what does a picture of a triangle with two obtuse angles look like?
356323
Absurd... A triangle cannot have more than 1 obtuse angle, since the sum of all angles of a triangle is 180 degrees.
715400
Angle of line from horizontal position.
159884
It looks like a 90 degree angle:     |_
1106173
The direction (angle) of |u| and |v|?
611646
Simply bisect the obtuse triangle once and only once through point at which the A side and B side intersect through the middle of C. \n\nSince a triangle, by defintion, has 3 angles that equal 180 degrees, the smallest angles would be, using whole numbers for clarity, 178deg, 1deg and 1 deg. Bisecting the 178deg angle makes two 89deg angles, which is under the definition of an acute angle. \n\nSince an acute angle is defined as an angle less than 90deg, and a triangle is defined as having 3 angles with a sum of 180deg, we know that the 180deg can not be just one angle (180deg, when bisected, creates 2 right angles) and still be a triangle. Therefore, we know that for any triangle ABC with an obtuse angle at AB, bisecting the angle once, and only once, through AB to the midpoint of C will create two right triangles with angles of 90deg, 90deg - angle#1deg, and 90deg - angle#2deg.
669147
number of sides, congurency?, size of angles, distinguishing characteristics
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"
815746
It looks as if there's one big triangle, but in fact the smallest angle of the green triangle isn't the same as the smallest angle of the red triangle, and so although it looks as if there's one big triangle cut up, it isn't so.  And rearranging it makes the hole appear.  I can do a diagram showing just where the extra area unit is, but not on here.\n\nDo you know trigonometry? If so, notice the tangents of the two angles I mentioned are different.  One is 2/5 and the other is 3/7.\n\nIf you don't know trigonometry, just notice that the green triangle has the sides about the rightangle in the ratio 2:5, the red triangle has them in the ratio 3:7.  It's the swapping of the positions of those two triangles that takes up the extra area.  The green triangle has the smaller angle, so when it's the lower triangle the "hypotenuse" of the big triangle has an inward kink.  When the red triangle is the lower one it has an outward kink.  The space between these two "hypotenuses" is one square unit of area.
997563
You're looking for the angle that has a sine of 1/2.\nDrawing a triangle representing this is very helpful.  You will see that it is a 30-60-90 triangle.  \nIf you can find it in degrees, then you can convert it to radians by multiplying it by pi/180.\nsine(30) = 1/2\n30 degrees = 30(pi/180) = pi/6
37762
Obtuse angle is an angle more than 90 degrees.
1107584
|\n  | \n  |  A\n  |\n  |_______\n......B\n\nThis is what "A" perpendicular to "B" looks like.  I hope this helps!
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?
723693
I'm pretty sure you are correct.  I am also working thru Calc III (at the moment, actually).  It helps to picture the gradient as perpendicular to the tangent plane.
570600
Ok, first just so you get the figure right, draw dots n name them like i have n at the places like m showing them below\n      A                B\n      \n    E                   F     \n\n   D                      C             G\n\nNow, join AB, D-C-G, EF , A-E-D and B-F-C  also.. join\n A-F-G .. GET IT?? and you'll have AE equal to ED as well as BF equal to CF as the median of the trapezium joins the mid=points of the non-parallel sides...\n\nNOW IN triangle BFA and triangle CFG ,you have\nBF congruent to CF ----   Proved tht F midpoint of BC\nangle BFA congruent to angle CFG--vertically opposite angles\n and as AB is parallel to CD (opposite sides of a trapezoid) it is also parallel to CG ( construction was to join D,C and G , remember??\n\ntherefore, abgle ABF congruent to angle FCG --- alternate angles for parallel lines AB and CG ..\n\nSO YOU NOW HAVE triangle ABF congruent to triangle CFG --- Angle-Side-Angle Test of congruency...\n\ntherefore, AF congruent to FG --- corresponding sides for congruent triangles\n\nNOW, IN THE BIG TRIANGLE ADG,\n  E is the midpoint of side AD - by construction\n  F is the midpoint of AG  --- Proved AF congruent to FG\n \ntherefore... EF parallel to DG --- By the Midpoint Theorem which states that-- A line passing through the midpoints of any two sides of the triangle is parallel to the third side (and has half the length of the third side )\n\nSO WE'VE NOW PROVED EF // DG\n .\n. . , EF IS ALSO // TO DC --- D,C AND F LIE ON THE SAME LINE.. I.E. THEY ARE COLLINEAR\n\n SO, EF//DC\nBUT DC IS PARALLEL TO AB ---- OPPOSITE SIDES OF THE TRAPEZIUM ABCD\n \n .\n. . , EF IS ALSO PARALLEL TO AB\n\n .\n. ., WE CAN PROVE THAT ----> AB // EF // DC  (REASON- both AB and DC are parallel to the same linE EF)  \n                     \n\n .\n. . WE HAVE PROVED THAT THE MEDIAN OF THE TRAPEZIUM (in this case GE)  IS PARALLEL TO BOTH THE BASES (i.e AB and CD )
746093
It's straight ahead geometry.\nDraw a circle with center O.  Draw a diameter horizontally through the circle. Label the left intersect as A and the right intersect as B.  Now draw a vector from O to a point P on the circle.  What tdo you know about triangles AOP and BOP in terms of the angle the vector makes with the diameter?  Use law of sines, law of cosines, trig, etc. to show that angle APO and angle BPO add up to 90 degrees and your done.\n\n\nDoug
1159510
the four angles add up to 360 degrees
733210
You basically have 2 options. Use a protractor and measure them (The internal angles total 180 degrees).\n\nThe other option is using the trig functions (Sin, cos and tan), however you must have some info on the triangle eg. length of sides to apply any of these.