4-36 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \theta_1=30.0\degree\\ \theta_2=50.0\degree\\ \theta_3=80.0\degree\\ d_1=5.00\ut{m}\\ d_2=9.20\ut{m}\\ d_3=13.5\ut{m}\\ \end{cases} $$ $$ \begin{aligned} \phi_2 &= 180\degree-(\theta_2-\theta_1)\\ &=180\degree-(50\degree-30\degree)\\ &=160\degree\\ \end{aligned} $$ $$ \begin{aligned} \phi_3 &= \phi_2-\theta_3+180\degree\\ &=160\degree-80\degree+180\degree\\ &=260\degree \end{aligned} $$ $$ \begin{cases} \vec d_1 &= 5\cos30\degree\i+5\sin30\degree\j\\ \vec d_2 &= 9.2\cos160\degree\i+9.2\sin160\degree\j\\ \vec d_3 &= 13.5\cos260\degree\i+13.5\sin260\degree\j\\ \end{cases} $$ $$ \begin{cases} \vec d_1 &= \cfrac{5\sqrt3}{2}\i+\cfrac{5}{2}\j\\ \vec d_2 &= -\cfrac{46}{5}\cos20\degree\i+\cfrac{46}{5}\sin20\degree\j\\ \vec d_3 &= -\cfrac{27}{2}\sin10\degree\i+\cfrac{27}{2}\cos10\degree\j\\ \end{cases} $$ $$ \begin{aligned} \Sigma \vec d&\approx -6.659295490811724\i\\ &~~~~~+18.94148998426096\j\\ &\approx -6.66\i+18.9\j\\ \end{aligned} $$