3-20 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec r = 4.8\hat j\\ \vec a:(7.8,50^\circ)\\ \vec a+\vec b=\vec r\\ \end{cases} $$ $$ a= (7.8\cos50^\circ)\hat i + (7.8\sin50^\circ)\hat j$$ $$ \begin{aligned} \vec b &= \vec r - \vec a\\ &= (-7.8\cos50^\circ)\hat i + (4.8-7.8\sin50^\circ)\hat j\\ \end{aligned} $$
(a)$b=?$ $$ \begin{aligned} b&=\sqrt{(-7.8\cos50^\circ)^2+(4.8-7.8\sin50^\circ)^2}\\ &=\frac{3}{5} \sqrt{233-208 \cos(40 {}^{\circ})}\\ &\approx 5.2\ut{m}\\ \end{aligned} $$
(b)$\theta_b=?$ $$ \begin{aligned} \theta_b&=\tan^{-1}\(\frac{4.8-7.8\sin50^\circ}{-7.8\cos50^\circ}\)\\ &=\tan^{-1}\left(\cot40^\circ-\frac{8}{13} \csc40^\circ\right)-\pi \\ &\approx -2.9\ut{rad} \end{aligned} $$