3-43 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \vec A &= 7.00\ut{km}\j\\ B&=10.5\ut{km}\\ \theta_B&=60\degree\\ \end{cases}$$ $$\begin{aligned} \vec B &=10.5\cos60\degree\i+10.5\sin60\degree\j\\ &=\frac{21}{4}\i+\frac{21}{4}\sqrt{3}\j \end{aligned}$$ $$\begin{aligned} \vec C &=-\vec B+\vec A\\ &=-\(\frac{21}{4}\i+\frac{21}{4}\sqrt{3}\j\)+7\j\\ &=-\frac{21}{4}\i+\(7-\frac{21}{4}\sqrt{3}\)\j\\ \end{aligned}$$ $$\ab{a}$$ $$\begin{aligned} C&=\sqrt{\(-\frac{21}{4}\)^2+\(7-\frac{21}{4}\sqrt{3}\)^2}\\ &=\frac{7}{2} \sqrt{13-6 \sqrt{3}}\ut{m}\\ &\approx 5.651925834942238\ut{m}\\ &\approx 5.65\ut{m}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} \theta &= \tan^{-1}\frac{7-\cfrac{21}{4}\sqrt{3}}{-\cfrac{21}{4}}\\ &=-\pi -\tan ^{-1}\left(\frac{4}{3}-\sqrt{3}\right)\\ &\approx -2.762192390864914\ut{rad}\\ &\approx -2.76\ut{rad}\\ \end{aligned}$$