Triangle ABC is an equilateral triangle. Angle C= x+10. Find x??. Please help, ill leave u good points!!?
952714
x is 50 degrees. in an equilateral triangle all angles r same, so each angle is 60 because all angles in a triangle equal 180.\n\nc=x+10\n60=x+10\n-10   -10\n50=x
878828
here's a non-rigorous fiddle for\nthis one-the full proof is too\nlong-winded\n\nlet s=(a+b+c)/2 ie s=half the perimeter of\na triangle\na,b and c, are the sides of triangle ABC\n\nstandard formulae:\n\nsin(A/2)=sqrt{(s-b)(s-c)/bc}\ncos(A/2)=sqrt{s(s-a)/bc}\n\nthe area of ABC\n=1/2bcsinA  (sine rule)\n=1/2bc*2sin(A/2)*cos(A/2)\n(standard trig identity)\n=bc*sqrt{(s-b)(s-c)/bc}\n      *sqrt{s(s-a)/bc}\n=bc*sqrt{s(s-a)(s-b)(s-c)}\n             /sqrt{(bc)^2}\n=(bc/bc)sqrt{s(s-a)(s-b)(s-c)}\n=sqrt{s(s-a)(s-b)(s-c)}  as required\n\nmany a good tune has been\nplayed on an old maths fiddle \n\ni hope that this helps
713469
Can't u prove it ?\nuse simple geometrical angle laws.\n\nif we have a triangle : \nupside-- the inclined plane\ndownside--- the horizontal\nright side--- the gravitational force (vertical)\n\nDRAW IT........ as u know \n\nand then draw the perpendicual component on the plane     \n\nuse this: sum of the two angles inside the triangle    =180-90=90\n\nsum of angles at the gravitional point= 180-90=90 also\n\nu will prove the equality
966422
The area of an equilateral triangle is\nA = (1/4)sqrt(3) * a^2, where a is the length of the side.\nSince the triangle is equilateral, you know that the length of each side is 10/3.\nThe area is\nA = (1/4)sqrt(3) (10/3)^2\nA = (1/4)sqrt(3) (100/9)\nA = (25/9)sqrt(3)
563121
a right angled isosceles triangle\n\nplease give me 10 points now
841879
Impossible. The angle of an equilateral triangle is 60° and the angles get larger from there (square 90°, pentagon 108°, .... , n-gon (180 - 360/n)°)\n\nHowever there are two possibilities. \n1. If you meant that the interior angle of each triangular slice (from vertices to centre is 40, then\n360°/n = 40° so n = 9 ie it is a nonagon (9-gon)\n\n2. If you meant the exterior angle of the polygon is 40 then sum of exterior angles is 360\n\nie n * 40° = 360° so n = 9 ... again a 9-gon.
1094272
In your problem, (CE)(ED)=(AE)(EB). \n\nSubstituting, you get (6)(5)=(2)(EB)\nSimplifying, you get 30=2EB\nDividing 2 from each side, you get 15=EB\n\nEB is 15 units.\n\nFor proof of (CE)(ED)=(AE)(EB), connect CB, and AD,\nmeasure of angle BCD=1/2 the measure of arc BD\nmeasure of angle BAD=1/2 the measure of arc BD\nmeasure of angle ADC=1/2 the measure of arc AC\nmeasure of angle ABC=1/2 the measure of arc AC\nmeasure of angle BCD=measure of angle BAD\nmeasure of angle ADC=measure of angle ABC\nBy AA Similarity Theorem, \ntriangle CBE is similar to triangle DAE\nBy the definition of similarity,\nED/EB=AE/CE\nBy the cross product theorem,\n(CE)(ED)=(AE)(EB)
56351
a^2 + b^2 = c^2 where c is the hypotonuse of the triangle.\n               :&#92;\n          a  : &#92; c\n               :__&#92;\n                b\n\na and b are interchangable but the key is the c must be the site opposite the right angle.\n\nThen you just plug in the numbers if you are given a or b\nsay a=2 and c=3\n\nthen 2^2 + b^2 = 3^2\n4 + b^2 = 9\nb^2 = 9-5 = 4\nb = square root of 4 or 2\n\nIf you are given a and b it is somewhat easier\na=3 b=4\n3^2 + 4^2 = c^2\n9+16 = c^2\n25 = c^2\nc = square root of 25 \nc = 5
315798
I think what you're trying to explain is that all points A,B,C are on the cirumference of a circle, with the line AC being a diameter of the circle.\n\nThis forms a right angled triangle with the right angle at point B. That's just a rule of circle geometry when 1 side of a triangle in a circle is a diameter. By angle sum of a triangle, the angle at C must be 62 degrees.\n\nDraw in the line 0B, forming an iscoceles triangle OBC, since OB and OC are equal radii. In this triangle, angles at B and C are both 62 degrees, so angle at centre is 56 degrees. This is the angle that subtends the arc you want to find.\n\nIf you want the arc length you need the radius and use the formula\nl = 2 * pi * r * 56 / 360
880763
The coordinates of the vertices of the 3 rectangles, by symmetry, would be of the form, (0, +- 1, +-x), (+-1, +- x, 0), and (+-x, 0, +-1), where x is unknown.  (0,1,x), (0,1,-x), and (1,x,0) forms one equilateral triangle. Therefore we know that 4x^2 = 1+(x-1)^2 +x^2 .  This reduces to x^2 + x -1 = 0.  Solving for x, we have x =  1/2(-1 + sqrt(5)).   From this, we can arrive at the ratio of rectangle sides at 1/2(1 + sqrt(5)), which is the golden ratio.
253427
An equilateral triangle is a triangle where the sides all have the same length.  Also, all the angles are equal, and since the angles of a triangle add to 180, this means the angles are all 60 degrees.\nAn isosceles triangle is a triangle that has two equal side lengths, and the angles opposite these sides are also equal.  All equilateral triangles are also isosceles triangles.\nA scalene triangle has no sides of the same length and no angles of the same degree.
743138
join AC\nnow youhavetwo triangles ABC and ADC\nin triangle ABC <CAB+<B+<ACB=180*.............(1)<s of a triangle\nin triangle ADC <CAD+<D+<ACD=180*............(2)--------do--------\nadding (1) and (2)\n<ABC+<ADC+<B+<D+ACB+<ACD=360*\n=><DAB+<B+<BCD+<D=360*\nhence proved
966174
a triangle with angles 30, 60,90\nhas sides in ratio 1: (SR 3) : 2\nThe equilateral triangle can be formed from\ntwo 30,60,90 triangles.\neach of those can be formed from 3 smaller 30,60,90 triangles\nso the equilateral triangle can be formed from six 30,60,90 triangles.\nIf radius of circle is 1 unit,\neach of the six small 30,60,90 triangles has \nsides (1/2) ((SR 3)/2) 1\nso the equilateral triangle has side SR(3) units\nso if the equilateral triangle has side x, the circle has radius x /(SR(3)),\nand the circle circumference is 2 pi x /((SR 3))
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"
729289
what are you supposed to prove?\n\nedit:\nyou mean why the three vertices of an equilateral triangle are concyclic? i am unable to make out theproblem
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....
621583
What type of triangle is it?\nIf it is an equilateral triangle, all sides will be the same length, so p=(1.25 X 3)m\n\nHope this helps