1. 150 ° = ( π 180 ) ( 150 ) = 5 π 6 150 ° = ( π 180 ) ( 150 ) = 5 π 6
2.
+ +
− −
+ +
3. sin 205 ° = − sin ( 205 ° − 180 ° ) = − sin 25 ° sin 205 ° = − sin ( 205 ° − 180 ° ) = − sin 25 °
4. x = − 9 , y = 12 r = ( − 9 ) 2 + 12 2 − − − − − − − − − − − √ = 15 sin θ = 12 15 = 4 5 sec θ = 15 − 9 = − 5 3 x = − 9 , y = 12 r = ( − 9 ) 2 + 12 2 = 15 sin θ = 12 15 = 4 5 sec θ = 15 − 9 = − 5 3
5. r ω v = = = = 2.80 ft 2200 r / min = ( 2200 r / min ) ( 2 π rad / r ) 4400 π rad / min ω r = ( 4400 π ) ( 2.80 ) = 39 , 000 ft / min r = 2.80 ft ω = 2200 r / min = ( 2200 r / min ) ( 2 π rad / r ) = 4400 π rad / min v = ω r = ( 4400 π ) ( 2.80 ) = 39 , 000 ft / min
6. 3.572 = 3.572 ( 180 ° π ) = 204.7 ° 3.572 = 3.572 ( 180 ° π ) = 204.7 °
7. tan θ = 0.2396 θ ref = 13.47 ° θ = 13.47 ° or θ = 180 ° + 13.47 ° = 193.47 ° tan θ = 0.2396 θ ref = 13.47 ° θ = 13.47 ° or θ = 180 ° + 13.47 ° = 193.47 °
8. cos θ = − 0.8244 , csc θ < 0 ; cos θ negative , csc θ negative , θ in third quadrant θ ref = 0.6017 , θ = 3.7432 cos θ = − 0.8244 , csc θ < 0 ; cos θ negative , csc θ negative , θ in third quadrant θ ref = 0.6017 , θ = 3.7432
9. θ θ is in quadrant IV since sin θ < 0 sin θ < 0 and cos θ > 0. cos θ > 0. Since sin θ = − 6 7 , sin θ = − 6 7 , we will use y = − 6 y = − 6 and r = 7. r = 7. Thus, x = 7 2 − ( − 6 ) 2 − − − − − − − − − − √ = 13 − − √ . x = 7 2 − ( − 6 ) 2 = 13 . So tan θ = y x = − 6 13 − − √ . tan θ = y x = − 6 13 .
10. s θ A = = = = r θ , 32.0 = 8.50 θ 32.0 8.50 = 3.76 rad 1 2 θ r 2 = 1 2 ( 3.76 ) ( 8.50 ) 2 136 ft 2 s = r θ , 32.0 = 8.50 θ θ = 32.0 8.50 = 3.76 rad A = 1 2 θ r 2 = 1 2 ( 3.76 ) ( 8.50 ) 2 = 136 ft 2
11. A = 1 2 θ r 2 A = 38.5 cm 2 , d = 12.2 cm , r = 6.10 cm 38.5 = 1 2 θ ( 6.10 ) 2 , θ = 2 ( 38.5 ) 6.10 2 = 2.07 s = r θ , s = 6.10 ( 2.07 ) = 12.6 cm A = 1 2 θ r 2 A = 38.5 cm 2 , d = 12.2 cm , r = 6.10 cm 38.5 = 1 2 θ ( 6.10 ) 2 , θ = 2 ( 38.5 ) 6.10 2 = 2.07 s = r θ , s = 6.10 ( 2.07 ) = 12.6 cm
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