A shop keeper is making a triangular sign for his store front, but he must keep the sign under 20ft^2 to adhere to zoning laws. If the base of the sign is 2 ft, what is the maximum height of the triangular sign?
199535
The area of a triangle is\n\n1/2 * Base * Height\n\nSo it the max base is 2 ft, and the max area is 20 ft^2, then the max height would have to be 20ft
508171
Angle of elevation should be that raised from the ground until it is vertical (90 degrees), correct?  The wall which the ladder is placed against will have a vertical face (90 degrees).  The ladder, therefore, represents the hypotenuse of the triangle so formed.\nCosine is the length of the adjacent sided divided by the hypotenuse.  So cos 70 = X/40 or 40*(cos 70) = X.\nWith me so far?\nOnce you have X, you have the hypotenuse and 1 side of a right angle triangle and finally a use for Pythagoras's theorem the square on the hypotenuse equals the sum of the squares on the other 2 sides.\nCan you do the rest of the arithmetic?  I hope this problem was in Trigonometry, otherwise I have given you the wrong way to solve it. But it is the easier way.
458244
You need two right triangles, I've attached a diagram...\nhttp://img247.imageshack.us/img247/7576/bldgny3.gif\n\nThe bottom triangle has an angle of depression of 20°.  We know the opposite side is 30m (same height as the window, assuming a level street).  We want to figure the adjacent side 'A', being the distance across the street.\n\nFrom the SOH-CAH-TOA mnemonic --> Tan(angle) = Opposite/Adjacent.\n\nSo tan(20°) = 30/A\n\nSolving for A:\nA = 30/tan(20°)\nA ≈ 82.4m\n\nNow we look at the top triangle.  The angle of elevation is 50°.  We now know the adjacent side is 82.4m.  We need to determine the opposite side 'O'.\n\nAgain, tan(50°) = O/82.4\n\nSolving for O:\nO = tan(50°) * 82.4\nO ≈ 98.2m\n\nThe total height of the building is the sum of the two heights (30m to the window and 98.2m to the roof).\n\nSo the building is approximately 128.2m tall.
672772
1) The triangle is isosceles.\nThe other two angles are equal.\nLet m<A = x, then m<B = x\nx+x+62 = 180\n2x = 180-62\n2x = 118\nx = 59\nThe largest angle is 62, so the side opposite that angle has the longest length.\nAnswer: c) AB\n2) Since the x-xoordinates are the same, you just have look at the y-coordinaes\n2 + 3 = 5\n5 + 3 = 8\nAnswer: (0,8)\n3) The sum of the interior angles of an n-sided polygon is \n180(n-2)\nSet this equal to 1260 and solve for n\n1260 = 180(n-2)\n4) \n5) Add all the angles together and set it equal to 180(4-2)\nx + x + (x+15) + (x+45) = 180(4-2)\nSolve for x.\n\nWhy are you removing your questions?  They were not short questions and people actually spent time answering them.
500350
Let x be the length of the corners.  Then the base will be (30 - 2x) by (30 - 2x) and the height will be x.\n\nYou want that to be a maximum, so create a formula for the volume, take the first derivative, set it to zero and you'll have the maximum.\n\nf(x) = x(30 - 2x)(30 - 2x)\nf(x) = x(900 -120x + 4x²)\nf(x) = 4x^3 - 120x² + 900x\n\nNow take the derivative\nf'(x)= 12x² - 240x + 900\n\nNow factor it:\nf'(x) = 12(x² - 20x + 75)\nf'(x) = 12(x - 5)(x - 15)\n\nAnd set it to zero:\n12(x - 5)(x -15) = 0\n\nSo x must be 5 or 15, but at x = 15, you have a minimum volume (0).  At x = 5 you have a maximum volume of 5 in x 20 in x 20 in = 2000 cu. in.
360180
Okay, you know that when you cut either side of a paralleogram and put it on the other side, it'll equal a rectangle and still have the same area, right?\nThe base and height will remain the same when the paralleogram is transformed into a rectangle. This rectangle can be cut perfectly into two triangles. Because the height and the base of the triangle and the rectangle are the same, then the rectangle's size will the twice the size of the triange.\nExample.\nTriangle, rectangle base: 3cm\nTriangle, rectangle height: 4cm\nArea of triangle: 3x4/2=12/2=6cm square\nArea of rectangle: 3x4=12cm square\nWe can see that the area of the rectangle is twice the area of the triangle.
