3-12 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec p_1 =\(40\ut{km} \)\hat i\\ \vec p_2 =\(30\ut{km} \)\hat j\\ \vec p_3:(25\ut{km},90^\circ-30^\circ)\\ \end{cases}$$

$$\begin{aligned} \vec p_3 &= (25\cos30^\circ )\hat i + (25\sin30^\circ )\hat j\\ &=\(\frac{25}{2} \)\hat i + \(\frac{25}{2}\sqrt3 \)\hat j \end{aligned}$$ $$\begin{aligned} \vec r &= \Sigma \vec p=\vec p_1+\vec p_2+\vec p_3\\ &=\(40+\frac{25}{2} \)\hat i + \(30+\frac{25}{2}\sqrt3 \)\hat j\\ &=\(\frac{105}{2} \)\hat i + \(30+\frac{25}{2}\sqrt3 \)\hat j\\ \end{aligned}$$
(a)$r=?$ $$\begin{aligned} r &= \sqrt{\(\frac{105}{2}\)^2+\(30+\frac{25}{2}\sqrt3 \)^2}\\ &=5 \sqrt{165+30 \sqrt{3}}\ut{km}\\ &\approx 74\ut{km} \end{aligned}$$
(b)$\theta_r=?$ $$\begin{aligned} \theta_r&=\tan^{-1}\(\frac{30+\frac{25}{2}\sqrt3}{\frac{105}{2}}\)\\ &=\tan^{-1}\left(\frac{4}{7}+\frac{5}{7\sqrt{3}}\right)\\ &\approx 0.78\ut{rad} \end{aligned}$$