a sail is in the shape of a triangle with sides of 10m and 13m find the largest angle of the sail thanks .>?
536681
for this if you are looking for the angle you dont need the lengths of the sides.  A sail would be a right triangle therefore the largest angle would be 90 degrees.\nIf you are looking for the largest SIDE then you should use\nthe Pythagorean Theorem.  a^2 + b^2 = c^2
943161
The faster way to solve this questions is using graphical method. Draw the respective triangles (choosing appropriate scale) and you will get answer.
1116580
Brian gave you the answer, but he didn't explain it thoroughly.\n\nPlace your vector at the origin of an x and y plane.  The x and y axes are perpendicular, so you can use the Pythagorean formula to get the magnitude which is the length of the hypotenuse of the right triangle formed by the points (0,0), (10,0), and (0,8). sqrt(10^2+8^2)\n\nNow you need direction which is the angle between the x axis and the vector. Since you have solved for all three lengths of the triangles sides, you can use any inverse trig function to find the angle.  Brian chose tangent.\n\nGood work, Brian!
825411
first side=x\nsecond side=2*x\nthird side=2*x+3\n1. + 2. + 3. => x+2x+2x+3=33 => x=6 \n\nfirst side: 6cm\nsecond side: 12cm\nthird side: 15cm\ntotal: 33cm
394622
A triangle in which all 3 sides are of different lenghts, with different angles.
800228
the angles of a triangle add up to 180 degrees\na right angled triangle means one of the angles is 90\nso you have 90+ 65+ X += 180\nX = 25\n25+65+90= 180
551883
The answer is C\nwell, one side's length of a triangle should be shorter than the other 2's plus (2 sides plus should be longer than one side)\nthen: 3x+4+8x+2>6x-1\n      8x+2+6x-1>3x-4\n      6x-1+3x+4>8x+2\n--> x>-7/5\n    x>3/11\n    x>-1\nThus, the answer should be x>3/11\n:)
748474
Draw a triangle. Side a is 35m. Side b is 5m, and Angle C (the angle between sides a and b) is 155 degrees (180-25).\n\nsides a and b are the runners path. The remaining side, c, is the total displacement. By the law of cosines, the following is true of a triangle:\n\nc^2 = a^2 + b^2 - 2ab cos C.\n\nc^2 = 35^2 + 5^2 - 350 cos (155)\n\nAnd solving for c will get apprx  39.59 meters.\n\nYou can then plug this back into the law of cosines to get the angle B:\n\nb^2 = a^2 + c^2 -2ac cos B\n\n5^2 = 35 ^ 2 + 39.59^2 - 2 * 39.59 * 35 cos B\n\nSolving for B will get you about 3.06 degrees.
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....
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where is the right angle in the triangle.  Between the right and bottom, between the bottom and left?  Without knowing this, we can't tell you the answer.\n\nBUT\nin a right triangle\nhypotenuse squared = side 1 squared + side 2 squared
110565
>>>> what is the altitude of a right triangle?\n\nI have several here, and I am at 12 metres above sea level.\nSeriously, you haven’t provided enough information. You need at minimum the length of 2 sides or a side and an angle as well as the orientation to be able to calculate height (altitude).\n\n>>>>>>if the equation is d=2r what is the d and the r what exectly \n>>>>>>>am I finding with this equation \n\nd = diameter. r= radius. You are finding either how far it is from one side of a circle to another (d), or how far it is form the edge of a circle to the centre (r).
458187
find the area of the base and multiply it by the height\n\n(1/2)*b*h1*h2  for a triangular prism h1 is height of the triangle, h2 is the height of the prism\n\npi*(r^2)*h for cylinder, h is height, r is radius of the base
184797
find the angle between two sides.
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
109605
Just use Cosinus Theorem: c^2=a^2+b^2-2abCos(gamma)\n1.\nc=sqrt(4+64-2*2*8*sqrt(3)/2))\nc=6.34\n\n2. Is not a right Triangle.. but anyway lets find the angle A:\n22^2=15^2+33^2-2*15*33*Cos(A)\nCos(A)=830/990\nA=ArcCos(83/99)\nA=33°1' 48.54"
579191
Area of a triangle:\n\nA = 1/2 (B)(A)\nwhere B = base which could be any of the three sides\nand A = altitude (or height), which is the perpendicular distance to the base B from the opposite vertex.
865033
okay. so. there's a special set of rules for a 30-60-90 triangle.\n\nthe sides are proportional to each other by the ratio \n\nx : v/( 3)x : 2x\n\nwhere the v/ is the square root.\n\nthat means that the greatest angle corresponds to the greatest length.\n\nbecause the longest side is 7.5 meters, that means that is 2x.\n\ndivide by two for x, and that is the 30* length. (because its the smallest angle, therefore is opposite the smallest length)