45387
A man 6.25 feet tall casts a 5-foot shadow. At the same time, a nearby tree casts a 20-foot shadow. How tall is the tree?\n============================================================\n\nSimple ratio/proportion problem:\n\n6.25 feet : 5-foot shadow :: x feet : 20-foot shadow\n\n6.25/5 = x/20\n20(6.25) = 5x\n4(6.25) = x\n25 = x\n\nThe tree is 25 feet tall.
483936
1)One leg of a right triangle is 12 cm long. The diference between the other two sides is 2 cm.\nFind the sides of the triangle. (Pythagorean Theorem)\n\nWhat we can take from this problem is that one LEG is 12 cm long, while the other LEG is unknown (we will call it X) and the HYPOTENUSE -- the longest side of a right triangle -- is also unknown (we will designate it X+2, since it differs 2 cm from the unknown leg).\n\nSetting up Pythagorean Theorem as always:\n\n "Leg squared" + "Leg squared" = "Hypotenuse squared"\n\n  12^2 + x^2 = (x+2)^2\n\nExpanding:\n\n  144 + x^2 = x^2 + 4x + 4\n\nCollecting like terms:\n\n  4x = 140\n\nDivide by 4:\n\n   x = 35\n\nThe unknown leg is 35 cm, while the hypotenuse is 37.  Thus, we can also call the triplet (12,35,37) a "Pythagorean Triple" (of which there are infinitely many!).\n\n\n2) The hypotenuse of a right triangle is 58 cm long. The diference between the other two sides\nis 2 cm. Find the sides of the triangle. (Pythagorean Theorem\n\nIn this case, we know the hypotenuse, but do not know the lengths of the legs.  For convenience, we will again use the expressions X and (X+2) for each side's length.\n\nBy P.T.:\n\n x^2 + (x+2)^2 = 58^2\n\nExpanding:\n\n x^2 + x^2 + 4x + 4 = 3364\n\nCollect like terms:\n\n 2 x^2 + 4x - 3360 = 0\n\nDivide by the GCF (2):\n\n  x^2 + 2x - 1680 = 0\n\nCan we factor this trinomial easily?  We could, if we knew integer factors of -1680 (one positive, one negative) that combine to make 2.\n\nLuckily, -1680 can be factored into -2*2*2*2*3*5*7, and these can be grouped into two sets:\n      2*3*7 = 42\n -2*2*2*5 = -40\n\nFactor:\n\n (x+42) (x-40) = 0\n\nAlthough two solutions come from these factors (-42 and 40), we may only consider the positive answer, since lengths of triangle's legs are always positive!\n\n Therefore, x=40 (Leg 1), and Leg 2 is 42 cm.
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The length of the wire must be (2x5)+(2x7)=24cm.  So each side of the triangle must be 24/3=8cm. long.
