3-18 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec A + \vec B= \vec C\\ \vec A:(12.0,40^\circ)\\ \vec C:(16.0,200^\circ)\\ \end{cases}$$ $$\vec A = (12\cos40^\circ)\hat i + (12\sin40^\circ)\hat j$$ $$\vec C = (16\cos200^\circ)\hat i + (16\sin200^\circ)\hat j$$ $$\begin{aligned} \vec B &= \vec C - \vec A\\ &=(16\cos200^\circ-12\cos40^\circ)\hat i + (16\sin200^\circ-12\sin40^\circ)\hat j\\ \end{aligned}$$
(a)B=? $$\begin{aligned} B&=\sqrt{(16\cos200^\circ-12\cos40^\circ)^2+(16\sin200^\circ-12\sin40^\circ)^2}\\ &=4 \sqrt{24 \cos20^{\circ}+25}\\ &\approx 27.6\ut{m}\\ \end{aligned}$$
(b)$\theta_B=?$ $$\begin{aligned} \theta_B&=\tan^{-1}\(\frac{16\sin200^\circ-12\sin40^\circ}{16\cos200^\circ-12\cos40^\circ}\)\\ &\approx -2.64\ut{rad} \end{aligned}$$