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

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MAT-129 OL009 Written Assignment 2 Section 5.1: EX: 24
20x+10) + (3x+9) = 180 20x + 19 = 180 23x=180-19 23x=161 X= 161 23 X=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=90 9x=90+45 9x=135 X= 135 9 X=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: 48 180°-124°51’ 179° 60’ -124° 51’ 55° 09’ EX: 62 34°51’35” = 34° + 51 ° 60 + 35 ° 3600 1’= 1 ° 60 1= {1 °} over {3600 ≈ 34° + 0.85° + 0.009722° ≈ 34.8597° EX: 84 -203° 20’ -203° 20’ + 360° = 359° 60’ - 203° 20’ = 156° 40’ EX: 104 225° 225°+n.360 EX: 126 1 minute = 1000 times 60 seconds = 1000 times 2x30 seconds = 1000 times 2 seconds = 1000 30 times 2 seconds = 100 3 times = 100 3 x 360 degrees = 100 x 120 degrees = 12,000 degrees in 2 seconds.
Section 5.2 EX: 14 EX: 44 3x+5y=0; x≥0 3(5)+5y=o 15+5y=0 5y=-15 y=-3 r= x 2 + y 2 = ( 5 ) 2 + ( 3 ) 2 = 25 + 9
= 34 sin θ = y r = 3 34 = 3 34 X 34 34 = 3 34 34 cos θ = x r = 5 34 = 5 34 X 34 34 = 5 34 34 tan θ = y x = 3 5 csc θ = r y = 34 3 = 34 3 sec θ = r x = 34 5 cot θ = x y = 5 3 = 5 3 EX: 90 csc θ, given that sin θ = - 8 43 csc θ = 1 sin θ = 1 8 43 = 1 / ( 8 43 ) = 1 X ( 43 8 ) = 43 8 csc θ = 43 8 EX: 110 Since cos θ > 0 in quadrants I and IV, sec θ > 0 in quadrants I and IV. Both conditions are met in I and IV quadrants. The figure at the bottom of example 5 on page 512 in the text book assists in
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