The sides of a parallelogram have lengths of 60 ft and 80 ft and the smallest angle is 56 degress.  What is the lenght of the shorter and longer diagonal?
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Consider a triangle with sides a, b, c and angle A oppose to the side a.  Side a is the diagonal that you are going to find in each case.\n\nBy knowing 2 sides (b, c) and the angle (A) in between, then use Cosine rule:\n\na^2 = b^2 + c^2 - 2bc cos A\n\na^2 = 60^2 + 80^2 - 2 * 60 * 80 cos(56)\n\na = 68.06 ft  (shorter diagonal)\n\nConsider when A = 180 - 56 = 124, use Cosine rule again:\n\na^2 = 60^2 + 80^2 - 2 * 60 * 80 cos(124)\n\na = 123.97 ft (longer diagonal)
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since it is right angle triangle, the addition of square of two arms will be equal to the square of hypotenuse.\n\nNow consider the smaller leg is 'x' inches, thus the other leg would be 'x+14' inches and hypotenuse would be 'x+16' inches.\n\nFrom the property of right angle triangle, we have now\n\n(x+16)^2 = x^2 + (x+14)^2\n\nor, x^2+32x+256=x^2+x^2+28x+196\nor, -x^2+4x+60=0\nor, x^2-4x-60=0\nor x^2-10x+6x-60=0\nor x(x-10)+6(x-10)=0\nor (x-10)(x+6)=0\n\nso x can have 2 values -6 and 10. since it is a distance it can not be negative, thus the shortest length of the right angle triangle is 10 inches. The other arm would 24 inches and the hypotenuse will be 26 inches.... (answer is 10,24,26)\n\ncheers:)
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Rhombus\n\ns = side length of rhombus\nh = height of rhombus \nd1 = long diagonal of rhombus\nd2 = short diagonal of rhombus \n\nArea = hs\n         = s2 sin A\n         = s2 sin B\n         = (½) d1d2
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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.
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150 cm.\n\nExplanation: each turn has a length equal to the diagonal of a rectangle of 24 and 18 cm, where 18 = height / 5 turns.\nThe length of the diagonal is therefore 30 cm.\n\nSo the total length is 5x30 = 150 cm
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decagon=180*8=1440*\npentagon=180*3=540*
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Drop a perpendicular to side of length 10 from the vertex where the side of length 8 ends.\n\nA right angle triangle is formed,\n\nthe perpendicular is 8*sin 60 = 4sqrt(3)\n\nArea of ||gm is 10*4*sqrt(3) = 69.282 approx.
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C-120... Two supplementary angles must equal 180.... 60+120=180\nB-32 feet... 8*4=32
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The figure we are talking about is an isosceles trapezoid. The lower base is equal to the two diagonals and the upper base and the two sides are all equal in length.\n\nCall the trapezoid ABCD with AB being the lower base and cd bbeing the upper base. The two sides which are both equal to the upper base are BC and AD. Also notte that AB parallel to CD.\n\nThe diagonals are AC and BD. \n\nIt turns out that the two base angles are bisected by each of the diagonals.Since the base angles are equal then they are each divided into two equal angles. Call each of them angle 1. also By the parallel postulate, angle CDB and angle must also be equal to angle 1.\n\nNow for the base to be the same length as the diagonals, triangle\nABC and Triangle must both be isosceles. This will be true only if angle ADB=.twice angle 1 and angle ACB = twice angle 1.\n\nTherefore the two base angles add up to 4 angle 1s and the agles C and D are both = 3 angle ones. Thus the four vertex angles add up to 10 angle 1s. Since the total of all angles in a quadrilateral is 360 degrees, the angle 1 must 360/10 =36 degrees.\n\nThus each base angle = 2 angle 1s = 2*36 =72 degrees and each of the upper angles must be equal to 3 angle 1s =3*36=108 degrees.\n\nIn summary angle A and angle B = 72 degrees and\nangles C and D both = 108 degrees.\n\nHope this was not too confusing. It will help if you draw a picture.
