3-10 할리데이 11판 솔루션 일반물리학

$$\begin{cases} E=5.0\ut{m}, \theta_E=+0.9\ut{rad}\\ F=6.0\ut{m}, \theta_F=-75.0\degree\\ G=7.00\ut{m},\theta_G=+1.2\ut{rad}\\ H=4.00\ut{m}, \theta_H=-210\degree\\ \end{cases}$$ $$\begin{cases} \vec E =& 5\cos0.9\i+5\sin0.9\j\\ \vec F =& 6\cos(-75.0\degree)\i+6\sin(-75.0\degree)\j\\ \vec G =& 7\cos1.2\i+7\sin1.2\j\\ \vec H =& 4\cos(-210\degree)\i+4\sin(-210\degree)\j\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} \Ans =& \vec A \\ =& \vec E + \vec F + \vec G + \vec H\\ =& \bra{-2 \sqrt{3}+\frac{3 }{\sqrt{2}}\(\sqrt{3}-1\)+5 \cos\frac{9}{10}+7 \cos \frac{6}{5}}\i\\ &+\bra{2-\frac{3}{\sqrt{2}}\(1+\sqrt{3}\)+5 \sin \frac{9}{10}+7 \sin\frac{6}{5}}\j\\ \approx & 3.733366778167407\i+6.645353192173592\j\\ \approx & 3.7\i+6.6\j \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} A^2=& \bra{-2 \sqrt{3}+\frac{3 }{\sqrt{2}}\(\sqrt{3}-1\)+5 \cos\frac{9}{10}+7 \cos \frac{6}{5}}^2\\ &+\bra{2-\frac{3}{\sqrt{2}}\(1+\sqrt{3}\)+5 \sin \frac{9}{10}+7 \sin\frac{6}{5}}^2\\ A\approx & 7.62225337738492\ut{m}\\ \approx & 7.62\ut{m} \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} \vec A =& a_x\i+a_y\j\\ =&\bra{-2 \sqrt{3}+\frac{3 }{\sqrt{2}}\(\sqrt{3}-1\)+5 \cos\frac{9}{10}+7 \cos \frac{6}{5}}\i\\ &+\bra{2-\frac{3}{\sqrt{2}}\(1+\sqrt{3}\)+5 \sin \frac{9}{10}+7 \sin\frac{6}{5}}\j,\\ \end{aligned}$$ $$\begin{aligned} \theta_A &= \tan^{-1}\frac{a_y}{a_x}\\ &=\tan^{-1}\bra{\frac{2-\frac{3}{\sqrt{2}}\(1+\sqrt{3}\)+5 \sin \frac{9}{10}+7 \sin\frac{6}{5}}{-2 \sqrt{3}+\frac{3 }{\sqrt{2}}\(\sqrt{3}-1\)+5 \cos\frac{9}{10}+7 \cos \frac{6}{5}}}\\ &\approx 60.67267444351846\degree\\ &\approx 60.7\degree\\ \end{aligned}$$ $$\ab{d}$$ $$\begin{aligned} \theta_A &= \tan^{-1}\frac{a_y}{a_x}\\ &=\tan^{-1}\bra{\frac{2-\frac{3}{\sqrt{2}}\(1+\sqrt{3}\)+5 \sin \frac{9}{10}+7 \sin\frac{6}{5}}{-2 \sqrt{3}+\frac{3 }{\sqrt{2}}\(\sqrt{3}-1\)+5 \cos\frac{9}{10}+7 \cos \frac{6}{5}}}\\ &\approx 1.058937935030015\ut{rad}\\ &\approx 1.09\ut{rad}\\ \end{aligned}$$