construct a triangle ABC bc=3.5 b=60 deg and altitude ad=2.5.and draw its circumcircle?
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file:///c:/Documents%20and%20Settings/Kelley/My%20Documents/Word%20Documents/triangle2.mht\nCopy the code into notepad and save it with the name triangle.html, then open it in web window.
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First, you need a protractor. Create a right triangle ABC with C as the right triangle. Let angle B be your angle. Make rough measurements of the three sides of the triangle. now:\nsin B = AC/AB\n\nOnce you know its sine, use these formulae for the other trigonometric functions:\ncos B = sqrt(1 - sin² B)\ntan B = sin B/cos B\n\nSpecial cases are:\nsin 0 = 0\nsin 90 = 1\nsin 30 = 1/2\nsin 60 = sqrt3/2\nsin 45 = sqrt2/2\nFrom these, you can compute all angles.\nremember,\nsin B = sin (B - 360n) where n is an integer\nsin B = cos (90 - B)\ncos B = sin (90 - B)\ncos (-B) = cos B\nsin (-B) = -sin B\n\nexample:\ncalculate sin, cos and tan of -150.\nfirst, sin -150 = -sin 150\n-sin 150 = -cos(90 - 150)\n= -cos (-60)\n= - cos 60 = -sin(90 - 60) = - sin 30\n= -1/2\nsin -150 = -1/2\ncos -150 = cos 150 = sin(90 - 150) \n= sin (-60) = -sin 60\n= -sqrt3/2\ntan -150 = sin -150/cos -150\n= (-1/2)/(-sqrt3/2)\n= sqrt3/3\n\nJust use these sign rules:\nif angle is in Q1 = sin cos and tan are +\nif angle is in Q2 = sin is +,cos is -,tan is -\nif angle is in Q3 = sin is -,cos is -,tan is +\nis angle is in Q4 = sin is -,cos is +,tan is -\n\n^_^
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the side length is 2\nso the altitude should be 3^(1/2)\ngood luck
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Draw the two points (a,b) and (c,d) on a graph so that the segment connecting them is NOT horizontal or vertical. Then make a right triangle with one vertical leg and one horizontal leg where the connecting segment is the hypotenuse.  \n\nThe length of the horizontal leg is c-a\nThe length of the vertical leg is d-b\n\nSo the length of the hypotenuse, which is the distance from (a,b) to (c,d), is\n\nsquare root of (c-a)^2+(d-b)^2
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I wish I remembered terms for this stuff.\nFind the length of the base, divide that by two.\nUse this number to find the length of a line drawn from the point of the triangle to the base. Congrats, that number is the height.\nTriangle ABC.\nLength from point A to C is 3.\nFrom A to B is 4.\nSo B to C is 5.\nTake the base, we'll say line AC, and divide that by 2, so you get 1.5.\nUse that number, now, as the length from A to D.\nSo, 1.5 squared plus 4 squared is 18.25. Square root of 18.25 is 4.27, so your height is 4.27.\nPythagorean Theorem.\nA^2 + B^2 = C^2
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what you want to do is draw an isosoles triangle with the 2 congruent angles the size you want. \n\nStart with a base with a certain length (use the compas to measure, say 4 inches).  Next determine the midpoint by marking arcs nboth above and below your line from each endpoint using the same compass diameter, then drawing a line through the two intersections.  You now have a right abngle at the midpoint.  You can now determine how tall your triangle must be using the sin table and use your compass to mark the point on the vertical line to complete the triangle with the angle you want.
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Several ways. The first that comes to mind is to figure the area with Heron's formula (area = sqrt(s(s - a)(s - b)(s - c)), where s = (a + b + c)/2). The area can also be figured as 1/2 base times altitude, so you know the altitudes of the triangle. The altitudes give you right triangles where you know the hypotenuse (a side of the triangle), and a leg, so you can compute the trig functions of all the angles. Then the angles can be computed as arcsin or arccos of the angle (arcsin(altitude/side))
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You can draw a triangle with base of 6 and height of 8.\n\nAs the area of a triangle is one-half*base*height, as long as the product of your base and height is 48 units, use those dimensions for your triangle. For example, you could draw a right triangle with base of 6, height of 8, and hypotenuse of 10, but it doesn't *have* to be a right triangle. You can also draw an isoceles triangle with base of 6 and height of 8 - make your base, start your height line at the "3" mark, and then form the rest of the triangle by dropping two segments from the top of the "height" segment to each end of the "base" segment.
