40/sin A = 70/sin 90..find angle A?
928486
sin 90 = 1, right?\n\n40/sin A = 70, so\n\nsin A = 40/70 = 4/7\n\nA = arcsin(4/7)
1166171
The easy way to remember this is to memorize the \n\n3 4 5 \n\nright triangle.  You will notice that any two multiples of the above will give you the other leg as a multiple.  The theorem is\n\nA^2 + B^2 = C^2,  so  3^2 + 4^2 = 5^2,  hence  9 + 16 = 25\n\nnow you have\n\n2*3 2*4 thus your other leg is 2*5\n\n6^2 + 8^2 = 10^2,  which is 36 + 64 = 100\n\nas for the angles, use the law of sines\n\nsin(90)/5 = sin(a)/3 = sin(b)/4\n\nso a = asin(sin(90)*3/5) = ~37 deg\n\nhence angles of 90 53 37
932500
I don't know any good math websites but I do know how to solve this.  b would be the side adjacent to the 40 degree angle and a is the side opposite it.\n\nTangent = opposite leg/adjacent leg so\ntan 40 = 15/b\n\nb tan 40 = 15\n\nb = 15 / tan 40
997563
You're looking for the angle that has a sine of 1/2.\nDrawing a triangle representing this is very helpful.  You will see that it is a 30-60-90 triangle.  \nIf you can find it in degrees, then you can convert it to radians by multiplying it by pi/180.\nsine(30) = 1/2\n30 degrees = 30(pi/180) = pi/6
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
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
736970
Here are some real tough ones.(Looked tough to me)\n\nProve that cot(90 - theta)sin(90 - theta) = sin theta\n\nFind Cos(3x)\n\nFind Sin(3x)
774120
Sure,\nYou are adding two vectors:\n\nV1 has magnitue 0.45 m/s and in the W direction (head of vector faces W) \n\nThe 2nd vector is has magnitude 0.35 m/s and 30 degrees E of N (or 60 degrees from +ve x-axis) and its tail is added to the head of the first). So the angle betwen the two vectors is 60 degrees.\n\nUse the SAS rule for finding the magnitude of the resultant vector\n\nc = sqrt(a^2+b^2 -2ab cos(theta))\n\nVr = sqrt((0.45)^2 +(0.35)^2 - 2*0.45*0.35 cos60)\n   = 0.409 m/sec \n\nThe direction of Vr is from the tail of the first vector to the head of the 2nd.\n\nTo determine the angle, A, bewteen Vr and the first vector use the sine law.\n\n0.409/sin60 = 0.35/sinA\n\nSinA = 0.74 and A= 47.8 degrees\n\nSo the direction of the craft is 90-47.8 = 42.2 degrees W of N
249103
well it depends on the angle between quantities. \nat 0 degree, simply add the given magnitude\nat 90 degree, use the pythagorean theorem\nat 180 degree, simply subtract the magnitude\nat any angle, other than 0, 90 and 180, use the laws of sine and cosine
12429
Here are a couple i learned for double angle formulas.\n\nFor sin(a+b)= sin(a)cos(b)+cos(a)sin(b)\nSay out loud: "sine cosine cosine sine"\n\nFor cos(a+b)= cos(a)cos(b)-sin(a)sin(b)\nSay out loud: "cosine cosine SIGN sine sine"\nNotice the word "SIGN". The word SIGN is there to remind you that the SIGN of the rightmost term has the opposite SIGN of the input to the cosine on the left hand side of the equation (if its a+b, then subtract. if its a-b, then add).\n\nJust say it out loud like 3 times. It has a really nice rythm:\n\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n\nSee? Easy.
422360
A triangle is 180 degrees... so 28 + 58 = 86. so now solve this: 180-86=....94!! Tah Dah!
957436
By drawing a sketch diagram, you can work out the system in question can be represented as a right angle triangle, with the hypotenuse equal to 1 (kg) and the opposite side = 0.5 (kg).  the angle you are trying to calculate is between the hypotenuse and adjacent, and is given by the formula \n\nsin a = opp / hyp\n\ntherefore sin a = 0.5 / 1\nsin a = 0.5\n\na = arcsin(0.5)\n\na  = 30 degrees.\n\nAs both weigths are acting in opposition to each other, the angle will be positive, so your answer is option 2, 30 degrees.\n\nHope this helps!
908544
sin 2a / (1 + cos 2a)\n\n(2 sin a cos a) / (1 + (2 cos ^2 a - 1) )\n\n2 sin a cos a / (2 cos ^2 a)\n\nsin a / cos a\n\ntan a
599964
1.31 x sin A = sin (45)\nA = 32.7°\n\n1.31 x 0.54 = 1.33 sin B\nB = 32.1°
969737
sine 30° 15'\n\nCos ( 90 - A)\n\n89 60\n30 15\n- - - - -\n\n59 45\n\nThe answer is cos 59° 45'\n\n- - - - - - - -s-
835259
I'll give you a hint and let you try to figure out the rest yourself (unless someone else tells you the exact answer). =)\n\nUse your trig identities:\n\ntan A = sinA/cosA\n\ncot A = 1/tanA or cot A = cosA/sinA
833971
use the formula sinA/sinB.\nA is the angle of incidence,B is the angle of refraction.\nso in this case it is sin40/sin25 = 1.52(to 3sig. fig)