Transcript TRIGONOMETRY - Patchogue-Medford School District

Coterminal Angles
Angle measured in standard position.
Initial side is the positive x – axis which is fixed.
Terminal side is the ray in quadrant II,
which is free to rotate about the origin.
Counterclockwise rotation
is positive, clockwise rotation is negative.
Coterminal Angles: Angles that have the same
terminal side.
60°, 420°, and –300° are all coterminal.
Standard position and
reference angles
Angles in standard position are between 0 and 360 degrees
Reference angles are between 0 and 90 degrees
Drawing an angle in standard
position
• -2100
-2100
so it’s in the second quadrant
Coterminal angles
• Add
(multiples)
360
• Until we get an angle between 0 and 360
• Examples:
60°, 420°, –300°, 750°,
o
•
•
•
•
60° is already there
420 – 360 = 60°
-300 + 360 = 60°
750-360=390-369=30°
Drawing an angle in standard
position
• -2100
-2100
so it’s in the second quadrant and it is the angle -210 +360 = 150
Now the reference angles:
Quad I
Quad II
180-angle
Angle - 180
Quad III
360-angle
Quad IV
the reference angle for
150:
Quad I
Quad II
180-150=30
Quad III
Quad IV
The unit circle: (x,y) points are
(cosθ,sinθ)
• The 4 quadrantal points:
90
(0,1)
180
(1,0)
(-1,0)
(0,-1)
270
Quick guide for quadrantals
• sin and cos:
sin
cos
0
0
-1
1
-1
0
Using for sin and cos values
• Find sin 270, cos 180, sin 540
sin
cos
0
0
-1
1
-1
0
Using for sin and cos values
• Find sin 270, cos 180, sin 540
sin
cos
0
0
1
-1
-1
• Sin 270 = -1 cos 180 = -1
• sin 540=sin 180= 0
0
Do Now:
• List the ratios of the sides of a right
triangle for the following in terms of
• o,a,h (opposite, adjacent, hypotenuse)
sin  
cos  
tan  
sohcahtoa
• List the ratios of the sides of a right
triangle for the following in terms of
• o,a,h (opposite, adjacent, hypotenuse)
o
sin  
h
a
cos 
h
o
tan  
a
There are 3 more!
Reciprocal identities:
o
h
sin   csc  
h
0
a
h
cos  sec  
h
a
o
a
tan   cot  
a
o
Right Triangle Trig Definitions
B
c
a
C
•
•
•
•
•
•
b
A
sin(A) = sine of A = opposite / hypotenuse = a/c
cos(A) = cosine of A = adjacent / hypotenuse = b/c
tan(A) = tangent of A = opposite / adjacent = a/b
csc(A) = cosecant of A = hypotenuse / opposite = c/a
sec(A) = secant of A = hypotenuse / adjacent = c/b
cot(A) = cotangent of A = adjacent / opposite = b/a
Find the missing side..
• Find the side:
sinθ=2/3
4
θ
6
θ
Find the missing side
• Find the ratios:
sinθ=2/3
52
4
θ
6
2
3
θ
5
Find the cos,sec for each:
• 2 for each problem:
sinθ=2/3
52
4
θ
6
6
52
cos 
sec 
6
52
2
3
θ
5
5
3
cos 
sec 
3
5
Triangles formed by points:
Standard Positions
Quad I
Quad II
Quad III
Quad IV
Given a point: (3,-2)
Find the hypotenuse:
c  13
3
-2
Quad IV
Steps-Terminal sides
• Plot the point given and make a right
triangle.
• Use the pythagorean theorem to find the
missing side
• Set up the trig function proportions
Point: (3,-2)
Find the ratios:
h
13
csc  
o 2
h
13
sec  
a
3
a 3
cot  
o 2
3
13
-2
Quad IV
Special Right Triangles
Do Now: find the two ratios with your calculator, express
as a fraction
30°
45°
60°
cos(60o) 
45°
tan(45o) 
Special Right Triangles
30°
45°
2
2
3
1
3
cos(30 ) 
2
1
sin(30 ) 
2
3
tan(30 ) 
3