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In rt triangle ABC with B= 90degrees (draw figure)      \nIf C is theta, cosC=7/8(ignoring sign)\ni.e., BC = 7, CA = 8 (in simplest terms)\nusing pythagoras theorem, AB= sqrt(15)\n \nsince C=theta is in the second quadrant tan C will be negative\n\nso tan C = AB / BC \n\n  tan C= - { sqrt(15) } / 7\n\ntan(theta) = -0.5533
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Draw the triangle with the 20 degree angle.  Put a dot on the hypotenuse to represent the object.  Its mass goes straight down from the spot (gravity).  A perpendicular to the hill through the point represents the resultant.  The horizontal vector connects the bottom of the mass vector to the bottom of the resultant (perp.) vector making a 20-70-90 right triangle.  Tan (20) = horiz/vert (vert = mass) \nso horiz = mass times tan(20)
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1/2 bh or bh over2.\nType triangle is inconsequential.\nb=base  h=height\nMeasure both, multiply, divide in half.
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(Click on the link below for the diagram. Expand to get a clearer view)\n\nLet the point of intersection of altitude and base be P\n\nThe length BP and PC is x.\nThe length AB = AC = (40-x-x)/2 = 20-x\n\nThen we use pythagoras theorem to find x.\nx^2 + 8^2 = (20 - x)^ 2\n40x = 336\nx = 336/40 = 8.4\n\nHence the area of the triangle\n= 1/2(base*height)\n= 1/2(16.8*8)\n= 67.2
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It is best that you draw a triangle.\n\nLet x be the distance between fire and tower A.\nLet y be the distance between fire and tower B.\n\n(sin 34)/x=(sin(180-34-14))/30\n(sin 14)/y=(sin(180-34-14))/30\n\nx=22.574, y=9.766\n\n0.5*x*y*(sin 132)=0.5*d*30 (area of triangle)\nplug in the numbers and d=5.46km\n\nThere might be simpler methods, but this is the first thing to comes to my mind.
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5.196 (3 times root 3)\nDive the triangle into 3 equal triangles.  Then divide each one of those triangles into 2 triangles.  The hypotnuse of each triangle has a length of 2, and by trig. (it is a 30-60-90 triangle) the base has a length of the square root of 3 and the height is 1. Area of a triangle =(base times height)/2=(radical 3)/2.  There are 6 smaller triangles total, so multiply the area by 6=total area=3 times radical 3 (5.196)\nHope this helps, it would be a lot easier if I had a picture that I could post and it would make more sense.
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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
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if you dont have one already, make a nice big copy of the entire circle with everything labeled.  repetition and review is really just the best way to remember anything.  when i try to think of the unit circle values for 30-60-90 triangles, i just think of it in terms of the long side and the short side.  the short side is always 1/2 and the long side is always (sq. root of 3)/2.  if i need to remember the values for, say, a 120 degree angle, i always draw a quick little sketch- just enough to show which are the short and long sides.  the horizonal is short, so the cosine is 1/2.  the vertical is long, so the sine is (root 3)/2.  this method works for all angles that are multiples of 30 degrees.  45, 135, 225, and 315 degrees are easy because you only have to remember one value for all the sine and cosines: (root 2)/2\n\njust dont forget to look at your sketch to see which of the values is negative! (and if the picture method doesnt work for you, theres a trick to knowing those too . . .\n\nhere are the quadrant numbers: (imagine the axis in between)\n\n2 1\n3 4\n\nuse this acronym to remember what trig functions are positive in each quadrant:\n\nALL---------> ALL of them are positive\nStudents----> Sine is positive\nTake--------> Tangent is positive\nCalculus----> Cosine is positive
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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....