P(-6,7) lies on the terminal arm of an angle in standard position. What is the value o the principal angle a to the nearest degree?

1 answer

To find the principal angle \( \theta \) in standard position given that the point \( P(-6, 7) \) lies on its terminal arm, we use the inverse tangent function to determine the angle.

First, we calculate the angle \( \theta' \) formed with the positive x-axis using the coordinates of the point \( P(-6, 7) \):

\[ \tan(\theta') = \frac{y}{x} = \frac{7}{-6} \]

This yields:

\[ \theta' = \tan^{-1}\left(\frac{7}{-6}\right) \]

Calculating this value gives us a negative angle, if we consider the principal value using a calculator (make sure your calculator is in the correct mode for degrees).

Since \( P \) is in the second quadrant (where \( x \) is negative and \( y \) is positive), we need to adjust the angle we obtained by adding \( 180^\circ \) to it:

\[ \theta = 180^\circ + \theta' \]

Now calculating \( \theta' \):

\[ \theta' \approx \tan^{-1}\left(-\frac{7}{6}\right) \approx -48.37^\circ \]

Thus:

\[ \theta \approx 180^\circ + (-48.37^\circ) \approx 180^\circ - 48.37^\circ \approx 131.63^\circ \]

Rounding this value to the nearest degree gives:

\[ \theta \approx 132^\circ \]

Thus, the value of the principal angle \( a \) to the nearest degree is:

\[ \boxed{133^\circ} \]