so u have this square triangle ABC with A=the square angle. B is 30 degrees. BM (M being on AC)is the line that splits angle B in 2. AB is 2a times square root of 3 and u have 2 know tan of angle MBA  .
124576
first draw a right triangle and label the angles and sides. now angle B cut in half is 15 degrees and angle A is 90 degrees.  lets assign AM a value of x.  so the tangent function is equal to the opposite leg over ajacent leg=> tan(15degrees)=x/(2*sqrt(3)).  take the tan of 15 and then multiply that by 2*sqrt of 3 and that will be your answer for the length of AM which is approx=.9282  to find MC you must first find the length of AC with the same tangent function using 30degrees as your angle so lets assign AC a value of y. so tan(30degrees)=y/(2*sqrt(3)) which is = 2 so MC=AC-AM= 1.0718
619857
A = 45 degrees or (pi / 4) radians\n\n1/sin^2A-1/cos^2A-1/tan^2A-1/cot^2A-1/sec^2A-1/csc^2A = -3\nIf i understand the way you wrote the problem above, it's 1/ each term, subtract each other right?\n1/ ((sinA)^2) - 1/ ((cosA)^2) - 1/ ((tanA)^2) - 1/ ((cotA)^2) - 1/ ((secA)^2) - 1/ ((cscA)^2) = -3\nI'm going to drop the A part, i assume you know that it will still be officially added in the final version\n\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cot^2)] + [-(tan^2)] + [-(cos^2)] +  [-(sin^2)] = -3\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cos^2) / (sin^2)] + [-(sin^2) / (cos^2)] + [-(cos^2) - (sin^2)] = -3\ncombine denominators\n[( 1 - (cos^2) ) / (sin^2)] + [( (-1) - (sin^2) ) / (cos^2)] + (-1) [(cos^2) + (sin^2)] = -3\n[(sin^2) / (sin^2)] + (-1)[( 1 + (sin^2) ) / (cos^2)] + (-1) [ 1 ] = -3\nTrig identity [(cos^2) + (sin^2)] = 1\n[ 1 ] + (-1)[( 1 + (sin^2) ) / (cos^2)] + [ -1 ] = -3\n(-1)[( 1 + (sin^2) ) / (cos^2)] = -3\n[( 1 + (sin^2) ) / (cos^2)] = +3\n1 + (sin^2) = 3 * (cos^2)\nTrig identity 1 = [(sin^2) + (cos^2)] =\n[(sin^2) + (cos^2)] + (sin^2) = 3 * (cos^2)\n2 * (sin^2) = 2 * (cos^2)\nsin^2 = cos^2\nsinA = cosA\nA= 45 degrees of (pi/4) radians
522054
a)A=41°20’ b=17.2cm\ncos 41°20'=17.2/h\nh=17.2/cos 41 1/3°=22.9cm\n\nb) 1) A=48°40’ a=21.0cm\nsin 48 2/3°=21/h\nh=21/sin 48 2/3°=28.0cm\n\nsolve TRIANGLE ABC if \nA=50° B=81° a=95cm\n\nC=180-50-81=49°\nlaw of sines\n95/sin 50=b/sin 81=c/sin 49\nb=95 sin 81/sin 50=122.5cm\nc=95 sin 49/sin 50=93.6cm
494951
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
459431
Without a picture or more information, your question isn't answerable, but I'm guessing that angles x and a are parts of a right angle, in which case the answer would be 41 degrees (49 + x = 90).
315582
Pick an angale, not the 90 degree angle and call it Theta.  The short side that forms this angle is called the x side of the triangle and the long side that forms this angle is the hypotenuse or 'r'.  The third side is called the opposite side or 'y'.\n\n\n.... /|\n..../.|\n.../..|\n../...|\n./....|\n/___|  Bottom is x, slanting side is r, and right side is \n          the y side.  The angle at the bottom left is Theta.\n\n\nSin(Theta) will = y/r\nCos(Theta) will = x/r\nand\nTan(Theta) will = y/x side.\n\nThis all presumes you don't already know 2 of the sides.  If you know 2 sides already, you use the pythagorean thm.\n\nExampe: Theta = 30 degrees, r = 10 feet, what are the others?\n\nSin(Theta) = y/r  and Sin(30 degrees) happens to be 1/2.\n\nso, 1/2 = y/10.  Solving for y you get y=5.
