3-36 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \vec A &= 0\\ \vec B &= 55\ut{km}\i\\ r_1&=24\ut{km}, \theta_1=-25\degree\\ \vec r_2 &= 8.0\ut{km}\j \end{cases} $$ $$ \begin{aligned} \vec C &= \(\vec A + \vec r_1 + \vec r_2\)-\vec B\\ &=\(0 + 24\cos(-25\degree)\i+24\sin(-25\degree) + 8\j\)-55\i\\ &=(24\cos25\degree-55)\i+(8-24\sin25\degree)\j\\ \end{aligned} $$ $$ \begin{aligned} C &= \sqrt{(24\cos25\degree-55)^2+(8-24\sin25\degree)^2}\\ &=\sqrt{-384 \sin (25 {}^{\circ})-2640 \cos (25 {}^{\circ})+3665}\ut{km}\\ &\approx 33.31759339620459\ut{km}\\ &\approx 33\ut{km}\\ \end{aligned} $$