Find the measure of the hypotenuse of a 30-60-90 triangle whose longer leg has a measure of 6 meters.?
440893
the ratio of the sides of a 30-60-90 triangle is\n\n1 : √3: 2 \n\nThus the ratio of the longer side to the hypoteneuse is √3 : 2\n\nso set up a ratio expression:\n\n6/x = √3/2\n\nand solve:\n\ncross multiply:  \n12 = √3x\nx = 12/√3\nx = 4√3
1086757
If you realize that this would make a 30-60-90 triangle the answer is really easy.  The ratio of the sides of a 30-60-90 triangle are 1:2:sqrt(3)  In your example, that means that the height off the ground would be half the length of the bed.  Half of 86" is 43", so it will be 43" inches off the ground.\n\nYou can also use the sin(30) formula, but sin(30) is simply 0.5, so you get the same answer:\n\nn = 86 sin(30)\nn = 86 * 0.5\nn = 43\n\nSo the answer either way is 43 inches off the ground.
187445
hypotenuse=36cm=c\none leg=20cm=b\nother leg=?=a\n\nFormula:\na^2=c^2-b^2\na^2=36^2-20^2\na^2=1296-400\na^2=896\na=square root of 896.\nby using a calculator extracting the square root, we arrive at an answer of\na=29.9332590941 or 30cm\n\nThe other leg's measure is 30cm by the Pythagorean Theorem.
753210
The rope is the hypotenuse of the triangle you need to analyse so it's going to be sin60 or cos60 times 75N. An easy way to decide which: as a test, consider if the angle were 90. The horizontal component would be 0. Is sin90 or cos90 = 0? Cos, right? So the component of the force parallel to the floor is cos60 times 75N.
87753
Just add up the lengths of the three sides. Works for any triangle - wether it's equilateral, isosceles or obtuse.
821495
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
967287
The triangle will have 2 sides of length x, and the remaining side will be length P-2x making the parameter x + x + P - 2x = P.\nBegin by drawing a line from the vertex where the two sides of length x meet, to the center of the side which is length P-2x. The line should be perpendicular to the side P-2x thereby splitting the triangle in half, into 2 right triangles. One of the right triangles has hypotenuse = x and one leg = (1/2)(p-2x). So the remaining leg has length S.R. (x^2 - (p/2 -x)^2). We calculate the area of this right triangle, and double it since there are two triangles:\nArea = (2)(1/2){(1/2)(p-2x)S.R.(x^2 - (p/2-x)^2)\n         = (1/2)(p-2x)S.R.{px - (p^2)/4}\n         = (1/4)(p-2x)S.R.(4px-p^2)\nNote that your given answer is wrong - it has an x^2 where it should have a p^2....
54081
Relative to the x-axis (East in the positive direction), the angle is 90-28.6 = 61.4 deg.\n\nVector 1:\nx = 303*cos(61.4 deg) = 145.04 m\ny = 303*sin(61.4 deg) = 266.03 m\n\nVector 2:\nx = 199 m\ny = 0 m\n\nVector 1+2:\nx = 145.04 + 199 = 344.04\ny = 266.03\n\nA) Final displacement\n\nd = sqrt(344.04^2 + 266.03^2) = 434.90 m\nangle = arctan(266.03/344.04) = 37.7 deg\n\nB) Difference: walked - displacement\n\nwalked = 303 + 199 = 502 m\ndisplacement = d = 434.9 m\n\ndifference = 502 - 434.9 = 67.1 m
185495
The Pythagorean Theorem can only be applied to Right Triangles only.\n\nIn a right triangle, the side opposite the 90 degree angle is called the HYPOTENUSE. The other two sides are called LEGS.\n\nTHEOREM:\nThe square of the hypotenuse is equal to the sum of the squares of the legs. \n\nEQUATION: Supposing,\nc = hypotenuse\na = first leg\nb = second leg\n\nThen:\n\nc^2 = a^2 + b^2\n\nThis is the conventional way of assigning the letters to the sides of the right triangle. However, if other letters will be assigned, then determine which letter will represent the hypotenuse so that its square can be placed on the left side of the equation. The others will be placed on the right side.
