in a triangle pqr pq=5cm qr=4cm angleqpr =40?
816840
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
150244
(50p9q5)/(10pq2)\n\n= (50p/10p)*(9q/q)*(5/2)\n= 5*9*(5/2)\n= 225/2\n= 112.5
685032
plz tell me wat is meant by x.wat angle is it.as i understood ur question QT is a line outside the figure & perpendicular to PQ.hence the angle should be 180 degrees
1023518
Use the Law of Sines.\n\nsin A/a = sin B/b\n(sin 36)/24 = sin B/47\nsin B = (47 sin 36)/24\nsin B = 1.151079452\nAt this point you have a problem--it is impossible for the sine of an angle to have a value greater than 1.\n\nSo either you copied the problem wrong, or you are being asked to recognize that a triangle cannot be formed with these dimensions.
1078871
you have not given the distance on screen D\nso i took D=5cm\n \nso displacement y=m*wavelenght*D/d\n                             =0.4152 cm\n\nand angle q=4.7469 degree\n\nok if you don understand then go on this link and put the values
1153228
If the two triangles are similar, then all the angles are the same.  So tan(Vert/Horiz) = the top left angle.  Tan 1.5/1.6 = 43.15   Tan 43.15 = Vert/Horiz\n\n1.5/1.6 = x / 4\n\nx = 4.267 m
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
866906
q = (5r-3) / (4r-1) multiply by 4r-1\n4qr-q=5r-3 add q\n4qr=5r-3+q    subtract 5r\n4qr-5r=q-3 factor out r\nr(4q-5)=q-3   divide by 4q-5\nr=(q-3)/(4q-5)
367069
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
801171
From the law of signs\nPQ/sin R=PR/sin Q\nPQ/sin 60=PR/sin 50\nPQ/PR=sin 60/sin 50=1.13\nPQ=1.13*PR\ntherefore PQ does not equal PR\nPQ>PR
967287
The triangle will have 2 sides of length x, and the remaining side will be length P-2x making the parameter x + x + P - 2x = P.\nBegin by drawing a line from the vertex where the two sides of length x meet, to the center of the side which is length P-2x. The line should be perpendicular to the side P-2x thereby splitting the triangle in half, into 2 right triangles. One of the right triangles has hypotenuse = x and one leg = (1/2)(p-2x). So the remaining leg has length S.R. (x^2 - (p/2 -x)^2). We calculate the area of this right triangle, and double it since there are two triangles:\nArea = (2)(1/2){(1/2)(p-2x)S.R.(x^2 - (p/2-x)^2)\n         = (1/2)(p-2x)S.R.{px - (p^2)/4}\n         = (1/4)(p-2x)S.R.(4px-p^2)\nNote that your given answer is wrong - it has an x^2 where it should have a p^2....
722996
Drop a perpendicular to side of length 10 from the vertex where the side of length 8 ends.\n\nA right angle triangle is formed,\n\nthe perpendicular is 8*sin 60 = 4sqrt(3)\n\nArea of ||gm is 10*4*sqrt(3) = 69.282 approx.
1030265
a=c*sin(30)=.5c since it is apposite the 30 degree angle\nb=sqrt(c^2-a^2)= .866c\nor b=c*cos(30)=.866c\n\na=b=c*sin(45)=.707c=
878228
Insufficient information the included angle or a right angle somewhere is needed.\n\nAssuning these are right angled triangles there are two possibiliies:\n\nEither the left side is a hypotenuse or the right side is a hypotenuse\n\nCase 1;\nBy Pythagoras' Rule\n\n12² = 8² + x²\nSo x² = 80\nx = √80\n= √(16 * 5)\n=4√5\n\nCase 2\nBy Pythagoras' Rule\n\nx² = 8² + 12²\nSo x² = 208\nx = √208\n= √(16 * 13)\n= 4√13\n\nIf neither case is true and the included angle is θ\n\nThen, by the cosine rule\n\nx² = 8² + 12² - 2 * 8 * 12*cosθ\nSo x = √(208 - 16 * 12 * cosθ)\n= √(16 * 13  - 16 * 12 * cosθ)\n=4√(13 - 12 * cosθ)\n\nNOTE: This solution satisfies the inequalities given above as in any triangle 0° < θ < 180°\ncos 0° = 1, cos 180° = -1\nSo 4√(13 - 12 * 1) < x < 4√(13 - 12 * (-1)1)\nie 4√1 < x < 4√25\nwhence 4 < x < 20
161223
law of cosines : a^2 = b^2 + c^2 -2bc * cos(A)\nwhere a, b,c are side lengths and A is the angle oppisite side a.
766258
The length of the wire must be (2x5)+(2x7)=24cm.  So each side of the triangle must be 24/3=8cm. long.
485045
Let a(P,Q) be the direction angle of the vector going from point P to Q measured counterclockwise from AB (so that a(A,B)=0). Then the direction angles of the bisectors going through B and C respectively are:\n\na(B,C)/2\n(a(B,C)-pi+a(A,C))/2\n\nand the angle between the two (A1)\n\nA1 = a(B,C)/2 - (a(B,C)-pi+a(A,C))/2 = pi/2 - a(A,C)/2\n\nIn terms of the original angle at A (which is a(A,C)) this is:\n\nA1 = pi/2 - A/2\n\nThat is, the recursion relation is:\n\nA_(n+1) = pi/2 - A_n/2\n\nSubstituting A_n = pi/3 + X_n, gives:\n\nX_(n+1) = - (1/2)*X_n\n\nSo clearly X_n = (-1/2)^n * Constant\n-->\nA_n = pi/3 + (-1/2)^n * Constant\n\nSince A_0 = A , the constant is: A-pi/3\n\n--> answer 4\n\nTo guess the answer without doing any work, just note that A_0 must be A, so only 3 and 4 are eligible. It's easy to see A_n can't be a constant just by sketching a triangle with a very obtuse angle at A.