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Trigonometry
Angles & Degree Measure
Trigonometric Ratios
Right Triangle Problems
Law of Sines
Law of Cosines
Problem Solving
Angles and Degree Measure

Angles are always measured from the right xaxis to the terminal side of the angle.



Angles and Degree Measure
Angles are often expressed in units of degrees,
but can be into smaller increments called
minutes and seconds. These units are used to
rename the decimal part of a degree measure.
 1o = 60’
and 1’ = 60 “
 30.25o = 30o + (.25o)(60’/1o) = 30o 15’ 0”
 30.29o = 30o + (.29o)(60’/1o) = 30o 17.4’ the
0.4’ is then multiplied again to get seconds;
(.4’)(60”/1’) = 24”. This gives a final answer of
30o 17’ 24”

Angles and Degree Measure
Solving for degrees from minutes and seconds is
also possible.
 Example: 30o 27’ 45” = the sum of each of the
values expressed in degree units.
 30 + (27’)(1o /60’) + (45”)(1o /3600”) =
30 + 0.45 + .0125 = 30.46250 = 30.46o

Angles and Degree Measure

Coterminal angles are angles that have the same
starting and ending point but are expressed in
different values based on the direction of the
rotation or the number of rotations.

Going around CCW and returning to
the terminal side results in (360 +
). The rotations can be any integer
value. Even with increasing angle
measure, each angle is still equal to
the original angle.
Going around CW is (-) and results in
(-360 + ).
Angles and Degree Measure

Reference angles, denoted by , are always
measured from the x-axis of the quadrant that
the terminal sides is located back to the terminal
side. Reference angles are positive.


Trigonometric Ratios
c
a

b
Right Triangle




a – opposite side
b – adjacent side
c – hypotenuse
 - theta (reference )
Sine Function – sin 
 Sine ratio is the value
of the length of the
opposite side divided
by the hypotenuse of
a right triangle.
 sin  = a/c
 Values for sin 
oscillate between 0 to
1 to 0 to –1 to 0
Trigonometric Ratios
c
a

b
Right Triangle




a – opposite side
b – adjacent side
c – hypotenuse
 - theta (reference )
Sine Function – sin 
 Sine ratio works for
angle values between
0 and 360o and then
repeats.
 sin  = a/c is also
expressed as follows:
sin  = opposite side
hypotenuse
Trigonometric Ratios
c
a

b
Right Triangle




a – opposite side
b – adjacent side
c – hypotenuse
 - theta (reference )
Cosine Function – cos 
 Cosine ratio is the value of
the length of the adjacent
side divided by the
hypotenuse of a right
triangle.
 cos  = b/c
 Values for cos  oscillate
between 1 to 0 to -1 to 0
to 1
Trigonometric Ratios
c
a

b
Right Triangle




a – opposite side
b – adjacent side
c – hypotenuse
 - theta (reference )
Cosine Function – cos 
 Cosine ratio works for
angle values between
0 and 360o and then
repeats.
 cos  = b/c is also
expressed as follows:
cos  = adjacent side
hypotenuse
Trigonometric Ratios
Tangent Function – tan 
c
 Tangent ratio is the value
a
of the length of the

opposite side divided by
the adjacent side of a
b
right triangle.
Right Triangle
 tan  = a/b
 a – opposite side
 Values for tan  oscillate
 b – adjacent side
between  and -  within
 c – hypotenuse
o increment
each
180
  - theta (reference )
Trigonometric Ratios
c
a

b
Right Triangle




a – opposite side
b – adjacent side
c – hypotenuse
 - theta (reference )
Tangent Function – tan 
 Tangent ratio works for
angle values between 0
and 360o and then
repeats.
 tan  = a/b is also
expressed as follows:
tan  = opposite side
adjacent side
Trigonometric Ratios
c
a

b
Right Triangle




a – opposite side
b – adjacent side
c – hypotenuse
 - theta (reference )
Sine Function change in values
as  changes:
 Sin 0 = 0.0000
 Sin 30 = 0.5000
 Sin 45 = 0.7071
 Sin 60 = 0.8660
 Sin 75 = 0.9659
 Sin 90 = 1.0000
 Sin 120 = 0.8660
 Sin 135 = 0.7071
 Sin 150 = 0.5000
 Sin 180 = 0.0000
 Afterwards, sin repeats with
negative values
Trigonometric Ratios
c
a