386417
Where c is the hypotenuse, and a and b are legs of the triangle:\nc² = a² + b² --- Pythagoren Therom\nc² = (12)² + (20)² --- Replace what you know...\nc² = 144 + 400 --- Find 12² and 20²...\nc² = 544 --- Add them together...\nc = √544 --- Square root of 544 equals c\nc ≈ 23.32"
374150
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
349546
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
390165
This is an impossible figure as by your own figures ΔABC is equilateral with AB = 2AL = 10\nBC = 2BD = 10 and\nAC = 2AE = 10\nso AT = 2/3 * BT = 2/3* 5√3\n= (10√3)/3\n≠10
41456
There is many answers\na=1 b=2 c=3 is true\nalso a=b=c=sqrt(3) is true but not integers\n\nHow we can get them? \na,b,c must satisfy abc=a+b+c\n\nWhy?\natan(a) + atan(b) + atan (c) =pi\nlet atan(b)+atan(c) = y     (eq1)\n\natan(a) + y =pi\ntaking tan to the 2 sides\n\ntan [atan(a) + y] =tan(pi) = 0\n[tan(atan(a)) +tan (y) ]/[1- tan(atan(a))*tan (y) ]=0\n\nso a + tan(y)=0                       (eqn2)\nbut tan(y) = tan (atan (b) +atan(c))    from eqn1\n=(b+c)/(1-bc)\n\nsubstitue in eqn2\na+(b+c)/(1-bc)=0\nmultiply by (1-bc)\na(1-bc)+b+c=0\na+b+c-abc=0\n***********\n*a+b+c=abc*\n***********\n\na=1 b=2 c=3 is true since 1+2+3 =1*2*3 =6\na=b=c=sqrt(3) is true since 3sqrt(3)=srqt(3)^3
60919
If you know the total sum of all the angles in the figure, then subtract the value of all the given angles from the total and this should give you the missing angle. \n\nFor example- you are given a triangle. In it, two angles are supplied for you- one angle measures 35 degrees and another measures 70 degrees. Now you know that the sum of all the angles in a triangle is 180 degrees. So subtract 70 and 35 from 180. This will give you the value of the thrid angle of the triangle. \n\nHope this helps!
513971
Use the fact that the sum of the angles of a triangle add up to 180 degrees. Let A, B, C be the angles of the triangle. The m<A + m<B + m<C =180. If you know m<A and m<B, you can find m<C.\n\nWhen I write m<A, I mean the measure of angle A, not m is less than A.
429030
Use the sine law.\n A/sin(a) = B/sin(b) = C/sin(c)\nwhere angles a, b, c are opposite sides A, B, C respectively.\n\n   In your example  5/sin(prq)=4/sin(40)\nsin(prq)=5(sin(40))/4=\n(5/4)(.6428)=.803\nsin^(-1)(.803)=53.4degrees  and the sine is positive in the\n2nd quadrant as well as the first so angle qrp may also be\n180-53.4 = 126.6\n\nSo if it's 53.4 then angle pqr is 180-(53.4+40)=86.6\n\nAgain use the sine law to solve for pr.\npr/sin(86.6)=4/sin(40)\npr=4(sin(86.6))/sin(40)\npr=4(.998)/.6428\npr=6.21\n\nAnd if it's 126.6  angle pqr is 180-(126.6+40)=13.4\npr/sin(13.4)=4/sin(40)\npr=4(.232)/.6428\npr=1.44
177203
Draw a picture, it's easier. \n\nSo the picture shows that from the triangle with point A, tan40° = x÷100\nso:\nx = 100tan40° = 83.91\n\nand in the triangle with point B, \ntan20° = 83.91÷y\nso:\ny = 83.91÷(tan20°) = 230.54\n\nthe triangle on the ground, containing  point A, point B, and the base of the tower, is a right triangle because one side points south and the other one points east, so just use pythagorean theorem to get the length of the hypotenuse, which is the length from A to B.
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tanA + tanB + tanC = tanA tanB tanC is true for any non-right angle triangle\n\nrefer the link below . adequately solved
459347
The midpoint of A and B is (15+7)/2 = 11.\nIf C = 11 then AC =BC.\nTherefore AC < BC  for 11<C<15