3-25 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} A&=18.0\ut{m}\\ \theta_A&=60\degree\\ \vec B&=12.0\ut{m}\i+8.00\ut{m}\j\\ \phi &= 30.0\degree\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} A_2&=A_1=18\\ \theta_{A2}&=\theta_{A1}-\phi\\ &=60\degree-30\degree\\ &=30\degree \end{aligned} $$ $$ \begin{aligned} \vec A_2&=18\cos30\degree\i+18\sin30\degree\j\\ &=\(9\sqrt3\i+9\j\)\ut{m}\\ &\approx \(15.58845726811989\i+9\j\)\ut{m}\\ &\approx \(15.6\i+9.00\j\)\ut{m}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} B_2 &= B_1 \\ &=\sqrt{12^2+8^2}\\ &=4\sqrt{13} \end{aligned} $$ $$ \begin{aligned} \theta_{B2}&=\theta_{B1}-\phi\\ &=\tan^{-1}\frac{8}{12}-30\degree\\ \end{aligned} $$ $$ \begin{aligned} \vec B_2&=B_2\cos\theta_{B2}\i+B_2\sin\theta_{B2}\j\\ &\approx \(14.39230484541326\i+0.9282032302755093\j\)\ut{m}\\ &\approx \(14.4\i+0.928\j\)\ut{m}\\ \end{aligned} $$