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, we multiply the y component by the x component and then take the inverse of that result.

This tells us thevector's angle with the x axis.

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

Here,

= 90-34º=56 (Clockwise)

A_x = 16 meter

Here,

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

Here, we will usenegative sign because vector is in second quadrant.

A_y = 16 tan 56

= 23.7 meter

Hence, the y coordinate is 23.7 meter which can be rounded off to 24 meter approx..

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

$A=\sqrt{{\left(A_x\right)}^2+}{\left(B_x\right)}^2$

=$\sqrt{{\left(23.7\right)}^2+{\left(16\right)}^2}$

=28.6 meter

Hence, the magnitude of the vector is 28.6 meters and the x coordinate is 23.7 meter .