Memorize!
1
60°
1
2
2
2
sin(45 ) 
2
tan(45 )  1
1
2
3
sin(60 ) 
2
tan(60 )  3
cos(45 ) 
cos(60 ) 
.8660 
3
2
AND .7071 
2
2
45°
Special Right Triangles
30°
45°
2
2
3
1
1
60°
sec 60 
csc 60 
cot 60 
Memorize!
1
sec 45 
csc 45 
cot 45 
.8660 
3
2
AND .7071 
2
2
45°
Special Right Triangles
30°
2
3
1
60°
sec 30 
A chart for memory

0
sin
0
1
2
cos
1
3
2
tan
0
3
3
30
45
2
2
2
2
1
60
90
3
2
1
2
3
1
0
und
The unit circle: (x,y) points are
(cosθ,sinθ)
• Find x and y:
90
(0,1)
(x,y)
1200
180
(-1,0)
(0,-1)
270
(1,0)
Do Now:
• What part of an hour is 45 minutes?
• If you are to stay in your room for a quarter
of an hour, how many minutes is that?
• Using degree mode, find tan1 3
Degrees,minutes,seconds
• DMS
• Find the tan 300 12’ in the calculator (4
decimals)
• Enter it using the angle menu:
= .5820
Degrees,minutes,seconds
• DMS
• Find the sec 300 13’ in the calculator (4
decimals)
• Enter it using the angle menu:
Degrees,minutes,seconds
• DMS
• Find the sec 300 13’ in the calculator (4
decimals)
• Enter it using the angle menu:
•
= 1.1572
Finding an angle:
• Find the smallest angle when…
• sinθ = .3825
(be in degree mode)
• Use sin-1
Finding an angle:
•
•
•
•
Find the smallest angle when…
sin θ= .3825
(be in degree mode)
Nearest degree = 22
Nearest minute= 220 29’
“arc” means to find an angle:
•
•
•
•
Find the smallest angle when…
arccos (.2395)
(be in degree mode)
Nearest degree =
Nearest minute=
Finding an angle:
•
•
•
•
Find the smallest angle when…
cos θ= .2395
(be in degree mode)
Nearest degree = 76
Nearest minute= 760 9’
34 seconds is more than
½ a minute.
Find the angle to the nearest
minute:
• Find angle A:
7
A
5
Find the angle to the nearest
minute:
• Find angle A:
5
cos1 ( )  44.4153
7
 44 0 24 '55"
7
 44 0 25 '
A
5
Conversions-degrees to
radians or radians to degrees
a.2.m.1 & 2
Degrees to radians: Multiply angle by

180 
.
Ex. Convert 60 degrees to radians


60 
 radians

3
180

Radians to degrees: Multiply angle by
 180

 45
4

Note: 1 revolution = 360° = 2π radians.
180

.
Practice:
• Convert to radians:
• 30, 90,45, degrees
Practice:
• Convert to radians:
• 30, 90,45, degrees
30 
90 
45 
• Convert to degrees:
3  13
,
,
4 9
6

180

180

180




6

2

4
Practice:
• Convert to degrees:
45
3 180

 135
4

20
 180

 20
9  30
13 180

 390
6

More practice:
• Convert to radians: (approximate to
hundredths
• 42 degrees
• Convert to degrees: (to nearest tenth)
• 2 radians
Approximating:
• Convert to radians: (approximate to
hundredths
• 42 degrees

7
42 

 .73 radians
180 30
• Convert to degrees: (to nearest tenth)
• 2 radians
2
180


360

 114.6 o
arc length
Arc length = central angle x radius, or
s  r.
s
θ
r
Note: The central angle,θ, must be in radian measure.
S=rθ
• Given θ = 2.5 and the radius =10, find the
intercepted arc length.
S=rθ
• Given θ = 2.5 and the radius =10, find the
intercepted arc length.
S = 2.5(10)
S = 25
S=rθ
• Given θ = 500 and the radius =12 cm, find
the intercepted arc length.
But here we don’t have the angle in radians, so we will need to
Convert first….
S=rθ
• Given θ = 500 and the radius =12 cm, find
the intercepted arc length to the nearest
hundredth.

