your problem! I'm a 10th grader in geometry and it's the most lovely subject!
866713
The initial poster didn't read the entire question.  This is not as simple as the pythagorean theorem idea he/she had.  You need to use vectors to solve this problem.  Assume X is east-west and Y is north-south, with positive being east and north.\n\nThe first vector is 5.0 km at 11.5 degrees north of due east.  You need to find the X and Y components of this vector.  Use SOH CAH TOA, and you have the H, and you have the angle.  So to get the value of the X component, you need to use sin:\n\ncos (11.5) = X/5.0, so the X component is 5 cos (11.5)\n\nsimilarly for the Y component, use sin:\n\nsin (11.5) = Y/5.0, the Y component is 5 sin (11.5)\n\nGreat, so the vector for the first distance is \n<5 cos(11.5), 5 sin(11.5)>\n\nThe second distance is 1.0 km due south.  Well, this is only in the Y direction, so the vector would be <0, -1>  Use the negative sign because we assigned north as positive, and south as negative.  Now we can add the components of the vectors to get a resultant vector.\n\n<5 cos(11.5), 5 sin(11.5) -1>\n\nThis says that the resultant vector has an X component of \n5 cos(11.5) and a Y component of 5 sin(11.5) -1.\nUsing the Pythagorean theorem, the magnitude of this resultant vector would be\nsqrt [ (5 cos (11.5))^2 + (5 sin(11.5)-1)^2 ]\n\nNow to get the direction, we need to use trigonometry again.  We have the opposite, and the adjacent, which is TOA.\n\ntan(theta) = Y/X = (5 sin(11.5)-1)/(5 cos(11.5))\n\nTake arctan of both sides:\ntheta = arctan[(5 sin(11.5)-1)/(5 cos(11.5))]\n\nSo the answer?\nThe resultant displacement vector is:\n\nsqrt [ (5 cos (11.5))^2 + (5 sin(11.5)-1)^2 ] at an angle of \narctan[(5 sin(11.5)-1)/(5 cos(11.5))]
567245
simple x=r*cos(theta), y=r*sin(theta), r^2=x^2+y^2\nsubstitude, not hard
880255
I can just give you a link to help you solve it.\n\nhttp://www.ticalc.org/pub/83plus/basic/math/geometry/
1147435
(tangent 30 degrees= 1km/x) - (tangent 60 degrees= 1km/y)\n\nSolve for the two variables, then subtract y from x.  That is how far the plane traveled in 10 seconds.
780884
As with many problems, it helps to draw a diagram.  I made a triangle, and then put in each of the altitudes, since they are my only known values.  With these in place, we now have 6 principal right triangles, of which we know the value one leg (the altitude).  In each of these triangles, the hypoteneuse is a complete side of the original triangle, and the angle opposite the known leg is also of the original triangle.  Personally, I chose to work with cosines and tnagents, but you could just as easily have used sines (though I think it might add a step or two).  Remember that the cosine of an angle in a right triangle is merely the ratio of the side opposite the angle to the hypoteneuse.  Don't worry that we don't know the value of any of these angles; I went through the entire problem without ever needing it.\n\nSo, since each angle, in its entirety, is in two of our right triangles, they give us two ratios to compare.  That is:\n\nIf the sides of the original triangle are x,y,z\ncos a1 = 20/x = 15/y\ncos a2 = 20/z = 12/y\ncos a3 = 12/x = 15/z\n\nWith a little cross multiplication and rearrangement of terms, we find that:\n12z = 15x = 20y.  \n\n(Your x, y, z may be in a different order depending on how you labeled your triangle, but if you manipulate your triangle a bit, you should be able to get it to match.  This will keep your coefficients identical to mine.  But even if you don't match it up, in the end you will get the same answer.  That's the beauty of math.)\n\nThe tangents are a little trickier, because they involve a fraction of the original side.  This involves the addition of a new variable on each, such that one portion of the line is "a", and the remainder is x - a; and then b and y - b; c and c - z.  The tangents then follow:\n\ntan a1 = 20/(y - b) = 15/a\ntan a2 = 12/(z - c) = 20/b\ntan a3 = 15/(x - a) = 12/c\n\nRearranging each gives us the three equations:\n20a = 15(y - b)\n12b = 20(z - c)\n15c = 12(x - a)\n\nFrom there, you can solve for either a, b, or c.  It's your choice.  I opted to solve for a.  Substituting the third equation into the second, and then that new equation into the first, we get:\n\n20a = 15(y - (20/12)(z - (12/15)(x - a)))\nwhich reduces to\n8a = 4x + 3y - 5z\n\nNow, side a is in a right triangle with hypoteneuse y and second leg 15, meaning that:\na^2 + 15^2 = y^2   or\na = sqrt(y^2 - 225)\n\nUsing this equation, plus the ratios we received from the cosine functions, we solve for y:\n\n8 * sqrt(y^2 - 225) = (16/3)y + 3y - (25/3)\n\nConveniently, the right half is equal to zero, so dividing by 8 and squaring:\n\ny^2 - 225 = 0\ny^2 = 225\ny = 15\n\nEnter this into our ratio:\n\n15x = 20y\n15x = 20*15\nx = 20\n\nand\n\n12z = 20y\n12z = 20*15\nz = 25\n\nSo finally we have the lengths of all three sides, and adding them together, the perimeter is\n15 + 20 + 25 = 60\n\nThis turns out to be a right triangle, but that is merely happenstance.  This process can be used to solve for any triangle with three known altitudes.\n\nHope this helps.
850990
In this case involving trigonometric operation on angles, you will need to carry the error over to your operation results. What I mean by that is:\n\nYou have an angle known only to 1 degree of precision, i.e.\n\nθ = 32° ± 1°\n\nTherefore, when you compute Cos (32°) = 0.848 \n\nYou will also need to perform the trig operation including the error:\n\nCos(31°) = 0.857 and Cos(33°) = 0.839\n\nSo your measurement is 0.848 ± 0.009 or 0.85 ± 0.01 depending on how many significant digits you want to keep.
466401
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!