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# Understand Quadrant and Angle

### Quadrant and xy-plane

### Positive Angle and Negative Angle

# Definition of Any Angle

### Angle in the Quadrant I

### Angle in Quadrant II

### Angle in Quadrant III

### Angle in Quadrant IV

In the figure shown above, the point O is the origin of the xy-plane. There are four quadrants in the xy-plane. If a point lies in quadrant I, then its x-coordinate is positive and y-coordinate is positive. If a point lies in quadrant II, then its x-coordinate is negative and y-coordinate is positive. If a point lies in the quadrant III, then its x-coordinate is negative and y-coordinate is negative. If a point lies in quadrant IV, then its x-coordinate is positive and y-coordinate is negative.

In the figure above, ray OA overlap with the positive x-axis, its one end is the origin of the xy-plane, the other point is point A. ray OA counterclockwise rotate from OA to OB to form an angle.
The vertex of the angle is the origin, OA and OB are two sides of the angle. By definition, angles counterclockwise rotated is called a positive angle and angles clockwise rotated is called a negative angle.
If ray OA counterclockwise rotate a full circle, then the degree measure of the angle is 360^{o}

Since OB is the terminal side of the angle a, formed by rotating ray OA counterclockwise to OB, so the angle a is a positive angle.

Since OC is the terminal side of the angle beta, formed by rotating ray OA clockwise to OC, so the angle beta is a negative angle.

In the figure shown above, OA is the initial side of the angle a. The initial side of the angle overlap with the positive x-axis. One end point of the ray OA is the point O, which is the origin of the xy-plane. The ray OA counterclockwise rotate to the OB position. The OB is called the terminal side of the angle a. The vertex of the angle a is the point O, its two sides are OA and OB. Since OB lies on quadrant I, then the angle a is called the angle of quadrant I. Since OA counterclockwise rotate to OB to form the angle a, so, the angle a is a positive angle.

If the terminal side of an angle lies in quadrant I, then the angle is the angle of the quadrant I.

If OB overlap with OA, then the angle a = 0^{o}. If OB overlap with the positive y-axis, then the angle a is 90^{o}. In these conditions, the angle a is not a quadrant angle.

In the range of 0 to 360^{o}, since the angle a is the angle of the quadrant I, so, 0^{o} < a < 90^{o}.

In the figure shown above, the vertex of the angle a is the origin. The initial side of the angle a is OA that is overlapping with the positive x-axis. The terminal side of the angle a is OB that lies in the quadrant II. So, the angle a is the angle of quadrant II. Since OA counterclockwise rotate to OB, so, the angle a is a positive angle.

If the terminal side of an angle lies in quadrant II, then the angle is the angle of the quadrant II.

If OB overlap with the positive y-axis, then the angle a is 90^{o}. If OB overlap with the negative x-axis, then the angle a is 180^{o}. In these conditions, the angle a is not a quadrant angle.

In the range of 0 to 360^{o}, since the angle a is the angle of the quadrant II, so, 90^{o} < a < 180^{o}.

In the figure shown above, the vertex of the angle a is the origin. The initial side of the angle a is OA that is overlapping with the positive x-axis. The terminal side of the angle a is OB that lies in the quadrant III. So the angle a is the angle of quadrant III. Since OA counterclockwise rotate to OB, so, the angle a is a positive angle.

If the terminal side of an angle lies in quadrant III, then the angle is the angle of the quadrant III.

If OB overlap with the negative x-axis, then the angle a is 180^{o}. If OB overlap with the negative y-axis, then the angle a is 270^{o}. In these conditions, the angle a is not a quadrant angle.

In the range of 0 to 360^{o}, since the angle a is the angle of the quadrant III, so, 180^{o} < a < 270^{o}.

In the figure shown above, the vertex of the angle a is the origin. The initial side of the angle a is OA that is overlapping with the positive x-axis. The terminal side of the angle a is OB that lies in the quadrant IV. So the angle a is the angle of quadrant IV. Since OA counterclockwise rotate to OB, so, the angle a is a positive angle.

If the terminal side of an angle lies in quadrant IV, then the angle is the angle of the quadrant IV.

If OB overlap with the negative y-axis, then the angle a is 270^{o}. If OB overlap with the positive x-axis, then the angle a is 360^{o}. In these conditions, the angle a is not a quadrant angle.

In the range of 0 to 360^{o}, since the angle a is the angle of the quadrant IV, so, 270^{o} < a < 360^{o}.

**Example 1**

In xy-plane, point P_{1} has the coordinate (-2, 3), point P_{2} has the coordinate (2, -3) and point P_{3} has the coordinate (-3, -2). Find which quadrants these three points lie in.

Solution:

Look the figure above, locate the point P_{1} on the xy-plane, its x-coordinate is -2 and y-coordinate is 3. Since x < 0 and y > 0, so, the point of P_{1} lies in quadrant II.

locate the point P_{2} on the xy-plane, its x-coordinate is 2 and y-coordinate is -3. Since x > 0 and y < 0, so, the point of P_{2} lies in quadrant IV.

locate the point P_{3} on the xy-plane, its x-coordinate is -3 and y-coordinate is -2. Since x < 0 and y < 0, so, the point of P_{3} lies in quadrant III.

**Example 2**

In which quadrant, the following angles lie in? (1). 420^{o} (2). -570^{o} (3). 600^{o}

Solution:

(1). Look the figure below, the angle 420^{o} starts from OA and counterclockwise rotate a full circle to the OA position and continue rotate to the OB position.
Since 420^{o} = 360^{o} + 60^{o}, so, the terminal side of 420^{o} lies in quadrant I. That is, 420^{o} is the angle of quadrant I.

Note: Both angles 420^{o} and 60^{o} have the same terminal side that lies in quadrant I and they are the angles of the quadrant I.

(2). Look the figure below, the angle -570^{o} starts from OA and clockwise rotate two full circles then change direction to counterclockwise rotate to the OB position. Since
-570^{o} = -2 × 360^{o} + 150^{o}, so, the terminal side of -570^{o} lies in quadrant II. That is, -570^{o} is the angle of quadrant II.

Note: Both angles -570^{o} and 150^{o} have the same terminal side that lies in quadrant II and they are the angles of the quadrant II.

(3). Look the figure below, the angle 600^{o} starts from OA and counterclockwise rotate two full circles then change direction to clockwise rotate to the OB position. Since
600^{o} = 2 × 360^{o} - 120^{o}, so, the terminal side of 600^{o} lies in quadrant III. That is, 600^{o} is the angle of quadrant III.

Note: Both angles 600^{o} and -120^{o} have the same terminal side that lies in quadrant III and they are the angles of the quadrant III.

As mentioned, if the terminal side of the angle a lies in quadrant I, then k × 360^{o} < a < k × 360^{o} + 90^{o}, in which k is integer.

If the terminal side of the angle a lies in quadrant II, then k × 360^{o} + 90^{o} < a < k × 360^{o} + 180^{o}.

If the terminal side of the angle a lies in quadrant III, then k × 360^{o} + 180^{o} < a < k × 360^{o} + 270^{o}.

If the terminal side of the angle a lies in quadrant IV, then k × 360^{o} + 270^{o} < a < k × 360^{o} + 360^{o}.