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© The Visual Classroom x y Day 1: Angles In Standard Position  terminal arm initial arm standard position  initial arm 0 º terminal arm x y non-standard.

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Presentation on theme: "© The Visual Classroom x y Day 1: Angles In Standard Position  terminal arm initial arm standard position  initial arm 0 º terminal arm x y non-standard."— Presentation transcript:

1 © The Visual Classroom x y Day 1: Angles In Standard Position  terminal arm initial arm standard position  initial arm 0 º terminal arm x y non-standard position  

2 © The Visual Classroom x y Positive angles rotate counterclockwise  ex: 80º x y Negative angles rotate clockwise  ex: –120º

3 © The Visual Classroom x y Quadrants I II III IV 0º 90º 180º 270º quadrantangle I II III IV 0º <  < 90º 90º <  < 180º 180º <  < 270º 270º <  < 360º

4 © The Visual Classroom x y Example: 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  1 is in quadrant II. x y P(x, y)  2 is in quadrant IV. 22 90º <   < 180º 270º <   < 360º

5 © The Visual Classroom x y 33 P(x, y)  3 lies in the negative x-axis.  3 = 180º

6 © The Visual Classroom 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1 The related acute angle lies between 0º and 90º.

7 © The Visual Classroom x y   x y   x y    = 150º  = 180º – 150º = 30º  = 220º  = 220º – 180º = 40º  = 325º  = 360º – 325º = 35º

8 © The Visual Classroom 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   = 180º + 39º  = 219º 

9 © The Visual Classroom 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 y P(–6, 7) 7 6   = 180º – 49.4º  = 130.6º 


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