3-9 할리데이 11판 솔루션 일반물리학

$$\begin{cases} A&=8.00\\ \theta_A&=130\degree\\ \vec B&=-4.31\i-6.60\j\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} \theta_{Ay-}&=\theta_A+90\degree\\ &= 220\degree\\ &=\frac{11}{9}\pi\ut{rad}\\ &\approx3.839724354387525\ut{rad}\\ &\approx3.84\ut{rad}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} \vec A = 8\cos130\degree\i+8\sin130\degree\j\\ \end{aligned}$$ $$\begin{aligned} \vec C&=\vec A\times\vec B\\ &=\bra{\frac{264}{5} \sin 40\degree+\frac{862}{25} \cos 40\degree}\k\\ \end{aligned}$$ $$\therefore \theta_{Cy-}=90\degree$$ $$\ab{c}$$ $$\begin{aligned} \vec D =&\vec A\times\(\vec B + 3.00\k\)\\ =&\bra{24 \cos 40\degree}\i+\bra{24 \sin 40\degree}\j\\ &+\bra{\frac{264}{5} \sin40\degree+\frac{862}{25} \cos 40\degree}\k \end{aligned}$$ $$\begin{aligned} \theta_{Dy-}&=\cos^{-1}\frac{\vec D\cdot(-\j)}{D\cdot1}\\ &=\cos^{-1}\frac{-24\sin 40\degree}{\frac{1}{25} \sqrt{2 (-249839 \sin (10\degree)+568920 \cos (10{}^{\circ})+801361)}}\\ &=\cos ^{-1}\(-300 \sin40\degree \sqrt{\frac{2}{-249839 \sin10\degree+568920 \cos10\degree+801361}}\)\\ &\approx 1.81061060093211\ut{rad}\\ &\approx 1.81\ut{rad} \end{aligned}$$