5
50 
S=rθ
S = 12(.8727)
180
10.47

18
 .8727 radians
S=rθ

• Given θ = 3 and the arc length =12, find
the radius to the nearest hundredth.
S=rθ
12  r(

3
)
Now, solve for r:
S=rθ

• Given θ = 3 and the arc length =12, find
the radius to the nearest hundredth.
12  r(
12

S=rθ
12  r(

3

r
r
3
)

3

3
)

Remember to “flip” and multiply!
3
36

 11.5
Do Now:
• How many hours and minutes is 4.5
hours?
• How many is 3.75 hours?
• What decimal part of an hour is 15 min?
Do Now:
• How many hours and minutes is 4.5
hours?
4hr. 30 min
• How many is 3.75 hours?
•
3 hr. 45 min.
• What decimal part of an hour is 15 min?
•
.25
Degrees to DMS
• Degrees minutes seconds:
• Multiply the portion of a degree by 60 for
the minutes and the decimal part left by 60
for the seconds.
• Example:
0
20.36 
20 0  .36(60 )
 20 0 21.6
 20 0 21  .6(60 )
20 0 21 36
DMS to degrees
• Put the minutes/60 and seconds/3600
1101440 
14
40
11 
60 3600
 11.24 0
Do Now: given a sector of a
circle:

• θ = 3 , the arc length =12, find the
radius in terms of 
Use: S = r θ
Do Now: given a sector of a
circle:

• θ = and 3 , the arc length =12, find the
radius in terms of 
S=rθ
12  r(

3
)
Now, solve for r:
Cofunctions
•
•
•
•
sin 40 = cos 50
sec 20= csc 70
tan 30 = cot 60
What’s the pattern?
Cofunctions
•
•
•
•
•
sin 40 = cos 50
sec 20= csc 70
tan 30 = cot 60
What’s the pattern?
Cofunctions are complimentary
Cofunctions
• sin x = cos (90 – x)
• sec x= csc (90 – x)
• tan x = cot (90 – x)
•
•
•
•
Cofunctions are complimentary
Example: if sin (x+30) = cos (2x)
2x +x + 30 = 90
x=20
Cofunctions
• 1. If sin A = cos 30
Then sin A = ?
2. If tan A = 1.3 then Cot (90-A) = ?
3. If x is an acute angle and sec
csc (4x + 15), solve for x.
x
2
=
Cofunctions
• 1. If sin A = cos 30
then sin A = ?
This means sin 60 which =
3
2
2. If tan A = 1.3 then Cot (90-A) = 1.3
x
2
3. If x is an acute angle and sec =
csc (4x + 15), solve for x
x
2 +4x + 15 = 90 4.5x = 75 x = 16
cofunctions
• 4. If sin 290 10’ = cos A, find A
• 5. If cot 2x = tan θ, find θ in terms of x
cofunctions
• 4. If sin 290 10’ = cos A
• 90 - 290 10’= 600 50’
• 5. If cot 2x = tan θ, find θ in terms of x
• 90 – 2x
Basic Trigonometric Identities
Quotient identities: tan(A) 
sin( A)
cos(A)
cos(A)
sin( A)
cot(A) 
Reciprocal Identities:
1
csc( A) 
sin( A)
1
sin( A) 
csc( A)
1
sec( A) 
cos(A)
1
cos(A) 
sec( A)
1
cot (A) 
t an(A)
1
t an(A) 
cot (A)
Pythagorean Identities:
sin 2 ( A)  cos2 ( A)  1
2
2
tan2 ( A)  1  sec2 ( A)
1  cot ( A)  csc ( A)
Arc Trig-finding angles…