360377
Start with what you know:\narea = 60 inches^2\nheight = 12 inches\n\nNeed to find base\nStart with known equation\nA = 1/2*b*h\nsolve for b = 2*A / h = 2 * 60 / 12 = 10 inches\n\nNow you know the base solve for the new height given the new area.\nArea = 60+20 = 80 inches^2\nh = 2*A / b = 2*80 / 10 = 16 inches\nso the new height is 16 inches\n\nCheck A = 1/2 * 10 * 16 = 80 inches^2
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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)
428717
The area of a triangle is given by the formula:\nA = 1/2 b * h\n\nLet b = h + 2\nA = 40\n\nSo:\n40 = 1/2 (h + 2)h\n40 = (h² + 2h)/2\n80 = h² + 2h\nh² + 2h - 80 = 0\n(h + 10)(h - 8) = 0\n\nh + 10 = 0\nor\nh - 8 = 0\n\nh = -10 \nor\nh = 8\n\nBut negative numbers don't make sense for lengths, so there is one answer:\n\nThe height = 8 ft\nThe base = 10 ft
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If the x and y axis are forming two of the sides of the triangle, the truangle is right angled so the area of the triangle will be 1/2 * base * height\n\nThe line through (3,2) therefore has to meet the x axis and the y axis to complete the triangle shape. \n\nConsider the gradient of the line = -1. The line would cut the x axis at (5,0) and the y at (0,5), making the triangle 0.5 * 5 * 5 = 12.5 units squared\n\nConsider gradient = -a. We'd have intersects at (3-2/a,0) and (0,2-3/a). The area of the triangle would be:\n\n0.5 * (3-2/a) * (2-3/a)\n= 0.5 * (6 - 4/a - 9/a + 6/a²)\n= 0.5 * (6 - 13/a + 6/a²)\n= 0.5 * (6a² - 13a + 6) ---- (multiply everything by a²)\n\nNow solve for minimum a by differentiating:\n\nd/da = 0.5 * (12a - 13)\n\nReaches a minimum when d/da = 0 :\n\n0 = 6a - 6.5\n6.5 = 6a\na = 1.086666\n\nNow put that back into our area equation, we get:\n\n= 0.5 * (3-2/a) * (2-3/a)\n= 11.556\n-------------
101113
Y'all didn't answer the question! You just found the length of the sides. :)\n\nThe largest square that will fit inside has it's 4 corners touching points on the circle.  If the radius is 10 feet, then the diameter is 20 ft.  The diameter is also the diagonal of the square with side length x, and is therefore the hypotenuse of a 45-45-90 right triangle.\n\nx^2 + x^2 = 20^2\n2x^2 = 400\nx^2 = 200\n\nThat's as far as you have to go, since the area of the square is length (x) times width (x), or x^2\n\n~*~*~So the area of the largest square is 200 sq feet.~*~*~
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Use the Law of Sines.\n\nsin A/a = sin B/b\n(sin 36)/24 = sin B/47\nsin B = (47 sin 36)/24\nsin B = 1.151079452\nAt this point you have a problem--it is impossible for the sine of an angle to have a value greater than 1.\n\nSo either you copied the problem wrong, or you are being asked to recognize that a triangle cannot be formed with these dimensions.
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Ok, you've got a good one here.\nFor a right triangle, height and base are the short sides; and area of the triange is (bh)/2. Since we know the area, and the length of the sides, we can plug it in to an equation, and find x. So...\nA = (bh)/2\n60 = ((5x)(3x-1))/2\n120 = (5x)(3x-1)\n120 = 15x^2 - 5x\n15x^2 - 5x -120 = 0\nDivide both sides by 5 to get\n3x^2 - x -24 = 0\nOk, there's a number of ways to do this, but the simplest way is to list the combinations of integers that will multiply to get -24 (the constant in the above equation). That gives us the combinations of (1, -24; 2, -12; 3, -8; 4, -6; 6, -4; 8, -3; 12, -2; and 24, -1). Now, we need to figure out the one combination that, when the second number is multiplied by three (the coefficient of x^2), you can add it to the first number to become (-1). It happens to be (8, -3) {-3*3=-9, -9+8 = -1}. There; you've factored the equation, leaving you with...\n(3x+8)(x-3)=0\nSince multiplication rules say that when ab=0, either a=0 or b=0, we can test by setting both numbers equal to zero; giving us...\n3x+8 = 0 ; x-3 = 0\nx = -8/3; x = 3\nSince x is an absolute length, and lengths can't be negative, we can disregard (-8/3) for now. That gives us x=3, and putting that back into the original lengths gives us sides of 15 and 8. To find the hypotenuse, we use Pythagorean Theorum: a^2 + b^2 = c^2, substituting in the known lengths gives us:\n15^2 + 8^2 = c^2\n225 + 64 = 289  = c^2\nc = √289 = 17\nSo, the hypotenuse of your triangle is 17.
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a)A=41°20’ b=17.2cm\ncos 41°20'=17.2/h\nh=17.2/cos 41 1/3°=22.9cm\n\nb) 1) A=48°40’ a=21.0cm\nsin 48 2/3°=21/h\nh=21/sin 48 2/3°=28.0cm\n\nsolve TRIANGLE ABC if \nA=50° B=81° a=95cm\n\nC=180-50-81=49°\nlaw of sines\n95/sin 50=b/sin 81=c/sin 49\nb=95 sin 81/sin 50=122.5cm\nc=95 sin 49/sin 50=93.6cm