Vector Ahas y-component Ay= +9.60 m. A makes an angle of 32.0 counterclockwise from the +y-axis. (a) What is the x-component of A ? (b) What is the magnitude of A ?

The vector's various coordinates are presented here. We know that in coordinates, the x component of a vector comes first, followed by the y component. To calculate the angle of the vector with respect to the x axis, just multiply the y component by the x component and then take the inverse of that result.

This will tell you thevector's angle with the x axis.

= tan^-1 (y/x) is one way to express it.

Here

$$\begin{gathered} \theta ={32}^{°} \\ {A}_y=9.6meters \end{gathered}$$

Here,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ \tan \left(\theta \right)=\left(\frac{A_y}{A_x}\right) \\ { \\ }_{A_x=\frac{A_y}{\tan \left(\theta \right)}} \end{gathered}$$

Here we will use negative sign because vector is in second quadarant

$$\begin{gathered} A_x=\frac{13}{\tan \left({32}^{°}\right)} \\ =-19.66meters \\ Hencethexcoordinateis-19.66meterwhichcanbe20meterapprox.. \end{gathered}$$

The magnitude of the vector can be calculated by the sum of the square of the magnitude of the individual coordinates

Hence the magnitude of the vector is 21.87 meters and the x coordinate is -19.66 meter .