construct a triangle ABC bc=3.5 b=60 deg and altitude ad=2.5.and draw its circumcircle?
595970
draw a line maesuring 2.5 cm..this is bc.then using a protractor make an angle of 60 degrees n on the angle measure the other side of 3.5 cm.join the end points and u get ur triangle.The circuncircle is a circle that touches all 3 vertices.then draw the perpendicular bisector of any two sides..the intersectin point becomes ur centre for the circle.now with the radius equal to the distance between the centre n the vertex draw a circle..tada!
811153
Hi, \n\nThe answer lies in the pythagorean theorem. \n\nFor any right (one of the angles is 90 degrees) triangle with sides a,b and c where c is the side opposite the 90 degree angle:\n\na^2 + b^2 = c^2\n\nso for your first excercise:\n\na = 4\nb = 4 Sqrt(3)\nc = ?\n\na^2 + b^2 = c^2\n4^2 + {4 Sqrt(3)}^2 = c^2\n16 + 4^2 * 3 = c^2\n16 + 16 * 3 = c^2\n64 = c^2 \nc = +- 8 (since we're dealing with a triangle, -8 doesn't make sense, so just ignore it.)\n\nc = 8. \n\nFollow that same pattern, and you should be all set. \n\nHope that helps, \nMatt
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"
644806
if the triangle is a regular triangle\n\na = (1/2)bh\n\na = (1/2)(6)(6 * sin60)\n\na = (1/2)(6)(3√3)\n\na = 9√3 = 15.588\n\n\nfor a regular polygon, \narea = (n*t^2) / (4*tan(π/n))\n\nwhere n = number of sides & t = length of side
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
459122
the side length is 2\nso the altitude should be 3^(1/2)\ngood luck
1091583
The altitude of the triangle is 5 times the square root of  3, or 5 times 1.7.\nArea of a triangle = 1/2 bh, so Area = 1/2 times (10)  times (5 times 1.7) = 42.5 or more accurately 43.30127019 using the correct value of the square root of 3.
1029855
D\n\nSeriously, what class are you in? Algebra? Geometry? You chould know this!! It's the pythagorean theorm- [a2 +b2 =c2] (2's = squared).  You have a and b, the two adjacent sides of the triangle.  Therefore, it is the square root of 50, or approx. 7.07.  Thats SUPER important, make sure you remember it!
918147
You have a right triangle with legs 106 and 88, a & b respectively. By the Pythagorean Theorem, the hypotenuse is square root of 18980. Now, use related rates. (dc/dt)2c=(da/dt)2a+(db/dt)2b. That becomes (dc/dt)*2*sq. rt of 18980=19(212)+22(176). dc/dt=28.67 knots per hour.
320776
The area S of a triangle is S = ½bh, where b is the length of any side of the triangle (the base) and h (the altitude) is the perpendicular distance between the base and the vertex not on the base.
328129
just draw a triangle has right angle which dimensions 1,sqrt(3),and hypothenus 2. if you want it, i can send to you
539310
After 2.5 hrs, Boat A has gone 50 km, and Boat B 80 km. Let the distances of A and B be two sides of a triangle, a 70 degree angle connecting them.\n\nThus we have a triangle with two known sides and one known angle.\n  \nLabel it A, B and C starting from the bottom left angle. Label each angle's opposite side its respective lowercase (a,b, and c). Fill in for b and c, and angle A. We don't know if this is a right triangle or not, so we're going to need to use the law of cosines:\n\na^2 = b^2 + c^2 - 2*b*c*cos(a)\nSubstitute for each known variable:\na^2 = (50)^2 + (80)^2 - 2(50)(80)cos(70 degrees)\na^2 = 2500 + 6400 - 8000 * cos (70 degrees)\na^2 = 8900 - 8000 * .34202\na^2 = 6163.84\na = 78.5101\n\nThus, using the law of cosines, the distance (a) from the two boats is approximately 78.5101 km. It is not exact as there is no exact value for cos (70 degrees).
366178
Take the average between the side of an inscribed square\nand that of an inscribed hexagon. With a hexagon you\nget an isosceles triangle with base as a side and a 60\ndegree angle at the apex, leaving 120 degrees to be split\nbetween the equal pair which are then also each 60 degrees.\nSo you have an equilateral triangle whose base is equal\nto the radius of the circle and of course the base is a side\nwhich has a length of 1. \n  Now inscribe a square. There are now 4 isosceles\ntriangles with a 90 degree angle at the apex and a pair\nof 45's at the base. Draw an altitude from the apex, \nperpendicular to the base, thus bisecting the 90 at\nthe apex making another 90, 45, 45 triangle whose\nside is 1/2 the side of that of the square. The hypotenuse\nis a radius of the circle so the side of the last mentioned\ntriangle is 1/2^2 (.707). This makes the side of the\nsquare 1.414. So averaging 1 and 1.414 you get 1.207.\n Using trig we find the real answer to be 1.1755.\nSo the percentage of error is 2.67 percent from 1.207.\nI wouldn't win an engineering prize with that kind of\naccuracy, but I think I could build a pretty good\ncatapult.
701162
Draw the triangle on an x-y plot with the right angle at 0,0 and the other two points at (12,0) and (0,5). First write an expression for the hypotenuse line.  The y-intercept is B = 5 and the slope is:\nm = rise/run\nm = -5/12\n\nthe equation of the line would then be:\ny = mx + B\ny = (-5/12)x + 5\n\nnow calculate the area of a rectangle:\nA = x*y\nA = x*((-5/12)x + 5)\nA = (-5/12)x^2 + 5x\n\nto get the maximum you need to take the derivative and set it to 0:\ndA/dx = 2*(-5/12)x + 5\n0 = (-5/6)x + 5\n(5/6)x = 5\nx = 6\ny = (-5/12)x + 5\ny = (-30/12) + 5 = 2.5\n\nYou should be able to do part b on your own now.\n\nFor part c just replace the numbers with variables:\na will be the length and b will be the height\n\nThe y-intercept is B = b and the slope is:\nm = rise/run\nm = -b/a\n\ny = (-b/a)x + b\n\nnow calculate the area of a rectangle:\nA = x*y\nA = x*((-b/a)x + b)\nA = (-b/a)x^2 + bx\n\nto get the maximum you need to take the derivative and set it to 0.  Treat b and a just as if they were actual numbers:\ndA/dx = 2*(-b/a)x + b\n0 = (-2b/a)x + b\n(2b/a)x = b\nx = a/2\ny = (-b/a)x + b\ny = (-b/a)(a/2) + b\ny = (-b/2) + b\ny = b/2
120928
ok, let the triangle be ABC with sides AB, BC, AC.\ndraw a typical line, get too point on in with the distance of BC, name them B and C. then draw an arc with the center of B and the length of AC and another one eith the center of C and the length of AB. if they had a common point, it would be A and your triangle is there, but if not, there is no triangle with those three sides.
155419
the base of a triangle times the height times 2\n\nb*h*2
919521
Imagine the center of the triangle. Draw a line segment from it to a vertex, and then one from the center to a midpoint of a side. This makes a 30-60-90 triangle, so the line segment is (p/2) sqrt(3). This is a radius of the cylinder, so we multiply it by 2 to get the diameter.\n\nd = p sqrt(3)\n\nor, as the question wants it\np = d/sqrt(3) = d sqrt(3)/3