3-17 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \overrightarrow{AC} &=18.0\ut{m}\\ \overrightarrow{AD} &=-13.0\ut{m}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} AB &= \overrightarrow{AC}+\overrightarrow{AD}\\ &=\sqrt{18^2+13^2}\ut{m}\\ &=\sqrt{493}\ut{m}\\ &\approx22.20360331117452\ut{m}\\ &\approx22.2\ut{m}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Ans &= AB\sin\phi\\ &= \sqrt{493}\sin48\degree\\ &\approx 16.50049290765208\ut{m}\\ &\approx 16.5\ut{m}\\ \end{aligned} $$