In triangle ABC, m<A= 30* and m<C =90*.  Find AB if BC = 8?
485132
B = 180º - 90º -30º = 60º\n\ncos B = 1 / 2 = BC / AB = 8 /AB\n\nSo AB = 16
843483
using (x+y)^3 = x^3+3x^2y+3xy^2+y^3\n\nwe get (2a+b)^3 = 8a^3+12a^2b+6ab+b^3\n\n(ab+c)^3 = a^3b^3+3a^2b^2c+3abc^2 + c^3\n\n(6ab-c)^3 = 216a^3b^3-108a^2b^2c+18 abc^2 - c^3
747277
5a+6b-3a+b=?\nSimplified= (2a+7b=?)  or (2a = -7b) or (7b = -2a)\nThey are all the same\n\n15x-3y-7x-4y=?\nSimplified = (8x-7y=?)  or (8x = 7y) or (-8x = -7y)\n\n15m-17n-23m+25n=?\nSimplified -8m+8n=? or (8n=8m) or (n=m)
122942
Any answer with B>90 is acceptable satisfying:\nA=2B-180\nand C=180-B-A\n\nfor example: (A,B,C)=(20,60,100) or (40,30,110)\n\nNote that ABC and DBC are similiar.
459896
you need to use the cosine rule.\n\nA^2 = B^2 + C^2 - 2BC*cos(a)\n\nwhere Capitals are the lengths of sides, and lower case is an angle opposite the side of the corresponding letter.
273778
The lengths of a right triangle are a^2 + b^2 = c^2 (Pythagorean theorem).  For 6, 7, 8, this would mean that 36 + 49 = 64, which is false.  For 10, 15, 20 you would need 100+225 = 400, also false.\n\nThe one common ration that works is 3, 4, 5  (9+16 = 25).  Multiples also work, such as 15, 20, 25 (225 + 400 = 625).
857682
The midpoint of A and B is (15+7)/2 = 11.\nIf C = 11 then AC =BC.\nTherefore AC < BC  for 11<C<15
708770
angle cdb = ancle cbd   (since traingle cbd is an isosceles)\n\nangle bcd =  180  -  (2*65)   =   50\n\nangle  acb =   180 - 68 - 30 =  82\n\n\n\nUsing sine rule,  (considering traingle  abc)\n\n(ab/sin 82) =  (bc / sin 68)\n\ngiven: ab= 5  ,  therefore  bc = 4.68   (2 decimal places)\n\n\n\n\nusing sine rule again,   (for triangle bcd)\n\n(bd/sin 50) =  (bc / sin 65)\n\nbc= 4.68   ,  therefore  bd = 3.96
723227
<R= 52* 180-83-45=52. because 3 <s of a triangle is 180.\n\nLaw of Sins:\nSin A / a = Sin B / b = Sin C/ c\n\n(sin 52)/ 18 = (sin 45)/ q\ncross multiply\n[ 18 x (sin 45) ] / (sin 52) = q\nq= 16.1 so about 16\n\n(sin 52)/ 18 = (sin 83)/ p\ncross multiply\n[ 18(sin 83) ] / (sin 52) = p\np=22.67 so about 23\n\nthere!\n<R = 52*\nq= 16\np= 23
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
92305
a= 34\n\na=b\n\ntherefore, b=34.\n\nunless this is a trick question.
705715
ab(a + b - 2c) + bc(b + c - 2a) + ca(c + a - 2b) >=? 0\n\na²b + ab² + b²c + bc² + ac² + a²c - 6abc >=? 0\n\nabc(a/c + b/c + b/a + c/a + c/b + a/b) - 6abc >=? 0\n\nabc(a/c + b/c + b/a + c/a + c/b + a/b - 6) >=? 0\n\nNow look at your terms inside.  You have:\na/c + c/a\nb/c + c/b\na/b + b/a\n\nThese are reciprocals.  Regardless of what numbers you pick for a, b, c (which must be 1 or higher), the sum will be no smaller than 2.  (1/1 + 1/1 = 2).\n\nAt a minimum you have 6, but it can never result in something smaller than 6, therefore the amount in the parentheses will be 0 or greater.  The product of this with a positive amount (abc) will also be 0 or greater.\n\nAlternatively,\na²b + ab² + b²c + bc² + ac² + a²c - 6abc >=? 0\na(b² + c² - 2bc) + b(a² + c² -2ac) + c(a² + b² - 2ab) >=? 0\na(b-c)² + b(a-c)² + c(a-b)² >=? 0\n\nHere you have positive integers and squares (which are always 0 or greater), so the sum will be greater or equal to zero.
710964
ab=6\n\nit is an equation because of =
477890
Consider a triangle with sides a, b, c and angle A oppose to the side a.  Side a is the diagonal that you are going to find in each case.\n\nBy knowing 2 sides (b, c) and the angle (A) in between, then use Cosine rule:\n\na^2 = b^2 + c^2 - 2bc cos A\n\na^2 = 60^2 + 80^2 - 2 * 60 * 80 cos(56)\n\na = 68.06 ft  (shorter diagonal)\n\nConsider when A = 180 - 56 = 124, use Cosine rule again:\n\na^2 = 60^2 + 80^2 - 2 * 60 * 80 cos(124)\n\na = 123.97 ft (longer diagonal)
728989
we have in a triangle\n\nsin B/b = sin C/c = k (say)\n\nsin B= bk  ...1\nsin C = ck  ...2 \nadditionally\n\nb^2 = c^2+a^2-2ac cos B\nso cos B = (c^2+a^2-b^2)/2ac ...3\n\ncos  C = (a^2+b^2-c^2)/2ab ...4\nfrom1 and 3 by deviding 3 by 1\ncot B = (c^2+a^2-b^2)/2abck\n\nor 2abck = (c^2+a^2-b^2) tan B  ..5\nsimilarly from 2and 4\n2abck =(a^2+b^2-c^2) tan C\n\nfrom above 2 we get the result as both are same
339354
(2x + b) * (a + 3c)\n\n\nsteps:\n2ax + 6xc + ba + 3bc \n1) group first two, and last two\n(2ax + 6xc) + (ba + 3bc)\n\n2) factor out 2x from the first, and then b from the second group\n2x*(a + 3c) +  b*(a + 3c)\n\n3) since (a + 3c) is common, combine 2x and b\n(2x + b)*(a + 3c)
890630
+---------65km--------------->\n|\n|\n|\n35km\n|\n|\n\nNotice that this is a right (90 degree) triangle.Formula for a triangle with sides a, b, c is:\na*a+b*b=c*c  where c is the hypotenuse\n\n35*35+65*65=c*c\n\nc= Square root(35*35+65*65)  \n=73.8