• Warm up: find the sec
3
Note: We never “flip” angles!
• Be sure you are in the correct mode and
0 to the nearest ten
find
sec 24
thousandth.
Finding exact angles:
• In degree mode, find the sin -1
• Another way to say that is, find
 when
2
sin  
2
2
2
Finding exact angles:
• Always in degree mode, find the arcsin
• Another way to say that is, find
 when
2
sin  
2
  45
But when else is sin positive?
2
2
Finding exact angles:
• In degree mode, find the arcsin
  45, 135
 3
 ,
180 -45
4 4
sin is positive in quadrants I and II
2
2
Finding exact angles:
• In degree mode, find the arctan
  60
180 + 60
tan is positive in quadrants I and III
3
Finding exact angles:
• In degree mode, find the arctan
  60, 240
tan is positive in quadrants I and III
3
Finding 2 angles…
• Find two angles for each of the following:
• cos θ= -.3100
• tan θ = -1.4180
• Find the first one in the calculator…
Finding 2 angles…
• Find two angles for each of the following:
• cos θ= -.3100
  108
• tan θ = -1.4180
  55  305
• Hint: Ask yourself when cos and tan are
negative!
Finding 2 angles…
• Find two angles for each of the following:
• cos θ= -.3100
  108, 252
• tan θ = -1.4180
  305,125
Using special triangles
• Set up a reciprocal and find:
2 3
csc  
3
Unknown angle AND reciprocal function: “flip” sides!
Using special triangles
• Set up a reciprocal and find:
2 3
csc 
3
3
sin  
2 3
sin1 (3 / (2 3))  60
Special angles:
• Chart:
θ
0
sin
0
1
cos
1
0
tan
0
30
45
1
60
90
und
Special angles:
• Chart:
θ
0
30
sin
0
1
2
cos
1
3
2
tan
0
3
3
45
1
2
2
2
2
60
90
3
2
1
1
2
0
3
und
Using the unit circle.
• Look at the unit circle to help find the
answer to # 62 on page 385.
All Students Take Calculus.
Quad II
Sin +
Tan +
Quad III
Quad I
All +
Cos +
Quad IV
All Students Take Calculus.
Quad I
Quad II
Sin +
Tan +
Quad III
cos(A)<0
sin(A)>0
tan(A)<0
sec(A)<0
csc(A)>0
cot(A)<0
cos(A)>0
sin(A)>0
tan(A)>0
sec(A)>0
csc(A)>0
cot(A)>0
cos(A)<0
sin(A)<0
tan(A)>0
sec(A)<0
csc(A)<0
cot(A)>0
cos(A)>0
sin(A)<0
tan(A)<0
sec(A)>0
csc(A)<0
cot(A)<0
All +
Cos +
Quad IV
Reference Angles-θ
Quad I
Quad II
θ’ = 180° – θ
θ’ = θ
θ
θ
θ’ = 180°+ θ
Quad III
θ
θ’ = 360° – θ
Quad IV
Reference Angles
Quad I
Quad II
θ’ = 180° – θ
θ’ = θ
θ’ = π – θ
θ’ = θ – 180°
θ’ = θ – π
Quad III
θ’ = 360° – θ
θ’ = 2π – θ
Quad IV
Using the unit circle• (cos x, sinx) are the (x,y) values on the
unit circle for the given angles.
• Decide the quadrant the angle is in
• Decide – or + for cos and sin
• Use special right triangles for the
proportion.
Unit circle
•
•
•
•
•
•
•
Radius of the circle is 1.
x = cos(θ)
 1  cos( )  1
y = sin(θ)
 1  sin( )  1
2
2
x

y
1
Pythagorean Theorem:
2
2
cos
(

)

sin
( )  1
This gives the identity:
Zeros of sin(θ) are n where n is an integer.

