Chapter 8

1. $150 ° = (\frac{π}{180}) (150) = \frac{5 π}{6}$
2.
1. $+$
2. $-$
3. $+$
3. $\sin 205 ° = - \sin (205 ° - 180 °) = - \sin 25 °$
4. $\begin{array}{l} x = - 9, y = 12 \\ r = \sqrt{{(- 9)}^{2} + 12^{2}} = 15 \\ \sin θ = \frac{12}{15} = \frac{4}{5} \\ sec θ = \frac{15}{- 9} = - \frac{5}{3} \end{array}$
5. $\begin{array}{rcl} r & = & 2.80 ft \\ ω & = & 2200 r / min = (2200 r / min) (2 π rad / r) \\ = & 4400 π rad / min \\ v & = & ω r = (4400 π) (2.80) = 39, 000 ft / min \end{array}$
6. $3.572 = 3.572 (\frac{180 °}{π}) = 204.7 °$
7. $\begin{array}{l} \tan θ = 0.2396 \\ θ_{ref} = 13.47 ° \\ θ = 13.47 ° or \\ θ = 180 ° + 13.47 ° = 193.47 ° \end{array}$
8. $\begin{array}{l} \cos θ = - 0.8244, \csc θ < 0; \\ \cos θ negative, \csc θ negative, \\ θ in third quadrant \\ θ_{ref} = 0.6017, θ = 3.7432 \end{array}$
9. $θ$ is in quadrant IV since $\sin θ < 0$ and $\cos θ > 0.$ Since $\sin θ = - \frac{6}{7},$ we will use $y = - 6$ and $r = 7.$ Thus, $x = \sqrt{7^{2} - {(- 6)}^{2}} = \sqrt{13} .$ So $\tan θ = \frac{y}{x} = - \frac{6}{\sqrt{13}} .$
10. $\begin{array}{rcl} s & = & r θ, 32.0 = 8.50 θ \\ θ & = & 32.0 ... \end{array}$

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Basic Technical Mathematics, 11th Edition by Allyn J. Washington, Richard Evans