b
Right Triangle




a – opposite side
b – adjacent side
c – hypotenuse
 - theta (reference )
Cosine Function change in
values as  changes:
 Cos 0 = 1.0000
 Cos 30 = 0.8660
 Cos 45 = 0.7071
 Cos 60 = 0.5000
 Cos 75 = 0.2588
 Cos 90 = 0.000
 Cos 120 = -0.5000
 Cos 135 = -0.7071
 Cos 150 = -0.8660
 Cos 180 = -1.0000
 Afterwards, cos  repeats
values
Trigonometric Ratios
c
a

b
Right Triangle




a – opposite side
b – adjacent side
c – hypotenuse
 - theta (reference )
Tangent Function change in
values as  changes:
 Tan 0 = 0.0000
 Tan 30 = 0.5773
 Tan 45 = 1.0000
 Tan 60 = 1.7321
 Tan 75 = 3.7321
 Tan 90 = undefined
 Tan 120 = -1.7321
 Tan 135 = -1.0000
 Tan 150 = - 0.5773
 Tan 180 = 0.0000
 Afterwards, tan  repeats
values
Trigonometric Ratios
c
a

b
Right Triangle




a – opposite side
b – adjacent side
c – hypotenuse
 - theta (reference )
Note that with each
function there is a
relation between an
angle measure and a
ratio of sides of the right
triangle.
The function of an angle is
equal to the decimal
value of the specific
ratio of sides.
Trigonometric Ratios
10
6
Likewise, the specific ratio is
equal to the function of an
angle.

8
Right Triangle




a – opposite side
b – adjacent side
c – hypotenuse
 - theta (reference )
Ex. Let sin  = 6/10 = .6000
Then  = Sin –1 .6000 =
36.9o
Trigonometric Ratios
C=2
Likewise, the specific ratio is
equal to the function of an
a=1
angle.

b = 1.732
Right Triangle




a – opposite side
b – adjacent side
c – hypotenuse
 - theta (reference )
Ex. Let sin  = 1/2 = .5000
Then  = Sin –1 .5000 =
30.0o
Trigonometric Ratios
10
6

8
Right Triangle




a – opposite side
b – adjacent side
c – hypotenuse
 - theta (reference )
Likewise, the specific ratio is
equal to the function of an
angle.
Ex. Let cos  = 8/10 =
.8000
Then  = cos –1 .8000 =
36.9o
Trigonometric Ratios
2
1
Likewise, the specific ratio is
equal to the function of an
angle.

1.732
Right Triangle




Ex. Let cos  = 1.732/2 = .8660
Then  = cos –1 .8660 =
a – opposite side
b – adjacent side
c – hypotenuse
 - theta (reference )
30.0o
Trigonometric Ratios
r
We can also solve for missing

values if we know two of the
20
three variables in a ration.
x
Ex. sin 30o = 20/r
Then r = 20/sin30o = 20/0.5 =
30o
 is the
complementary
angle of  and
so  = 90 – 30
or 60o
40
Tan 30o = 20/x
Then x = 20/tan30o =
34.6
Right Triangle Problems
Any situation that includes a
right triangle, becomes solvable
with trigonometry.

Angle of Elevation

Angle of Depression
Right Triangle Problems
This diagram has an inscribed triangle
whose hypotenuse is also the diameter
of the circle.
A
B

C
Given:  = 40o, AB = 6 cm
Determine the area of the
shaded region.
Solution: Sin 40o = 6/AC; AC = 6/ sin40o = 9.33 cm
AC/2 = radius of the circle = 4.67 cm; BC = 6/tan 40o =7.15 cm
Ao = r2 = 68.4 cm2, A = ½bh = 21.5 cm2; A = 46.9 cm2
Right Triangle Problems
l
h
r

This diagram has a right triangle within a
cone that can be used to solve for the
surface area and volume of the cone.
Given:  = 55o, l = 6 cm
Determine the area and volume
of the cone.
Solution: Sin 55o = h/l; h = 6 x sin55o = 4.91 cm
tan 55o = h/r; r = h/ tan 55o = 4.91/ tan 55o = 3.44 cm
Acone = r(r + l) =  (3.44)(3.44 + 6)cm2; Acone = 102 cm2
Vcone =⅓ r2h = ⅓ (3.44)2(4.91)cm3; Vcone = 60.8 cm3
Right Triangle Problems
h

x
The angle of elevation of a
ship at sea level to a
neighboring lighthouse is 2o .
The captain knows that the
top of lighthouse is 165 ft
above sea level. How far is
the boat from the lighthouse?
Solution: tan 2o = h/x; x = h/tan 2o
x = 165/ tan 2o = 4725 ft; 1 mile = 5280 ft, so the ship is .89
mile away from the lighthouse.
Function Values – Unit Circle
The unit circle is a circle with a radius of 1 unit.
Sin  = y; Cos  = x; Tan  = y/x

x
y
•The
coordinate of P will lead to
the value of x and y which in
turn leads to the values for
sine, cosine, and tangent.
•Use
the reference angle in each
quadrant and the coordinates to
solve for the function value.
Function Values – Unit Circle




