Can you have a triangle with two obtuse angles?why or how?
199438
No I just tried it it's like impossible.
464777
okay. so. there's a special set of rules for a 30-60-90 triangle.\n\nthe sides are proportional to each other by the ratio \n\nx : v/( 3)x : 2x\n\nwhere the v/ is the square root.\n\nthat means that the greatest angle corresponds to the greatest length.\n\nbecause the longest side is 7.5 meters, that means that is 2x.\n\ndivide by two for x, and that is the 30* length. (because its the smallest angle, therefore is opposite the smallest length)
119907
Your question is a little bit generic. You can draw a lot of "angles and triangles" with a compass.\nThe simplest procedure that comes to my mind is this:\n1) Trace an horizontal segment from a point A to another point B.\n2) Now point your compass in A, with radius equal to AB and trace an arc over the segment.\n3) Then point in B with the same radius and trace another arc. The two arcs will intersect in a third point C.\n4) If now you join A with C and C with B, you have a triangle with all segments and angles equal.\nBe more specific if you need more help.
159608
Given: \n1) a right triangle\n2) hypotenuse = 2\n3) the other two angles are 70 and 20 degrees\n\nFind:\nLengths of the remaining sides\n\nNote: \nWe have to be more specific here.\n\nLet:\na = side opposite the 70 degree angle (side adj. 20 degrees)\nb = side opposite the 20 degree angle (side adj. 70 degrees)\nc = hypotenuse = 2\n\nSolution:\n\n1) For side a:\n\nsin 70 = opp./hyp. = a/c\nTherefore, a = csin70 = 2sin70 = 1.88 ANSWER\nOR\ncos 20 = adj./hyp. = a/c\nTherefore, a = ccos20 = 2cos20 = 1.88 ANSWER\n\n2) For side b:\n\ncos 70 = adj./hyp. = b/c\nTherefore, b = ccos70 = 2cos70 = 0.68 ANSWER\nOR\nsin 20 = opp./hyp. = b/c\nTherefore, b = csin20 = 2sin20 = 0.68 ANSWER\nOR\nPythagorean Theorem: a^2 + b^2 = c^2\nb^2 = c^2 - a^2\nb^2 = 2^2 - 1.88^2\nb^2 = 0.4656\nTherefore:\nb = 0.68 ANSWER
199251
All you have to use is the fact that for any triangle, the measures of the three interior angles add to 180 degrees.  So if you have Triangles ABC and DEF, then you know:\nangleABC + angleBAC + angleBCA = 180 degrees, and\nangleDEF + angleEDF + angleEFD = 180 degrees.\n\nIf two angles are congruent, then they have the same measure in degrees.  So, if angleABC is congruent to angleDEF, and angleBAC is congruent to angleEDF, then:\nangleABC = angleDEF, and\nangleBAC = angleEDF.\n\nUsing substitution:\nangleABC + angleBAC + angleBCA = angleDEF + angleEDF + angleEFD\nangleDEF + angleBAC + angleBCA = angleDEF + angleEDF + angleEFD\nangleDEF + angleEDF + angleBCA = angleDEF + angleEDF + angleEFD.\n\nthen simplify by subtracting (angleDEF + angleEDF) from both sides:\n\nangleBCA = angleEFD.\n\ntherefore, the third angles are congruent.
429027
You need one more datum, either the third side or any angle.  If it were a right triangle, the right angle would comprise the third datum.
389628
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.
339660
There's not enough information to answer the question. We need to know at least one angle as well as the length of two sides.
209753
The cosine laws apply to triangles which are NOT right angled. I think you have maybe read that wrong:\n\nhttp://www.themathpage.com/aTrig/law-of-sines.htm\n\ngives you a full guide to using the cosine laws to solve problems like these.
137296
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
531363
Whenever I come across questions like this one, I realize two things:  (1) geometry students often feel burdened by all the strangely-named 'rules' (axioms, postulates, etc.); and (2) proofs can be intimidating!\n\nI'd like to try making this one simple, citing as few theorems as possible.  First, I will just say it "in English" for you.\n\nWhen two lines cross to form an X, you easily see at least 4 angles.  The up-down pair of angles (I'll call them v and ^) are 'vertical angles' and so are the right-left pair (I'll call them < and >).  Someone could have come up with a better name than 'vertical' ... especially since 'vertical' makes you think of "standing up straight."  Here, 'vertical' means "of the vertex."  Since the lines cross and form a vertex, the angles get called vertical.\n\nNow, why would the angles v and ^ be the same measure (congruent)?  The easiest answer is "Because they both have the same supplement."\n\nWhich supplement do they have in common?  You could say either < or >, it doesn't matter.\n\nOf course, I'm being a bit lazy by not using geometry-standard terms, so now we must label everything correctly and formally:\n\nGiven two intersecting lines, AB and CD, whose intersection is at the point X.  Prove that angle AXC is congruent to DXB.  (Hopefully your drawing will match my setup!)\n\nPROOF:\n\n(1) Angle AXC is supplementary to angle CXB.\n Reason: Angles AXC and CXB form a straight angle.\n\n(2) Angle DXB is also supplementary to angle CXB.\n Reason: (Same as Step 1)\n\n(3) Angle AXC is congruent to angle DXB.\n Reason: Two angles that are supplementary to the same third angle, are congruent.\n\nHope this sets you on your way!
