if angle a is 62.4 what is the supplement of angle A?
362096
Supplements are 180 degrees munis your original angle.  So your supplement here would be 117.6.
930936
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.
457981
a/b = c/d\n\nthen ad = bc\n\nso 24x = 6x9\n\nx = 54/24\n\nx = 9/4
466510
33.7 degrees\n\nIt is calculated using vectors  6 straight ahead and 4 to one side.  The hypotenuse is sqrrt(4^2 + 6^2) = sqrrt(52).\n\nSo the angle the boat will veer is the angle that has a sin equal to 4/sqrt(52) = .5547 = 33.7 degrees
943973
Tangent 25° 56' 30" = cot (90 - A)\n\n89 59 60\n25 56 30\n- - - - - - -\n\n64 3 30\n\nCotangent 64 degrees 3 minutes 30 seconds\n\nThe answer is. The complemenary angle of tangent 25° 56' 30" is cotangent  64° 3' 30 "\n\n- - - - - - - - - - - -s-
857791
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).
812747
a + b + c = 180\n52 + 34 + c = 180\nc = 94
705644
1) you mean angle between base and side is 52? \nthen height is 4*sin(52)\n\n2) Height * (1 - cubicRoot(1/1.618))\n\nof course, since 1.618 is golden section, 1/1.618=0.618
834268
When  two  angles are combined to total 90°, they are referred to as complimentary angles. Supplementary  angles  form when two angles combine to total 180°
1093073
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
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.
943820
The only angle for which that would be true is 0 degrees\nProof:\nx is the angle you're looking for\ny is the compliment\nz is the supplement\n\nx+y=90\nx+z=180\nz=2y\n\nthen\nx + 2y = 180\nsolve the compliment equation for y and substitute\nx + y=90\ny=90 - x\n\nx + 2 (90-x) = 180\nx + 180 - 2x = 180\n-x = 0
706305
R^2=A^2+B^2 -2(A*B)cos(180-acute angle)
82294
24x - 18 = -54\n24x = -54 + 18\n24x = -36\nx = -36 / 24\nx = 1.5
199013
add 2 other angle in degree and substract from 180\n ie. c=180 -(a+b)\nwhere a ,b are known angles \nand c is unknown angle
133118
There are formulas on the following page:\n\nhttp://www.acts.tinet.ie/trig.ratiosforacuteang_668.html
450833
join the 2 pointz frm da arc vit da radius.measure out da value of da angle forned here.then substitute da values 4 this formula:  2*pie*radius/da angle u measured\n\nhappy solvin!