In triangle ABC, angle ACB is 90 degrees.  If the altitude CD and the angle bisector AE meet at F prove that CE=CF
134019
Let CAB = x\nCAE = x/2\nEAB = x/2\n\nADF is a right angled triangle with 90 at D.\nDFA = 90 - x/2\nCFE = 90 - x/2 (vertically opposite angles)\n\nCAE = x/2. In triangle CAE, 90 at C.\nCEF = CEA = 90 - x/2\n\nIn CEF triangle, 2 angles are equal. Sides opposite to equal angles are equal. So CE = CF.
55614
Using the abc method… \n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nFe-->ax=2c------eq.1\nO-->ya+4b=12c+2d+e----eq.2\nH-->2b=2e----------eq.3\nS-->b=3c+d----------eq.4\n\nLet a=1\n\nFrom eq.1\nax=2c\nx/2=2/2c\nx/2=c-----eq.5\n\nForm eq.3\n2b=2e\n2/2b=2/2e\ne=b-----eq.6 \n\nFrom eq.2\nYa+4b=12c+2d+e\n1y+4b=12(x/2) +2d+b (Form eq.5&6)\nY+3b=6x+2d\n2d=y-6x+3b\nd=(y+3b-6x)/2------------eq.7\n\nFrom eq.4\nb=3c+d (but c=x/2)\nb=3(x/2) +d\nReplacing eq.4 by eq.7 we will get\nb=3x/2+(y+3b-6x) ½\nb= (3x+y+3b-6x) ½\n2[b=(y+3b-3x) ½]\n2b=y+3b-3x\n3x-y=b-------eq.8\n\nReplacing eq.6 by eq.8\ne=b\ne=3x-y\n\nReplacing eq.7 by eq.8\nd=(y+3b-6x)/2\nd=[y+3(3x-y)-6x]/2\nd=[y+9x-3y-6x]/2\nd= (3x-2y)/2\n\nFrom the above calculation we know that:\na=1, b=3x-y, c=x/2, d=(3x-2y)/2 & e=3x-y. \n\n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nWhen replacing the values we’ll get:\n[1]FexOy+[3x-y]H2SO4 -->[x/2]Fe2(SO4)3+[(3x-2y)/2]SO2 +[3x-y]H2O
291604
A circle has 360 degrees. Now imagine that from the icosagon you can make 20 equally-sized triangles. The angle made by each triangle is 360/20=18 degrees. The other two angles of each triangle are both 81 degrees.\n\nThe basic formula for the area of a triangle is 1/2 base x height. How do we find these? Well, imagine that from each of the 20 triangles we make two additional right triangles. In doing this we know several things:\n\n-The hypotenuse is equal to 6.5 cm (the radius).\n-The angles of the smaller triangle are now 9 degrees, 81 degrees, and 90 degrees.\n-The sine of 9 degrees is equal to the opposite side divided by the hypotenuse. If we multiply this value by 6.6 we get the length of the opposite side ("1/2 base").\n-The sine of 81 degrees multiplied by 6.5 gives us the "height".\n-Once you compute each value (1/2 base and height) you multiply them by each other and you get the area of the triangle.\n-Multiply the area of that triangle by 20, since you have 20 of them. Voila.
610642
C-120... Two supplementary angles must equal 180.... 60+120=180\nB-32 feet... 8*4=32
424744
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
561616
Draw a picture of the two congruent sections, which are connected at B. The exact angle between them doesn't matter so long as it's not greater than 90 degrees. You will see that if AD=CD, then angles A and C are equal. So what happens as angle A gets larger and angle C gets smaller? You should be able to see the answer for yourself.
596719
Let v1=C-B , v2=A-C , v3=B-A be the three displacement vectors making up the sides of the triangle. Define f1,f2,f3 to be the three forces X,Y,Z each rotated 90 degrees counterclockwise. Then f1 is parallel to v1, and so on. That is:\n\nf1 = s1*v1\nf2 = s2*v2\nf3 = s3*v3\n\nwhere s1,s2,s3 are three scalars. It's only necessary to show that s1=s2=s3 to prove that the three forces are proportional to the three opposite side lengths (|f1|/|v1|=|f2|/|v2|=|f3|/|v3|). This is easily done as follows:\n\nSince the forces are in equilibrium (f1 + f2 + f3 = 0):\n\ns1*v1 + s2*v2 + s3*v3 = 0\n\nMultiplying (v1+v2+v3)=0 by s1 and subtracting from the above equation:\n\n(s2-s1)*v2+(s3-s1)*v3 = 0\n\nSince v2 and v3 are linearly independent both coefficients must be zero\n\n-->\n\ns1 = s2 = s3\nQED.
