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$\begin{array}{l}{\text { Let the angle } \theta \text { be the angle that the vector } \overrightarrow{\boldsymbol{A}} \text { makes with }} \\ {\text { the }+x \text { -axis, measured counterclockwise from that axis. Find }} \\ {\text { the angle } \theta \text { for a vector that has the following components: }} \\ {\text { (a) } A_{x}=2.00 \mathrm{m}, A_{y}=-1.00 \mathrm{m} ;(\mathrm{b}) A_{x}=2.00 \mathrm{m}, A_{y}=1.00 \mathrm{m} ;} \\ {\text { (c) } A_{x}=-2.00 \mathrm{m}, A_{y}=1.00 \mathrm{m} ;(\mathrm{d}) A_{x}=-2.00 \mathrm{m}, A_{y}=-1.00 \mathrm{m}}\end{array}$

$\tan \theta=\frac{A _y}{A_ x}$

(a) الربع الرابع

$\tan \theta=\frac{-1}{2} \rightarrow \theta = \tan ^{-1}(-0.5)=360-26.6$

$\rightarrow \theta=333^{\circ}$

(b) الربع الأول

$\tan \theta=\frac{1}{2} \rightarrow \theta=\tan^{-1}(0.5)=26.6^{\circ}$

(c) الربع الثاني

$\tan \theta=\frac{1}{-2} \rightarrow \theta=\tan ^{-1}(-0.5)=180-26. 6^{\circ}$

$\rightarrow \theta=153^{\circ}$

(d) الربع الثالث

$\tan \theta=\frac{-1}{-2}=\theta=\tan ^{-1}(0.5)=180+26.6$

$\theta =207^{\circ}$

$\begin{array}{l}{\text { Vector } \vec{A} \text { is in the direction } 34.0^{\circ} \text { clockwise from the }} \\ {-y \text { -axis. The } x \text { -component of } \vec{A} \text { is } A_{x}=-16.0 \mathrm{m} . \text { (a) What is the }} \\ {y \text { -component of } \vec{A} ? \text { (b) What is the magnitude of } \vec{A} ?}\end{array}$

$\tan \theta=\frac{A_y}{A_ x}$

$\tan 34^{\circ}=\frac{|A _x|}{\left|A_{y}\right|} \rightarrow|A_ y|=\frac{|A_ x|}{\tan 34^{\circ}}=\frac{16}{\tan 34^{\circ}}=23.72 \mathrm{m}$

$∴ A _y=-23.72 \mathrm{m}$

$\rightarrow A=\sqrt{A_x^{2}+A_ y^{2}}=\sqrt{(-16)^{2}+(-23. 72)^{2}}=28. 6 \mathrm{m}$

$\begin{array}{l}{\text { A postal employee drives a delivery truck over the route }} \\ {\text { shown in Fig. E1.27. Use the method of components to determine }} \\ {\text { the magnitude and direction of her resultant displacement. In a }} \\ {\text { vector-addition diagram (roughly to scale), show that the resultant }} \\ {\text { displacement found from your diagram is in qualitative agreement }} \\ {\text { with the result you obtained using the method of components. }}\end{array}$

$\vec{A}=2 . 6 \mathrm{Km} \quad \overrightarrow{\mathrm{B}}=4 \mathrm{km} \quad \overrightarrow{\mathrm{C}}=3.1$

$R_{x}=A_{x}+B_ x+C_{x}=0+4+(3. 1 \cos 45)=6.2 k m$

$R_{y}=A_ y+B_ y+c_ y=2.6+0+(3.1 \sin 45)=4.8 \mathrm{Km}$

$\rightarrow R=\sqrt{R_{x}^{2}+R_{y}^{2}}=\sqrt{6.2^{2}+4.8^{2}}=7.8 \mathrm{Km}$

$\theta=\tan ^{-1}\left[\frac{4.8}{6.2}\right]=38^{\circ}$