maths problem-final.docx - Surname 1 Student Name Professor Name Course Date Introduction to Trigonometry 4.1 1 i 35 degrees complementary is 55degrees

maths problem-final.docx - Surname 1 Student Name Professor...

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Surname 1 Student Name Professor Name Course Date Introduction to Trigonometry 4.1 1. i) 35 degrees complementary is 55degrees, supplementary is 145 degrees ii) 60 degrees complementary is 30 degrees, supplementary is 120 degrees iii) π/3 complementary is 30 degrees, supplementary is 120 degrees iv) 2π/5 complementary is 18 degrees, supplementary is 108 degrees 2. a) (i) 5π/6 = 150 0 ii) 3π/4 = 135 0 iii) 5π/3 = 300 0 (b) i) 300 degrees = (300*π)/180 = 5π/3 ii) 75 degrees= (75* π)/180 = 5π/12 iii) 200 degrees= (200* π)/180 = 10 π/9 3. a) 750 degrees = 750+ (360)1= 1110 0 750+ (360)-1 =390
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Surname 2 b) -290 degrees = -290+ (360)1= 70 = -290 + (360)-1= -650 c) 14π/3= 840 +(360)1= 1200 = 840 + (360)-1= 480 4. Arc length= 2πrθ /360= (2*135*4.5*3.142)/360= 10.6043m (7*360/2*3.142*4)=θ = 100 degrees= (100* π)/180 = 5π/9 4.2 1. a) θ= π/3; sin π/3= -0.3048 cos π/3= -0.9524 tan π/3= 0.32 b) θ=60degrees; sin60= -0.3048 cos60= -0.9525 tan60= 0.32 c) θ= π/4; sin π/4= 0.8509 cos π/4= 0.5253 tan π/4= 1.6198 3 a) 360-300= 30; sin30= 1//2, cos 30= √3/2,tan30= 1/√3 b) 5π/4= 225; 225-180= 45; sin45= √2/2, cos 45= √2/2, tan 45= 1
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Surname 3 c) 3π/2= 270;270-180= 90; sin 90= 1, cos90= 0,tan 90= 0 4. a) -5 13 b) 2 -5 c)
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Surname 4 5. a) sin60= 1/2; cos60= √3/2 , cos30= √3/4, sin30= ¼, tan30= 1/2√3 b) secθ= 5, cos θ= 1/5, cotθ= 1/√24 and cot (90-θ)= 0.5774 6. Trigonometric functions is thought of as ratio of the side lengths in right triangles. Two similar
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