In Fig. $\mathbf{3}\mathbf{-}\mathbf{30}$, a vector awith $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}$magnitude of $\mathbf{17}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$is directed at angle $\mathbf{56}\mathbf{.}\mathbf{0}\mathbf{°}$counterclockwise from the +x axis. What are the components (a)and (b) ${\mathit{a}}_{\mathbf{y}}$of the vector? A second coordinate system is inclined by angle $\mathit{\theta }\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{°}$with respect to the first. What are the components (c)${\mathit{a}}_{\mathbf{x}}$and (d) ${\mathit{a}}_{\mathbf{y}}$in this primed coordinate system?

Using simple trigonometry, resolve the given vector to find the components of the vector.To find the components use sin and cos functions. There are two formulae used to calculate the angles.

$a_x=a\cos \theta$ (i)

$a_y=a\sin \theta$ (ii)

It is given that the magnitude of is $\overrightarrow{a}$and 17.0 m is directed at an angle $\theta =56.0°$in the counterclockwise direction.Also, the second coordinate system directed angle $\theta \text{'}=18.0°$with respect to the first system.

With $a=17$and $\theta =56°$, calculate the x component using equation (i),

localid="1656300610617" $$\begin{gathered} a_x=a\cos \theta \\ =17\times 0.559 \\ =9.51\mathrm{m} \end{gathered}$$

Thus, the x component of the vector is$9.51\mathrm{m}$

Calculate the y component using equation (ii).

$$\begin{gathered} a_y=a\sin \theta \\ =17\times 0.826 \\ =14.1\mathrm{m} \end{gathered}$$

Thus, the y component of the vector is $14.1\mathrm{m}$.

With $a=17$and with new coordinate system $\theta \text{'}=18°$, the new angle is$56°-18°=38°$

Calculate the x component using equation (i) for the new coordinate system.

$$\begin{gathered} a_x^{\text{'}}=a\cos 38 \\ =17\times 0.7880\& \\ =13.4\mathrm{m} \end{gathered}$$

Thus, the x component of the vector for the new coordinate system is $13.4\mathrm{m}$.

Calculate the y component using equation (ii) for the new coordinate system.

$$\begin{gathered} {a^{\text{'}}}_y=a\sin 38 \\ =17\times 0.6156 \\ =10.5\mathrm{m} \end{gathered}$$

Thus, the y component of the vector is $10.5\mathrm{m}$.