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Right Triangles

Angles in Standard Position

Sine Law & Ambiguous Case

Cosine Law

Problem Solving

Summary&Test

 

jdsmathnotes

 

 


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:

 

 

 

 

Text Box:
 

 

 

 

 

 

 


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.

 

 

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