3-28 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec a_1 = 0.50\ut{m}\hat i\\ \vec a_2:(0.70\ut{m},30^\circ)\\ \vec b_1:(1.6\ut{m},90-40^\circ)\\ \Sigma \vec a = \Sigma \vec b\\ \end{cases} $$ $$ \vec b_2 = \Sigma \vec a - \vec b_1$$ $$ \begin{aligned} \vec a_2 =& (0.7\cos30^\circ)\hat i + 0.7\sin30^\circ)\hat j\\ =&\(\frac{7 }{20}\sqrt3\)\hat i + \(\frac{7}{20}\)\hat j\\ \end{aligned} $$ $$ \begin{aligned} \vec b_1 =& (1.6\cos50^\circ)\hat i + 1.6\sin50^\circ)\hat j\\ \end{aligned} $$ $$ \begin{aligned} \vec b_2 =& \Sigma \vec a - \vec b_1\\ =&\(0.50+\frac{7 }{20}\sqrt3-1.6\cos50^\circ\)\hat i + \(\frac{7}{20}-1.6\sin50^\circ\)\hat j \end{aligned} $$
(a)$b_2=?$ $$ \begin{aligned} b_2=&\sqrt{\(0.50+\frac{7 }{20}\sqrt3-1.6\cos50^\circ\)^2+\(\frac{7}{20}-1.6\sin50^\circ\)^2}\\ \approx&0.88\ut{m} \end{aligned} $$
(b)$\theta_{b_2}=?$ $$ \begin{aligned} \theta_{b_2}=&\tan^{-1}\(\frac{\frac{7}{20}-1.6\sin50^\circ}{0.50+\frac{7 }{20}\sqrt3-1.6\cos50^\circ}\)\\ \approx&-1.5\ut{rad} \end{aligned} $$