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x

y

Day 1: Angles In Standard

Position

terminal a

rm

initial arm

standard position

initial arm 0º

term

inal

arm

x

ynon-standard position

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x

yPositive angles

rotate counterclockwise

ex: 80º

x

yNegative angles

rotate clockwise

ex: –120º

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x

yQuadrantsIII

III IV

0º

90º

180º

270º

quadrant angle

I

II

III

IV

0º < < 90º90º < < 180º

180º < < 270º

270º < < 360º

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x

yExample:

Let P(x, y) be a point on the terminal arm of an angle in standard position.

P(x, y)

Point P can be anywhere in the x-y plane.

1

1 is in quadrant II.

x

y

P(x, y)

2 is in quadrant IV.2

90º < < 180º

270º < < 360º

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x

y

3P(x, y)

3 lies in the negative x-axis.

3 = 180º

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The principal angle is the angle between 0º and 360º.

The related acute angle is the angle formed by the terminal arm of an angle in standard position and the x-axis. x

y

1

The related acute angle lies between 0º and 90º.

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x

y

x

y

x

y

= 150º

= 180º – 150º

= 30º

= 220º

= 220º – 180º

= 40º

= 325º

= 360º – 325º

= 35º

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Example 1: The point P(–5, –4) lies on the terminal arm an angle in standard position.

a) Sketch the angle.

b) Determine the value of the related acute angle.

c) Determine the principal angle .

x

y

P(–5, –4)

5

4

4tan

5

tan 0.8 1tan 0.8

39 = 180º + 39º

= 219º

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Example 2: The point P(– 6, 7) lies on the terminal arm an angle in standard position.

a) Sketch the angle.

b) Determine the value of the related acute angle.

c) Determine the principal angle .

x

yP(–6, 7)

7

6

7tan

6

tan 1.1666 1tan 1.1666

49.4 = 180º – 49.4º

= 130.6º