355681
Since the triangle is Isoceles, if you split the triangle down the middle you can turn in into a right triangle. You can use the Pythagorean theorem. (a^2 + b^2 = c^2.)\n\n13 will be the hypotenuse and 5 will be the base. To find the altitude you just fill everything in. \n\n5^2 + b^2= 13^2\n25 + b^2 = 169\nb^2 = 144\nb = 12
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180/7 °\nor 25 5/7 °\n\nSee picture:\nhttp://mudandmuck.com/str2/triangle2.JPG\n\n180-6x+2x = 3x since <A = <C (ΔABC is isoceles)\n180 = 7x\n180/7 = x
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.
19879
When two angles are supplementary, their sum is 180 degrees.  For example:\n\na + b = c\n100 + 80 = 180, therefore a and b are supplementary.\n\nIf a + b = 90, the angles are complementary.
144762
I assume you mean the altitude from the top point down to the "base" (bottom) of the triangle, since all triangles have 3 altitudes.\n\nTo find the altitude, you can cut the triangle in half, to make two right-angled triangles.  That gives you a triangle with A as one side (unknown length, the altitude) B as the other side (5--half of 10) and C as the third side (13).  Since these are right triangles we can follow Pythagorean's theorm of a^2+b^2=c^2 to solve.\n\nA^2+5^2=13^2\na^2+25 = 169\na^2= 144\na=12\n\nThe altitude is thus 12.
118985
Draw a triangle ABC such that AB is at the bottom and C is at the top. Assign corresponding sides: AB = c, BC = a, CA = b.\nNext, draw an altitude from C, call it CD.\nTherefore, there are now 2 right triangles, ADC and BDC.\n\nsin A = opp./hyp.\nsin A = CD/AC\nsin B = CD/BC\n\nThus,\nCD = AC sin A\nCD = BC sin B\n\nUsing transitive property:\nAC sin A = BC sin B\n\nDividing sin A sin B from both sides:\nBC/sin A = AC/sin B ----- (1)\n\nNext Draw a different altitude, from B, and call it BE.\nSo, there are again 2 new right triangles, CEB and AEB.\nTherefore,\n\nsin A = opp./hyp.\nsin A = BE/AB\nsin C = BE/BC\n\nThus,\nBE = AB sin A\nBE = BC sin C\n\nUsing transitivity again:\nAB sin A = BC sin C ----- (2)\n\nDividing sin A sin C from both sides:\nAB/sin C = BC/sin A\n\n\nWe now have 2 identities (1) and (2):\n(1) BC/sin A = AC/sin B\n(2) AB/sin C = BC/sin A\n\nUsing transitivity:\nBC/sin A = AC/sin B = AB/sin C\n\nbut AB = c, BC = a, CA = b, so\n\na/sin A = b/sin B = c/sin C\n\nThis is called the "Law of Sines"\n\n^_^
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The simplest solution to this problem can be done by using the "Sine Rule". If you do not know this rule you can find it in any standard textbook of trigonometry. The rule is stated below. \n\nIf the side BC, CA and AB of a triangle be represented by a, b and c respectively and if angles <BAC, <CBA and <ACB be represented by A, B and C respectively, then the area (say, A0) of triangle ABC is given by\n A0 = 0.5*a*b*sin(C) = 0.5*b*c*sin(A) = 0.5*c*a*sin(B).\n\nNow, in this case BF = (2/5)a and CF =(3/5)a [since BC = a; BF and CF are in the ratio 2:3]. Similarly, AE = (1/8)b; CE = (7/8)b; AD = (1/4)c; BD = (3/4)c.\n\nApply sine rule to triangles ADE, BDF and CEF and denote their respective areas as A1, A2 and A3. \n\nFor first triangle we can write\nArea of triangle ADE i.e. A1 = 0.5*(b/8)*(c/4)*sin(A)____(1)\nRecall, area of triangle ABC i.e. A0 = 0.5*b*c*sin(A)____(2)\nDividing equation (1) by equation (2) we get A1/A0 = 1/32.\n\nRepeating similar procedure with triangles BDF and CEF we obtain A2/A0 = 3/10 and A3/A0 = 21/40 respectively.\n\nNow, area of triangle DEF is equal to \n(A0 - A1 - A2 - A3) \n= {1-(1/32)-(3/10)-(21/40)}*A0\n= (23/160)A0\n\nAs A0 = 160 units so area of triangle DEF is 23 units.
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First we note that the two bases of the trapezoid are parallel and the segment joining the midpoints of the altitudes is parallel and equidistant from the bases. Therefore the vertical displacement of the midpoints of the bases from the midpoints of the altitudes is equal.\n\nIf we also contruct the segment joining the midpoints of the bases we see this intersects our other segment at its midpoint. We have now cut our quadrilateral into four triangles and it is an easy task to show that opposite triangles are congruent using SAS.
172941
you may be dealing with "right" triangles, but you don't identify the angle that is 90, so your question is bout impossible to answer. this type of question usually involves  \na squared + b squared = c squared (a^2 + b^2 = c^2) where a and b are the 'legs' of the triangle in question. (the two 'legs' of a right triangle make the 90 degree corner) :)
676750
a) the 90 degree angle