솔루션피아

(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} \cdo..

$$\vec x = 15\ut{m}\hat i + 15\ut{m}\hat j + 10\ut{m}\hat k$$ $$x=?$$ $$ \begin{aligned} x&=\abs{\vec x}\\ &=\sqrt{15^2+15^2+10^2}\ut{m}\\ &=5\sqrt{22}\ut{m}\\ &\approx 23.45207879911715\ut{m}\\ &\approx 23\ut{m}\\ \end{aligned} $$

$$\begin{cases} \vec r:(12\ut{m},30^\circ)\\ \vec r =r_x \hat i+ r_y \hat j\\ \end{cases} $$ (a)$r_x=?$ $$ \begin{aligned} r_x&=r\cos\theta\\ &=12\cos30^\circ\ut{m}\\ &=6\sqrt{3}\ut{m} \end{aligned} $$ (a)$r_y=?$ $$ \begin{aligned} r_x&=r\sin\theta\\ &=12\sin30^\circ\ut{m}\\ &=6\ut{m} \end{aligned} $$

$$\begin{cases} \vec a =a_x \hat i+ a_y \hat j\\ \vec a:(a,\theta)\\ a_x=\frac{1}{2}a\\ \end{cases} $$ $$ \tan\theta = ?$$ $$a_x=a\cos\theta=\frac{1}{2}a$$ $$ \cos\theta=\frac{1}{2}$$ $$ \theta = \frac{\pi}{3} $$ $$ \therefore \tan\theta = \tan\frac{\pi}{3}=\sqrt{3}$$