•Quadrant
I – Angle = 
•Quadrant
II – Angle = (180 - )
•Quadrant
III – Angle = (180 + )
•Quadrant
IV – Angle = (360 - )
The reference angle is always measured
in its quadrant from the x – axis.
Function Values – Unit Circle




•Quadrant
I – P (x,y)
•Quadrant
II – P (-x, y)
•Quadrant
III – P (-x, -y)
•Quadrant
IV – P (x, -y)
With the values of changing from (+) to
(-) in each quadrant, and with the
functions of sine, cosine, and tangent
valued with ratios of x and y, the
functions will also have the sign values
of the variables in the quadrants.
Function Values – Unit Circle
Degrees
Sine
0
30
45
60
90
120
135
150
0
½
2/2
3/2
1
3/2
2/2
½
Cosine
1
3/2
2/2
½
0
-½
Tangent
0
3/3
1
3
-
-3
-1
Degrees 180
210
225
240
270
300
315
Sine
0
-½
Cosine
-1
-3/2 -2/2
-½
0
½
2/2
3/2
Tangent
0
3/3
3
-
-3
-1
-3/3
-2/2 -3/2
1
-1
-2/2 -3/2
-3/2 -2/2
-3/3
330
-½
Function Values – Unit Circle
Example: Cos 240 = _______
-½
240
60
P (-½, -3/2)
Example: Sin 240 = -3/2
_______
Example: Tan 240 = _______
3
Law of Sines
Law of Sines – For a triangle with angles A, B, C and sides
of lengths of a, b, c the ratio of the sine of each angle and
its opposite side will be equal. Sin A = Sin B = Sin C
a
b
c
C
b
A
a
h
B
c
Proof:
Sin A = h/b; Sin B = h/a
h = b Sin A, h = a Sin B
b Sin A = a Sin B; Sin A = Sin B
a
b
Law of Sines
Proof, continued:
k
k is the height of the same
triangle from vertex A
C
b
k = b Sin C and k = c Sin B
B
c
A
Sin C = k/b and Sin B = k/c
a
Sin B = Sin C
bb
c
b Sin C = c Sin B;
Conclusion: Sin A = Sin B = Sin C
a
b
c
Law of Sines
Let’s take a closer look:
Suppose A < 90o
c
a
a
A
If a = c Sin A, one
solution exists
c
A
c Sin A
If a < c Sin A, no
solution exists
Law of Sines
Let’s take a closer look:
Suppose A < 90o
c
A
a
c Sin A
If a > c Sin A and
a  c, one solution
exists
c
A
a
a
c Sin A
If c Sin A < a < c ,
two solution exists
Law of Sines
Let’s take a closer look:
Suppose A  90o
a
c
A
If a < c, no solution exists
a
c
A
If a > c , one
solution exists
Law of Cosines
For any triangle given two
sides and an included angle,
A
c
B
b
a2 = b2 + c2 – 2bcCos A
h
a -x
D
a
PROOF:
x
C
b2 = a2 + c2 – 2acCos B
c2 = b2 + a2 – 2abCos C
b2 = h2 + x2; h2 = b2 - x2
Cos C = x/b ; x = b Cos C
c2 = h2 + (a-x)2; c2 = h2 + a2 –2ax + x2
Substitute and we get:
c2 = (b2 - x2)+ a2 –2a(bCos C) + x2
c2 = b2 + a2 – 2abCos C
Law of Cosines - example
B
Given: A = 50o; AB = 8 cm, AC = 14 cm
x
A
Find x.
C
x 2 = 8 2+ 14 2 – 2(8)(14)Cos50
x 2 = 260 – 144; x 2 = 116
x = 10.8 cm
Law of Cosines – Hero’s Formula
For any triangle, the area of the triangle can be
determined with Hero’s Formula.
s = ½(a + b + c) where s = semi-perimeter of
the triangle.
A =  s(s – a)(s – b)(s – c)
Law of Cosines – Hero’s Formula
Given: ABC; a = 7, b = 24, c = 25
7cm
25 cm
S = ½ (7 + 24 + 25) = 28
24 cm
A =  28(28 – 7)(28 – 24)(28 – 25) = 84 cm 2
Check: A = ½(b)(h) = ½(7)(24) = 84 cm 2