Transcript Trigonometry (Chapter 9)

2.1 Six Trig Functions for Right Triangles
sin () = Opposite
csc () = Hypotenuse
Hypotenuse
Opposite
cos () = Adjacent
sec () = Hypotenuse
Hypotenuse
[secant]
Adjacent
tan () = Opposite
cot() = Adjacent
Adjacent
Opposite
Note:
 is an Acute angle.

12
[cosecant]
[cotangent]
sin () = 12
13
csc() = 13
12
cos() = 5
13
sec() = 13
5
tan() = 12 / 5
cot() = 5 / 12
13

5
Trig Cofunction Identities
sin () = cos (90 - )
cos () = sin (90 - )
tan () = cot (90 - )
cot () = tan (90 - )
sec () = csc (90 - )
csc () = sec (90 - )
a
90-
c

b
2.2 Reference Angles
Evaluating trigonometric functions of angles greater than 90° as well as
negative angles is done by using a reference angle.
Let  be a nonacute angle in standard position that lies in a quadrant.
Its reference angle is the positive acute angle ’ formed by the terminal
side of  and the x-axis.
90
180°
90
0°
270°
180°
90
0°
270°
Angle: 120°
Reference Angle: 60°
Find exact values for
sin(120) =
cos (120) =
tan (120) =
Angle: 200° or -160°
Reference Angle: 20°
180°
0°
270°
Angle: 320° or -40°
Reference Angle: 40°
2.3 Finding Trig Functions with Calculators
•
Make sure the calculator is in ‘DEGREE’ mode
(Ti-83/84 – Use the ‘Mode’ Button)
Examples:
• Find sin (49º 12’)
12/60 = .2 so, find sin (49.2) = .756995
2. Find sec (97.977º)
Secant is the reciprocal of cosine.
Calculators do not have a ‘secant’ function so
sec (97.977) = 1/cos(97.977) = -7.205879
3. Sin  = .967709, Find 
Sin-1 (.967709) = 74.4º
2.4 Solving Right Triangles
B
25
x
50
C
y
A
Find the length of side BC
Sin (50) = x
---25
So, x = 25 Sin (50) = 25 (.766) = 19.15
How would you find side AC?
Think about using the sine, cosine or tangent function.
Angles of Elevation/Depression

20,000 ft

Ground
34,000 ft
Angle  is an angle of elevation from the ground to the sky.
Angle  is an angle of depression from the sky to the ground.
You can use trigonometric functions (sine, cosine, tangent) to find the angles & sides.
Tan () = 20,000 / 34,000
-1
 = tan (20,000/34,000) = 30
2.5 Bearing Application Problems
Bearings are expressed using one of two methods:
Method 1: (A single angle is given in
degrees)  220º
Method 2: (Directions and angle are
given)  S 40º W
-- Measure the angle in a ‘clockwise’
direction from North
-- Start with North/South Line
-- Use an acute angle to show direction
either east or west from the line.
220º
40º
Example: Bearing Problem
The bearing from A to C is S 52º E.
The bearing from A to B is N 84º E
The bearing from B to C is S 38º W.
A plane flying at 250 mph takes 2.4 hours to go from A to B.
Find the distance from A to C.
Finding distance from A to B
D= RT
96º
D= (250)(2.4) =600 miles
600 miles
A 84º
B
46º
38º
44º
52º
C
Finding distance from A to C
Sin (46) = AC
600
AC = 600Sin(46)
= 600 (.7193)
= 431.58 miles