Zeros of cos(θ) are 2  n where n is an
integer.
Quick guide for quadrantals
• Sin and cos:
sin
cos
0
0
-1
1
-1
0
Trig identity for tangent
• Identity:
sin 
tan  
cos 
• So….
cos 
cot  
sin 
Quadrantal angles
• Using the unit circle:
(cos  ,sin  )
sin 
tan  
cos 
• Find the trig functions for 270, 90, 540
degrees.
• (Use coterminal angles between [0 ,360)
solutions
•
•
•
•
•
•
•
270,90, 540:
Sin 270 = -1
Cos 270 = 0
Tan 270 = -1/0 = undefined
Sin 90 = 1
Cos 90= 0
Tan 90 = 1/0 = undefined
540 = 180 degrees
• Sin 540 = 0
• Cos 540 = -1
• Tan 540 = 0/-1 = 0
Using reference angles
• Use the coterminal angle between 0 and
360 degrees
Find the exact value of sec 570
Using reference angles
• Use the coterminal angle between 0 and
360 degrees.
Find the exact value of
sec 570  sec 210
210
o
Has reference angle of 30
 3
-
30
-1 2
Using reference angles
• Use the coterminal angle between 0 and
360 degrees.
Find the exact value of
sec 570  sec 210
 3
-
30
-1 2
2 3

3
 3
2
Find the exact value of…
• csc 2250
Find the exact value of…
• csc 2250
225 has a reference angle of 45 in the third quadrant where
sine is negative.
Using the calculator:
2
sin 45 
2
2
2 2
csc 45 

 2
2
2
csc 225   2
Graphs of sine & cosine
f ( x)  A sin(Bx  C )  D
g ( x)  A cos(Bx  C )  D
•
•
•
•
•
Fundamental period of sine and cosine is 2π.
Domain of sine and cosine is .
Range of sine and cosine is [–|A|+D, |A|+D].
The amplitude of a sine and cosine graph is |A|.
The vertical shift or average value of sine and
cosine graph is D.
• The period of sine and cosine graph is 2B .
• The phase shift or horizontal shift is CB .
Sine graphs
y = sin(x)
y = 3sin(x)
y = sin(x) + 3
y = sin(3x)
y = sin(x – 3)
y = sin(x/3)
y = 3sin(3x-9)+3
y = sin(x)
Graphs of cosine
y = cos(x)
y = cos(x) + 3
y = 3cos(x)
y = cos(3x)
y = cos(x – 3)
y = cos(x/3)
y = 3cos(3x – 9) + 3
y = cos(x)
Tangent and cotangent graphs
f ( x)  A tan(Bx  C )  D
g ( x)  A cot(Bx  C )  D
• Fundamental period of tangent and cotangent is
π.
• Domain of tangent is x | x  2  n  where n is an
integer.
• Domain of cotangent x | x  n  where n is an
integer.
• Range of tangent and cotangent is .

• The period of tangent or cotangent graph is .
B
Graphs of tangent and cotangent
y = tan(x)
Vertical asymptotes at
x

2
 n .
y = cot(x)
Vertical asymptotes at
x  n.
Graphs of secant and cosecant
y = sec(x)

 n .
Vertical asymptotes at x 
2
Range: (–∞, –1] U [1, ∞)
y = cos(x)
y = csc(x)
Vertical asymptotes at x  n.
Range: (–∞, –1] U [1, ∞)
y = sin(x)
Inverse Trigonometric Functions
and Trig Equations
y  sin 1 ( x)  arcsin(x)
Domain: [–1, 1]
  
Range:  , 
 2 2
0 < y < 1, solutions in QI and QII.
–1 < y < 0, solutions in QIII and QIV.
1
y  cos ( x)  arccos(x)
y  tan1 ( x)  arctan(x)
Domain: [–1, 1]
Domain: 
Range: [0, π]
 
Range:   , 
0 < y < 1, solutions in QI and QIV.
–1< y < 0, solutions in QII and QIII.

