Vector A is in the direction 34.0 clockwise from the -yaxis. The x-component of Ais Ax= -16.0 m. (a) What is the y-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.

$={\tan }^{-1}\left(\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\right)$is one way to express it.

Here

= 90-34º=56 (Clockwise)

A_x = 16 meter

=tan-1yx

Here,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A^y}{A^x}\right) \\ \tan \theta =\frac{A^y}{A^x} \\ {A}_x\tan \theta =A_y \end{gathered}$$

Here we will usenegative sign because vector is in second quadarant

A_y = 16 tan 56

= 23.7 meter

Hence the y coordinate is 23.7 meter which can be 24 meter approx..

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

$$\begin{gathered} A=\sqrt{{\left(A_x\right)}^2+{\left(B_x\right)}^2} \\ =\sqrt{{\left(23.7\right)}^2+{\left(16\right)}^2} \\ =28.6meter \end{gathered}$$

Hence the magnitude of the vector is 28.6 meters and the x coordinate is 23