Draw me a triangle with 3 90 degree angles?
539039
The only way a triangle can have three 90 degree angles is if it is drawn on the surface of a sphere.
436114
well, you could be an architect and be using a 30-60-90 triangle and you needed to know a particular side of it. you can use x times the square root of 3, x being the shorter leg, to find the length of the longer leg.
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.
699841
Three are three angles in a triangle which all add up to exactly 180 degrees.
899386
The x's will all be the same, and the y's will be the difference between 4 & the number:\n\nA=(2,5)\nB=(0,3)\nC=(5,2)\n\nAnother way to do this is to actually draw the triangle and the line y=4. This will give you a visual of what should happen.
454825
We can do this 2 different ways.\n\nFirst: Use the Pythagorean Theorem. Since all angles are right angles, the diagonal (measure from corner to corner) is the hypotenuse of the right triangle formed.\n\nBy the Pythagorean Theorem, 10^2 + 10^2 =x^2, where x=diagonal\n\nx^2=200\nx=sqrt(200)\nx=10 sqrt 2\n\n2nd way: use 45-45-90 Triangles\n\nWhen you draw any diagonal in a square, you get a 45-45-90 degree right triangle. This is because the 2 sides are the same, so the 2 base angles must be also. The top is 90, so the others must be 45 and 45 to equal each other.\n\nThe sides to a 45-45-90 right triangle are always in the ratio 1:1: sqrt(2) where sqrt(2) is the hypotenuse.\n\nYou know that the sides are 10 feet. Therefore the hypotenuse is 10 sqrt2 feet
905611
If u r saying that u have three sides of a triangle as a,b,c\nAnd u want to determine angle b/w them, u can use the formula\ncosA=(b^2+c^2-a^2)/2bc where A is the angle formed by the sides b and c.similarly if B is the angle formed by a and c, it can be calculated as\ncosB=(a^2+c^2-b^2)/2ac\nonce u have calculated A and B, subtract their sum from 180, u will get third angle.\nRemember u will have to use inverse table for cos to find A and B
889937
A triangle has 180 degrees.  a right angle is 90 degrees.  a triangle has 3 angles.
722736
the angles are just unknown if i get you correctly.\nthere are sin, cos, and tan proportions you can work out to find the rest. i think you only need three of the measurements to work out the rest of the triangle, like 2 sides and an angle or 2 angles and a side.  not positive.
324000
ok, we can call the three sides x,y,and z respectively. all angles of a triangle have to add up to 180. so\nx+y+z=180\n\nok. y=2x and z=y+10\n\nsubstitute\nx+2x+2x+10=180 then solve for x\n\n5x+10=180\n5x=170\nx=34 degrees
844799
Wallstream is correct -- start with a right triangle.\n\nLabel the interior angles a, b and c, with c being the 90-degree angle.\n\nNow notice that a and b are complementary angles that add up to 90-degrees.\n\nLet x = sina.  Since the angles are complementary, we also have that x = cosb.\n\nSo we can rewrite as follows:\n\nIf x = sina, then a = arcsinx\nIf x = cosb, then b = arccosx\n\nBut a+b = 90-degrees (aka pi/2).\n\nSo arcsinx + arccosx = pi/2\n\nHope that helped!
977003
Draw a straight line .now plot three points S,T and R on it ,taking ST=3.8 cm and TR=1.4 cm.\nWrite the distances 38 in betweenS and T and 14 between Tand R.your sketch is complete.\nNow ,rite the followin unde the sketch:-\nSR=ST+TR=38+14=52.
322891
The obvious way is by trisecting each of the triangle's 60 degree angles and projecting the lines to the circle's circumference.\n\nBut trisecting an angle is impossible with simple geometric tools.  \n\nSee this for an interesting discussion:  http://www.jimloy.com/geometry/trisect.htm\n\n\n\nn.b.  a compass is a magnetic navigational instrument.\n\nThe drawing instrument used for scribing arcs and circles is a pair of compasses.
563121
a right angled isosceles triangle\n\nplease give me 10 points now
48873
well i would like to pretend its a regular hexagon, in this case you would have to draw a line from one vertex to an opposite one, the line should meet at the center of gravity (the point where all 3 lines intersect)
763822
Divide the right angle into three equal parts, or three 30-degree angles.\n\nSounds like homework to me (which, hey, I'm cool with!), as if you owned this parcel this is NOT the way you'd want to divide it for real -- you wouldn't want three neighbors all bunched up like that at the confluence of the former right angle.\n\n.
934725
Hi. Don't you hate these? OK you have an unknown angle. Call it a. You have a complement, c, which is 90-a. c=3a-10. Any help?