2 2
0 < y < 1, solutions in QI and QIII.
–1 < y < 0, solutions in QII and QIV.
Trigonometric Identities
Summation & Difference Formulas
sin( A  B)  sin( A) cos(B)  cos(A) sin(B)
cos(A  B)  cos(A) cos(B)  sin( A) sin(B)
tan(A)  tan(B)
tan(A  B) 
1  tan(A) tan(B)
Solving trig equations
• Do now:
• Solve for both values of θ:
cos   .5
Solving trig equations
• Do now:
• Solve for both values of θ:
cos   .5
  120
0
• And in quadrant III, 180 +60 = 2400
• Now solve:
2 sin   1  2
solution
• Solving linear trig equations
2 sin  1  2
2 sin   1
1
sin  
2
1
  sin .5  30,150
Solving trig equations by gcf
factoring or square roots
• Do Now:
1. solve for x: 4x2 = 1
2. factor: 2xy + y
Solving trig equations by gcf
factoring or square roots
• Do Now:
1. solve for x: 4x2 = 1
1
x 
4
1
x 
2
2
2. factor: 2xy + y
y(2x +1)
Solving trig equations by gcf
factoring or square roots
1. solve for x: 4sin2 x = 1
1
sin x 
4
1
sin x  
2
2
2.2sinxcos x + sinx=0
sin x(2 cos x 1)  0
sin x  0, 2 cos x 1  0
Find all angles…
Solving trig equations by gcf
factoring or square roots
1. solve for x:
4sin2
x=1
2. 2sinxcos x + sinx=0
1
sin x 
4
1
sin x  
2
x  sin 1 .5  30,150
2
x  sin 1  .5  330, 21
sin x(2 cos x 1)  0
sin x  0, 2 cos x 1  0
x  0,180 cos x  .5
x  1200 ,2400
{0,120,180,240}
{30,150,210,330}
Solving by trinomial factoring:
• Do Now: solve for x:
c  5c  4  0
2
Solving by trinomial factoring:
• Do Now: solve for x:
c2  5c  4  0
(c  4)(c 1)  0
c  4,c  1
• Apply to trigonometry:
cos 2   5 cos  4  0
(cos  4)(cos 1)  0
cos  4, cos  1
reject,   0 0
Factor and solve
• For x:
[0,360)
2sin 2 x  sin x  3  0
• Think: 2x2 - x -3 = 0
Factor and solve
• For x:
[0,360)
2sin 2 x  sin x  3  0
(2sin x  3)(sin x 1)  0
3
sin x  , sin x  1
2
reject, x  270 0
Trigonometric Identities
Double Angle Formulas
sin(2 A)  2 sin( A) cos(A)
cos(2 A)  cos2 ( A)  sin 2 ( A)  1  2 sin 2 ( A)  2 cos2 ( A)  1
2 tan(A)
tan(2 A) 
1  tan2 ( A)
What to use for solving equations
Double Angle Formulas
sin(2A)  2 sin(A)cos(A)
cos(2A)  1  2 sin (A)
2
cos(2A)  2 cos (A) 1
2
Substitute and solve…[0, 360)
• sin 2x – sin x = 0
• Use the identity….sin 2x = 2sinxcosx
• 2sinxcosx-sinx=0 (use gcf)
Substitute and solve…[0, 360)
• sin 2x – sin x = 0
• Use the identity….sin 2x = 2sinxcosx
• 2sinxcosx-sinx=0 (use gcf)
•
sinx (2cosx – 1) = 0
• sinx = 0 2cosx – 1 = 0
• x=0,180
2cosx=1
•
cosx = .5
•
x = 60,300
Solve… [0, 360)
cos   2 sin 2
Remember to set = to 0 first!
Solve… [0, 360)
round to the nearest tenth
cos  2sin 2
cos  2sin 2  0
cos  2(2sin  cos )  0
cos  4 sin  cos  0
[0, 360)
Solve…
cos  2 sin 2
cos  2 sin 2  0
cos  2(2 sin  cos )  0
cos  4 sin  cos  0
cos (1  4 sin  )  0
cos  0 1  4 sin   0
  90, 270  4 sin   1
1
sin  
4
  14.5, 165.5
{14.5, 90, 165.5,270}
Cos2A…
• Choose right one!
cos 2  cos 
• Use the one that has cos and set = to 0!
Cos2A…
• Choose
cos 2  2 cos2  1
cos2  cos
2 cos  1  cos
2
2 cos   cos 1  0
2
Remember standard form, now factor and solve!
Cos2A…
• Choose
cos 2  2 cos2  1
cos 2  cos 
2 cos   1  cos
2
2 cos   cos   1  0
(2 cos   1)(cos   1)  0
2
Remember standard form, now factor and solve!
Cos2A…
• Choose
cos 2  2 cos2  1
cos 2  cos 
2 cos   1  cos
2
2 cos   cos   1  0
(2 cos   1)(cos   1)  0
cos  .5 cos  1
2
  120,240   0
Law of Sines & Law of Cosines
Law of sines
sin( A) sin(B) sin(C )


