 # MAT-129 WA#2.docx - MAT-129 OL009 Written Assignment 2...

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MAT-129OL009Written Assignment 2Section 5.1:EX: 24
20x+10) + (3x+9) = 18020x + 19 = 18023x=180-1923x=161X=16123X=7(20x+10) = (20 x 7 +10)=(140+10)=150°(3x+9) = (3 x 7 + 9)= (21+9)= 30°The two angles are 150° and 30°EX: 32(3x-5) + (6x-40) = 90°9x-45=909x=90+459x=135X=1359X=15(3x-5) = (3 x 15 – 5)=(45-5)=40°(6x-40) = 6 x 15 – 40)=(90-40)= 50°The two angles are 40° and 50°
EX: 48180°-124°51’179° 60’-124° 51’55° 09’EX: 6234°51’35”= 34° +51°60+35°36001’=1°601= {1 °} over {3600≈ 34° + 0.85° + 0.009722°≈ 34.8597°EX: 84-203° 20’-203° 20’ + 360° = 359° 60’ - 203° 20’= 156° 40’EX: 104225°225°+n.360EX: 1261 minute = 1000 times60 seconds = 1000 times2x30 seconds = 1000 times2 seconds =100030times2 seconds =1003times=1003x360degrees= 100 x 120 degrees= 12,000 degrees in 2 seconds.
Section 5.2EX: 14EX: 443x+5y=0; x≥03(5)+5y=o15+5y=05y=-15y=-3r=x2+y2=(5)2+(3)2=25+9
=34sin θ =yr=334=334X3434=33434cos θ =xr=534=534X3434=53434tan θ =yx=35csc θ =ry=343=343sec θ =rx=345cot θ =xy=53=53EX: 90csc θ, given that sin θ = -843csc θ =1sinθ=1843= 1 /(843)= 1 X(438)=438csc θ =438EX: 110Since cos θ > 0 in quadrants I and IV, sec θ > 0 in quadrants I and IV. Both conditions are met in

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Term
Winter
Professor
unknown
Tags
Euler s formula, tan
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