483936
1)One leg of a right triangle is 12 cm long. The diference between the other two sides is 2 cm.\nFind the sides of the triangle. (Pythagorean Theorem)\n\nWhat we can take from this problem is that one LEG is 12 cm long, while the other LEG is unknown (we will call it X) and the HYPOTENUSE -- the longest side of a right triangle -- is also unknown (we will designate it X+2, since it differs 2 cm from the unknown leg).\n\nSetting up Pythagorean Theorem as always:\n\n "Leg squared" + "Leg squared" = "Hypotenuse squared"\n\n  12^2 + x^2 = (x+2)^2\n\nExpanding:\n\n  144 + x^2 = x^2 + 4x + 4\n\nCollecting like terms:\n\n  4x = 140\n\nDivide by 4:\n\n   x = 35\n\nThe unknown leg is 35 cm, while the hypotenuse is 37.  Thus, we can also call the triplet (12,35,37) a "Pythagorean Triple" (of which there are infinitely many!).\n\n\n2) The hypotenuse of a right triangle is 58 cm long. The diference between the other two sides\nis 2 cm. Find the sides of the triangle. (Pythagorean Theorem\n\nIn this case, we know the hypotenuse, but do not know the lengths of the legs.  For convenience, we will again use the expressions X and (X+2) for each side's length.\n\nBy P.T.:\n\n x^2 + (x+2)^2 = 58^2\n\nExpanding:\n\n x^2 + x^2 + 4x + 4 = 3364\n\nCollect like terms:\n\n 2 x^2 + 4x - 3360 = 0\n\nDivide by the GCF (2):\n\n  x^2 + 2x - 1680 = 0\n\nCan we factor this trinomial easily?  We could, if we knew integer factors of -1680 (one positive, one negative) that combine to make 2.\n\nLuckily, -1680 can be factored into -2*2*2*2*3*5*7, and these can be grouped into two sets:\n      2*3*7 = 42\n -2*2*2*5 = -40\n\nFactor:\n\n (x+42) (x-40) = 0\n\nAlthough two solutions come from these factors (-42 and 40), we may only consider the positive answer, since lengths of triangle's legs are always positive!\n\n Therefore, x=40 (Leg 1), and Leg 2 is 42 cm.
350677
C - 3 possible answers\n\nIn any given right triangles with the two other angles being a and b,\n\nsin a = cos b AND sin b = cos a,\n\n\nas such, the first and fourth combination will give one answer, the other two will either give square trigonometric values of each angle i.e. sin^2 a and cos^2 a  OR sin^2 b and cos^2 b.\n\nThe only condition that will not satisfy this is when the angle is 45 degrees, which means a = b and contradicting the given statement of a being less than b.\n\nThus the answer is C.
516743
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?
672772
1) The triangle is isosceles.\nThe other two angles are equal.\nLet m<A = x, then m<B = x\nx+x+62 = 180\n2x = 180-62\n2x = 118\nx = 59\nThe largest angle is 62, so the side opposite that angle has the longest length.\nAnswer: c) AB\n2) Since the x-xoordinates are the same, you just have look at the y-coordinaes\n2 + 3 = 5\n5 + 3 = 8\nAnswer: (0,8)\n3) The sum of the interior angles of an n-sided polygon is \n180(n-2)\nSet this equal to 1260 and solve for n\n1260 = 180(n-2)\n4) \n5) Add all the angles together and set it equal to 180(4-2)\nx + x + (x+15) + (x+45) = 180(4-2)\nSolve for x.\n\nWhy are you removing your questions?  They were not short questions and people actually spent time answering them.
87192
You can't have 3 corners at 90 degrees without the fourth also at 90 degrees.\nYou can have two at 90 and the remaining add up to 90 degrees.\nA penta (5sided) gram, with even sides and angles, where the angle is 360/5 = 72 degrees.
356530
D.  The angles have to add up to 180.  If 2 of the angles are the same, then the 3rd has the same.  You have to get an enclosed side to be the same for the triangles to be congruent.