a
b
c
a
b
c


sin( A) sin(B) sin(C )
Use when you have a
complete ratio: SSA.
Law of cosines
c 2  a 2  b 2  2ab cos(C )
b 2  a 2  c 2  2ac cos(B)
a 2  b 2  c 2  2bc cos(A)
Use when you have SAS, SSS.
Law of Sines
Law of sines
sin( A) sin(B) sin(C )


a
b
c
a
b
c


sin( A) sin(B) sin(C )
Use when you have a
complete ratio: SSA or AAS
Law of Sines
Law of sines
sin(A) sin(B) sin(C)


a
b
c
AAS example: “Solve the triangle means, find:
Side b, side c and angle C.
A  36 ,RB  48 , a  8
0
0
Law of Sines
Find all sides and angles to the nearest whole:
sin(A) sin(B) sin(C)


a
b
c
A  360 ,RB  480 , a  8,RC  180  36  48  960
C
8
36
A
48
B
sin 36 sin 48

8
b
Law of Sines
Law of sines
sin(A) sin(B) sin(C)


a
b
c
A  360 ,RB  480 , a  8,RC  960 ,b  10
C
8
36
A
Now solve for c:
48
B
sin 36 sin 48

8
b
b sin 36  8 sin 48
8 sin 48
b
 10
sin 36
Law of Sines
Law of sines
sin(A) sin(B) sin(C)


a
b
c
A  360 ,RB  480 , a  8,RC  960 ,b  10
C
8
36
A
48
B
sin 36 sin 96

8
c
csin 36  8 sin 96
8 sin 96
c
 14
sin 36
Law of sines:
• Solve the triangle
• A=800, a= 14 ft. , b = 10 ft.
Steps:
B
14
800
A
10
C
1.Draw a triangle
2.Fill in the angles and side
3.Set up proportion and so
Law of sines:
• Solve the triangle
• A=800, a= 14 ft. , b = 10 ft.
B
14
10

sin 80 sin B
14
800
A
10
C
Law of sines:
• Solve the triangle
• A=800, a= 14 ft. , b = 10 ft.
14
10
B

sin 80 sin B
14
14
sin
B

10
sin
80
0
80
C
A
10 sin 80
10
sin B 
 .7034
14
1
o
B  sin (.7034)  45
Now: find angle C and side c:
Law of sines:
• Solve the triangle
• A=800, a= 14 ft. , b = 10 ft.
B
450
C= 180-(80 + 45) =550
14
800
A
10
C
14
c

sin 80 sin 55
14 sin 55
c
 12
sin 80
Law of sines
• Example:
A  26 0 , a  7, b  6
1. find sin B
(Round to 4 decimal places)
2. findRB
(To the nearest degree)
Law of sines
• Example:
A  26 0 , a  7, b  6
1. find sin B
(Round to 4 decimal places)
2. findRB
(To the nearest degree)
7
6
1.

