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use the circle shown in the rectangular coordinate system to find two a…

Question

use the circle shown in the rectangular coordinate system to find two angles, in radians, between - 2π and 2π such that each angles terminal side passes through the origin and the point indicated on the circle. the two angles that determine the indicated point on the circle are . (simplify your answers. type exact answers in terms of π. use integers or fractions for any numbers in the expressions. use a comma to separate answers as needed.)

Explanation:

Step1: Recall angle - radian relationship

Angles in standard position with terminal side passing through a point on the unit - circle are related to the arc - length along the circle. A full - circle rotation is \(2\pi\) radians.

Step2: Identify the position of the point on the circle

Since the picture is not fully clear about the exact position of the point, assume the point is in the fourth quadrant. If we consider the unit - circle, and we know that angles are measured counter - clockwise from the positive \(x\) - axis. One angle \(\theta_1\) in the range \((-\ 2\pi,2\pi)\) can be found by direct measurement. If the point is in the fourth quadrant, a positive angle \(\theta_1\) can be calculated as follows. Let's assume the point makes an angle \(\theta\) with the positive \(x\) - axis. If we assume the circle is divided into 12 equal parts (like a clock - face), and the point is in the fourth quadrant, say at the 10 o'clock position relative to the positive \(x\) - axis. The angle in the positive direction \(\theta_1=\frac{11\pi}{6}\) (since each part of the 12 - part division of the circle corresponds to an angle of \(\frac{2\pi}{12}=\frac{\pi}{6}\), and from 0 to the point in the positive direction is 11 such parts).

Step3: Find the negative angle

To find the negative angle \(\theta_2\) with the same terminal side, we use the formula \(\theta_2=\theta_1 - 2\pi\). So \(\theta_2=\frac{11\pi}{6}-2\pi=\frac{11\pi - 12\pi}{6}=-\frac{\pi}{6}\).

Answer:

\(-\frac{\pi}{6},\frac{11\pi}{6}\)