605775
Are you talkin about a trapezium or a trapezoid.\nif its a trapezium,here's how : \nlet the height be "h".\ndrop two perpendiculars equal to h on the base.the lengths b&#92;w the perps.is 6 units.therefore the rest of the length is 10.each individual length to the left of the perp is 5 units.\nin the triangle with the diagonal,the side adjacent to the angle with the base and diagonal is 6+5=11\ntherefore the angle is cos^-1(11/12)=a\ntherefore tan a=h/11\ntherefore h=11tan a\nanswer is height,h=4.795831523
1030296
for it to be a triangle the angles have to measure to 180, so if you have a 90 degree andlge off one, you could subtract 90 from that, take the remaining 90 and work with that, since you know your need a 1:1 of the legs , then you know you need an equal angle.. angle has to equal 90/2... and thats where i get stumped...\n      gl
498166
Draw a right triangle and let one of the angles be the arccosine of 1/2. That means the length of the adjacent leg divided by the length of the hypotenuse is 1/2, so you may write the length of the adjacent leg as 1*x and the length of the hypotenuse as 2*x, where x is some unknown scaling factor. \nThe sine of this angle is the length of the opposite side divided by the length of the hypotenuse, 2x. We get the length of the opposite side from the Pythagorean theorem: Call that length s. Then s^2 + x^2 = (2x)^2, so s^2 = 3x^2, so s = Sqrt[3]x, the positive square root since we're dealing with lengths. Now we have the sine ratio as opposite/hypotenuse = Sqrt[3]x/(2x) = Sqrt[3]/2.
1043340
What is n?  what is m?  What is q?  How can we come up with an answer without this data?\n\nI'm going to assume that q is the midpoint the segment or segments, and that m and n are also equidistant from q.\nThen you set up a right triangle, you're looking for the hypoteneuse length and the legs measure 10 and 16.\n\nIf that's so, then lm = √(10²+16²)\nlm = 2√(5²+8²)\nlm = 2√89
477890
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)
922431
Use pythagorean formula to solve:  a^2 + b^2 = c^2, where a and b are the legs of the triangle and c is the hypotenuese.\n\nIt's a little unclear what you mean by "right side is x" -- not sure if you mean it's one of the legs or the hypotenuese.  I will assume you mean hypoteneuse.\n\nSubstitute 51 and 45 for a and b in the formula to solve for c (or x as it's being called in your problem):\n\n51^2 + 45^2 = x^2\n\n2601 + 2025 = x^2\n\nx^2 = 4626\n\nx = √4626 = 3√514\n\n.
1166171
The easy way to remember this is to memorize the \n\n3 4 5 \n\nright triangle.  You will notice that any two multiples of the above will give you the other leg as a multiple.  The theorem is\n\nA^2 + B^2 = C^2,  so  3^2 + 4^2 = 5^2,  hence  9 + 16 = 25\n\nnow you have\n\n2*3 2*4 thus your other leg is 2*5\n\n6^2 + 8^2 = 10^2,  which is 36 + 64 = 100\n\nas for the angles, use the law of sines\n\nsin(90)/5 = sin(a)/3 = sin(b)/4\n\nso a = asin(sin(90)*3/5) = ~37 deg\n\nhence angles of 90 53 37
217108
You have a right triangle with legs a and b, and hypotenuse c.\na+b=8\nc=6\n\na^2 + b^2 = c^2 = 36\n\nSquare both sides of a+b=8\na^2 + 2ab + b^2 = 64\n\nSubtract previous equation:\n\n2ab = 64 - 36 = 28\n\nab = 14\n\n(1/2)*ab = 7\n\nArea = 7.