3-4 할리데이 10판 솔루션 일반물리학

(a) 20.0° $$ 20.0^\circ =20.0^\circ \cdot \frac{\pi\ut{rad}}{180^\circ}=\frac{\pi}{9}\ut{rad}\approx0.349\ut{rad}$$
(b) 50.0° $$ 50.0^\circ =50.0^\circ \cdot \frac{\pi\ut{rad}}{180^\circ}=\frac{5}{18}\pi\ut{rad}\approx0.873\ut{rad}$$
(c) 100° $$ 100^\circ =100^\circ \cdot \frac{\pi\ut{rad}}{180^\circ}=\frac{5}{9}\pi\ut{rad}\approx1.75\ut{rad}$$
(d) 0.330 rad $$ 0.330\ut{rad} = 0.330\ut{rad} \cdot \frac{180^\circ}{\pi\ut{rad}}=\frac{297}{5\pi}^\circ\approx18.9^\circ$$
(e) 2.30 rad $$ 2.30\ut{rad} = 2.30\ut{rad} \cdot \frac{180^\circ}{\pi\ut{rad}}=\frac{414}{\pi}^\circ\approx132^\circ$$
(f) 7.70 rad $$ 7.70\ut{rad} = 7.70\ut{rad} \cdot \frac{180^\circ}{\pi\ut{rad}}=\frac{1386}{\pi}^\circ\approx441^\circ$$