Given the vector components $x = -3.00 \, \text{cm}$ and $y = 8.00 \, \text{cm}$, we need to determine the angle it makes with the positive $y$-axis, measured counterclockwise.

Step-by-Step Process:

1. Determine the angle with the positive $x$-axis:

   • The tangent of the angle relative to the positive $x$-axis is given by: $\tan(\theta_x) = \frac{y}{x} = \frac{8.00 \, \text{cm}}{-3.00 \, \text{cm}} = -2.67$
   • Therefore, the angle relative to the $x$-axis is: $\theta_x = \tan^{-1}(-2.67)$ Using a calculator: $\theta_x \approx -69.44^\circ$

2. Adjust the angle relative to the positive $y$-axis:

   • The angle we found is measured clockwise from the negative $x$-axis, which is equivalent to the angle from the positive $y$-axis. The total angle in a full circle is $90^\circ - \theta_x$, as we are measuring counterclockwise from the $y$-axis.

   Thus, the angle relative to the $y$-axis is: $\theta_y = 90^\circ - (-69.44^\circ) = 90^\circ + 69.44^\circ = 159.44^\circ$

Thus, the vector makes an angle of 159.44° with the positive $y$-axis in the counterclockwise direction.

Would you like more details, or do you have any questions?

Related questions:

1. What would be the angle if the vector had a positive $x$-component?
2. How does the quadrant of the vector affect the calculation of the angle?
3. What is the magnitude of the vector with components $(-3.00 \, \text{cm}, 8.00 \, \text{cm})$?
4. What is the direction of the vector relative to the negative $y$-axis?
5. How would the angle change if both components were negative?

Tip: Always visualize vector direction when solving trigonometric problems to better understand angle orientation.