SG2 - [5.4/10 A B=90° therefore 90-68.5142° = 21.4858 = A sin 68.5142° = b/3579.42 therefore b = 3330.68 tan 68.5142°=3330.68/a therefore a

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I selected the following problems for module 2: 5.1/20, 5.2/10, 5.3/2, 5.4/10, 6.1/60, 6.2/8 and 6.3/8 [5.1/20] Given 6x-4 and 8x-12 (6x-4) + (8x-12) = 180° 14x-16=180° x=14 therefore [6(14)-4] = 80° and [8(14)-12] = 100°, therefore the answer is 80° and 100° [5.2/10] X=0 and Y=5 r=(x 2 +y 2 ) 1/2 = (0 2 +5 2 ) 1/2 = 5 sinθ=1/r = 5/5=1 cosθ=x/r=0/2=0 tanθ=y/x=5/0= undefined cotθ=x/y = 0/5 = 0 secθ=r/x = 5/0 = undefined cscθ=r/y=5/5=1 [5.3/2] sin A = 45/53 cos A = 28/53 tan A = 45/28
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Unformatted text preview: [5.4/10] A+B=90°, therefore 90-68.5142° = 21.4858 = A sin 68.5142° = b/3579.42, therefore b = 3330.68 tan 68.5142°=3330.68/a, therefore a =1311.04 [6.1/60]-3.47189 radians = -3.47189 (180°/π) -[1.98°+(.9246*60')] = -198° 55' [6.2/8] Since 5π/3 is in quadrant IV the reference angle is 300°, therefore the cos of 300° = 1/2 [6.3/8] y=2cosx 2cos (0) = 2 amplitude is 2 with a period of 2π, therfore the answer is C....
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This note was uploaded on 01/18/2012 for the course MATH 129 taught by Professor Nilanjanpaul during the Spring '11 term at Thomas Edison State.

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