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If you picture your trapezoid with the shorter parallel side and the two legs as sides 4, then the only unknown is the the long parallel side.  That side is at least 4, but with two extra parts on either side (if you drop perpendiculars from the short to the long side, you'd have a rectangle and two triangles).  So the long side is 4 + 2x, where x is the length of those two parts, which is also a leg of a triangle that we created in dropping those perpendiculars.\n\nThe height of the trapezoid is the same as the other leg of the triangle.  We don't know the length of either leg, but we do know that the hypotenuse is 4.  So through the Pythagorean theorem, we have\n\n4^2 = h^2 + x^2\nh^2 = 16 - x^2\nh = sqrt(16 - x^2)\n\nBack to the trapezoid - we know the area of a trapezoid is the average of the bases times the height, so we have\n\nA = (1/2)(4 + 4 + 2x)(sqrt(16 - x^2))\n= (4 + x)(sqrt(16 - x^2))\n= (4 + x)(16 - x^2)^(1/2)\n\nTo find the maximum area, take a derivative, set it to 0, and solve for x:\n\nA' = (4 + x)[(1/2)(16 - x^2)^(-1/2)](-2x) + [(16 - x^2)^(1/2)](1)\n= (-4x - x^2)(16 - x^2)^(-1/2) + (16 - x^2)^(1/2)\n\nIf we set this to 0, then\n0 = (-4x - x^2)(16 - x^2)^(-1/2) + (16 - x^2)^(1/2)\n-(-4x - x^2)(16 - x^2)^(-1/2) = (16 - x^2)^(1/2)\n4x + x^2 = 16 - x^2\n0 = 2x^2 + 4x - 16\n0 = x^2 + 2x - 8\n0 = (x + 4)(x - 2)\nx = -4 or 2\n\nSo your long side has to be 4 + 2(2) = 8 to maximize the area.
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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
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Similar triangles:\n\nLet's put all the units in inches first, so 57 ft 7.75 inches is:\n57 * 12 + 7.75 = 691.75 inches\n\n36 inch yard stick casts a 42.5 inch shadow\nx inches building casts a 691.75 inch shadow\n\n36 / 42.5 = x / 691.75\n\nCross multiply:\n42.5 x = 691.75 * 36\nx = (691.75 * 36) / 42.5\nx = 585.952941 inches\n\nDividing by 12, you get 48, with a remainder.\n48 ft = 576 inches, remainder 9.952941 inches\n\nSo all together your building is 48 feet 9.95 inches, or approximately 48 ft. 10 inches.
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The area of a parallelogram is equal to the base times the height.\nA = (b)(h)\n\nThe way to get maximum area is a square (which is a parallelogram). In a square, all sides are the same. Since 69/2 + 69/2 = 69...\n\nA = (69/2)(69/2) = 4761/4 = 1190.25 cm²
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In a 45-45-90 triangle, there is an important ratio. leg-leg-hypotenuse = 1-1-square root of two.\n\nTo find the length of the leg if the hypotenuse is given...\ndivide the hypotenuse by square root of two.\nFor example... if the hypotenuse were 5, the leg would be 5/square root of 2 = about 3.5
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Draw a picture.  You have 2 right triangles, with the satellite in the middle perpindicular to the ground.  Call this height "x"\n\nThen you have a right triangle coming off this height going off to the left with a lower-left angle of 37 degrees.  Another right triangle to the right with a lower-right angle of 57 degrees.  This will make one big triangle, with the height of the satellite coming down from the top vertex to somewhere along the bottom side.\n\nYou know that the entire length of this bottom side is 54, but you don't know how to divide it up.  Call the distance from the first observer (on the left) to the tower "y", and then the remaining distance will be "54-y"\n\nNow you have 2 variables, and enough information to create 2 equations:\n\ntan(31)= x/(54-y)\ntan(37)=x/y\n\nBy multiplying both sides of the second equation by "y" you get y*tan(37)=x; now you can substitute y*tan(37) back into the first equation\n\ntan(31)=(y*tan(37))/(54-y)  now it's just simple algebra.  But be careful, once you've solved it you have solved for y, so be sure to go back and find x.
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To determine if it is a parallelogram:\nIf a quadrilateral has one pair of opposite sides which are parallel and equal in length, then the figure is a parallelogram.\n\nTo determine if it is a rectangle:\nIf a parallelogram has two pairs of opposite sides which are parallel and equal in length, then the figure is a rectangle.\n\nGraph the points and determine the distance between the points, or, in your case, two points have an X of 0 and two have X of 3\n\nBetween the (0, 2) and (0, 6), there is a distance of 4\nBetween the (3, -2) and (3,2) there is a distance of 4\n\nBetween the (0,6) and (3, 2) there is a distance of 3 horizontally.\nBetween the (0,2) and (3, -2) there is a distance of 3 horizontally.\n\nYour figure has two pairs of opposite sides that are equal in length and parallel.  You have a rectangle.
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a) Assume the following:\n\nAngles a= 10 deg, b= 80 Deg, c=90 deg \nHypotenuse length = 2.5 m\nheight = 2.5 * cos 80 Deg= .434 m\nlength = 2.5 * sin 80 Deg= 2.462 m\nacceleration = "g" * Sin 10 Degree