3-26 할리데이 10판 솔루션 일반물리학

$$ \begin{aligned} \vec A=&(2.0\ut{m})\hat i + (3.00\ut{m})\hat j\\ \vec B:&(4.00\ut{m}, +65.0^\circ)\\ \vec C =& ( - 4.00 \ut{m})\hat i - (6.00 \ut{m} )\hat j\\ \vec D:&(5.00\ut{m}, -235^\circ)\\ \end{aligned} $$ $$ \begin{aligned} B=\(4\cos65.0^\circ\)\hat i + \(4\sin65.0^\circ\)\hat j\\ \end{aligned} $$ $$ \begin{aligned} D=\bra{5\cos(-235^\circ)}\hat i + \bra{5\sin(-235^\circ)}\hat j\\ \end{aligned} $$
(a)$\vec r=\vec A+\vec B+\vec C+\vec D=?$ $$ \begin{aligned} \Ans=&\vec A+\vec B+\vec C+\vec D\\ =&\bra{2+4\cos65.0^\circ-4+5\cos(-235^\circ)}\hat i\\ &+\bra{3+4\sin65.0^\circ-6+5\sin(-235^\circ)}\hat j\\ =&\bra{4\sin(25^\circ)-5\sin(35^\circ)-2}\hat i\\ &+\bra{4\cos(25^\circ)+5\cos(35^\circ)-3}\hat j\\ \end{aligned} $$
(b)$r=?$ $$ \begin{aligned} r=&\sqrt{\bra{4\sin(25^\circ)-5\sin(35^\circ)-2}^2+\bra{4\cos(25^\circ)+5\cos(35^\circ)-3}^2 }\\ =&\sqrt{-16 \sin (25 {}^{\circ})+20 \sin (35{}^{\circ})-24 \cos (25 {}^{\circ})-30 \cos (35{}^{\circ})+74}\\ \approx&5.69\ut{m} \end{aligned} $$
(c)$\theta_r=?$ $$ \begin{aligned} \theta_r=&\tan^{-1}\bra{\frac{4\cos(25^\circ)+5\cos(35^\circ)-3}{4\sin(25^\circ)-5\sin(35^\circ)-2}}\\ \approx&2.16\ut{rad} \end{aligned} $$