3-27 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} P &= 13.0\ut{m}, \theta_P=25.0\degree\\ Q &= 10.0\ut{m}, \theta_Q=(90.0+10.0)\degree\\ R &= 9.00\ut{m}, \theta_R=(90.0-20.0)\degree\\ S &= 8.00\ut{m}, \theta_R=(-90.0+40.0)\degree\\ \end{cases} $$ $$ \begin{cases} \vec P &= 13\cos25\degree\i+13\sin25\degree\j\\ \vec Q &= 10\cos100\degree\i+10\sin100\degree\j\\ \vec R &= 9\cos70\degree\i+9\sin70\degree\j\\ \vec S &= 8\cos(-50\degree)\i+8\sin(-50\degree)\j\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \Ans=&\vec P+\vec Q+\vec R+\vec S\\ \vec A=&\(-10 \sin 10\degree+9 \sin 20\degree+8 \sin 40\degree+13 \cos 25\degree\)\i\\ &+\(13 \sin 25\degree+10 \cos 10\degree+9 \cos 20\degree-8\cos 40\degree\)\j\\ \approx &18.26600162223048\ut{m}\i+17.67099297487253\ut{m}\j\\ \approx &18.3\ut{m}\i+17.7\ut{m}\j\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} A^2=&\(13 \sin 25\degree+10 \cos 10\degree+9 \cos 20\degree-8\cos 40\degree\)^2\\ &+\(13 \sin 25\degree+10 \cos 10\degree+9 \cos 20\degree-8\cos 40\degree\)^2\\ \end{aligned} $$ $$ \begin{aligned} A=&\sqrt{342+10 \sqrt{3}+117 \sqrt{6}}\ut{m}\\ \approx&25.41477538719004\ut{m}\\ \approx&25.4\ut{m}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \theta&=\tan^{-1}\frac{13 \sin 25\degree+10 \cos 10\degree+9 \cos 20\degree-8\cos 40\degree}{-10 \sin 10\degree+9 \sin 20\degree+8 \sin 40\degree+13 \cos 25\degree}\\ &\approx 0.7688426808855916\ut{rad}\\ &\approx 0.769\ut{rad}\\ \end{aligned} $$