     Home Right Triangles Angles in Standard Position Sine Law & Ambiguous Case Cosine Law Problem Solving Summary&Test UNIT 6  : BASIC TRIGONOMETRY WITH TRIANGLES

LESSON 3:  ANGLES IN STANDARD POSITION

Definition:   An angle is in standard position if it has its vertex at the origin and  initial arm along the positive x-axis.  The terminal arm is found by rotating the initial arm about the origin to a terminal position in one of the 4 quadrants.  The rotation is positive if it is in the counter – clockwise direction and negative if in the clockwise direction.    Example 1: Solution: x = 4 y = 3 r = 5  Example 2: Angles greater than 900 Solution:  x = -5 y = 12 r = 13 Note that sine is positive and cosine and tangent are negative for a second quadrant angle.

Example 3: Complementary angles (angles which add to 180o).

Using your calculator,  sin 120o = 0.8660254;   also  sin 60o = 0.8660254

Hence sin 120o = sin 60o  or

sin(120o) = sin (180o – 120o) = sin 60o

Note that sine is positive in both the first and second quadrants.

Note 60o is the reference(related acute)  angle relative to 120o.

Definition:

The reference (related acute) angle is the angle between the terminal arm and the x-axis.

In the above example (120o), it is found by subtracting from 180o.

Similarly,  cos 150o = -0.8660254;  also  cos 30 = 0.8660254

Hence cos 150o = -cos 300  or

cos(150o) = -cos (180o – 150o) = -cos 30o

Note that cosine is positive in the first quadrant and negative in the second quadrant.

Note 30o is the reference (related acute) angle relative to 150o (180o – 150o = 30o)

Result:  To find the trigonometric ratios of angles between 90o and 180o, use the following rules: Examples:

1.  Find cos 135o

Solution:

cos 135o = - cos (180o – 135o)

= - cos 45o                            ** 45o is reference (related acute) angle

= - 0.707106781                  ** Using your cacluator

2.  Find sin 114o.

Solution:

sin 114o = sin(180o – 114o)

= sin 66o

= 0.913545457

3.  Find tan 161o

Solution:

tan 161o = - tan(180o – 161o)

= - tan 19o

= - 0.344327613

Note that tangent is positive in the first quadrant and negative in the second quadrant.

4.  Find sin 123o using a calculator to 5 decimal places.

Solution: 5.  Find sec 149o using the calculator to 5 decimal places.

Solution:  Solution:  Solution:

Since csc A is positive, < A could be in either the first or second quadrant.  There are two solutions. 