sin 26 sin B
7 sin B  6 sin 26
6 sin 26
sin B 
 .3757
7
1
2. B  sin (.3757)  22
0
law of sines-ambiguous case
SSA-possibilities
• How many triangles possible?:
• Answer can be 0,1,or 2
• Ex: A=400, a= 4 ft. , b = 6 ft.
• Steps:
1.draw the triangle with dimensions
2.Set up law of sines and solve for B
law of sines• Solve the triangle for angle B:
• A=400, a= 4 ft. , b = 6 ft.
4
6

sin 40 sin B
B
a=4
400
A
C
b=6
law of sines-ambiguous case
• Solve the triangle:
• A=400, a= 4 ft. , b = 6 ft.
B
a=4
400
A
C
b=6
The supplement of 75 is 105….
4
6

sin 40 sin B
6 sin 40
sin B 
 .9642
4
1
B  sin (.9642)
 75
o
law of sines-ambiguous case
• How many triangles possible?
• A=400, a= 4 ft. , b = 6 ft.
Make a chart for 2 possiblities:
B
a=4
400
A
Triangle
1
A=40
B=75
C=65
Triangle
2
A=40
B=105
C=35
C
b=6
• 2 Triangles!
Solve the triangle
• A = 300, a=4, c= 10
Solve the triangle
• A = 300, a=4, c= 10
4
10

sin 30 sin c
Zero triangles!- there is an error message!
When only one triangle is
possible
• A=800, a= 14 ft. , b = 10 ft.
14
10

sin 80 sin B
14 sin B  10 sin 80
10 sin 80
sin B 
 .7034
14
B  sin 1 (.7034)  45 o
B’s supplement is 135…
Law of sines:
• Solve the triangle
• A=800, a= 14 ft. , b = 10 ft.
B
45
800
A
Only one triangle possible!
135
45
80
C
Too much already!
When the given angle is obtuse:
• Only 1 or zero triangles possible, just be
sure that the longer side is opposite the
largest angle.
Do now: set up the triangle:
• Given b=5 and c = 7 and A = 650, find a.
Law of Cosines
c  a  b  2ab cos(C)
2
2
2
b 2  a 2  c 2  2ac cos(B)
a  b  c  2bc cos(A)
2
2
2
Use when you have SAS, SSS.
Law of cosines for a side:
• Given b=5 and c = 7 and A = 650, find a.
a
5
650
7
Using the law of cosines
• Be sure that the side you start with is
opposite the cos of the angle.
Law of cosines for a side:
• Given b=5 and c = 7 and A = 650, find a.
a  5  7  2(5)(7)cos65
2
2
2
a  6.66
a
5
650
7
forces
• Using the law of cosines:
Two forces of 25 newtons and 85
newtons acting on a body form an
angle of 55°.
a. Find the magnitude of the resultant
force, to the nearest hundredth of a
newton.
b.Find the measure, to the nearest
degree, of the angle formed between
the resultant and the larger force.
Picture forces:
• Draw and then complete a parallelogram
25
55
85
Picture forces:
• Make a parallelogram
125
25
x
55
85
85
Picture forces:
• Make a parallelogram
125
85
25
x
55
x 2  25 2  85 2  2(25)(85)(cos125)
85
x  101.43
solve:
• b. Law of sines
125
85
θ
25
101.43
55
85
101.43 25

sin125 sin 
25sin125
sin  
101.43
0
  12
Law of cosines for an angle:
• Given b=7 and c = 8 and a =11, find C.
8
11
C
7
Law of cosines for an angle:
• Given b=7 and c = 8 and a =11, find C.
8 2  112  7 2  2(11)(7)cosC
64  121 49 154 cosC
Must be careful here! Do Not add the -154 to the 121+49
8
11
C
7
Law of cosines for an angle:
• Given b=7 and c = 8 and a =11, find C.
8 2  112  7 2  2(11)(7)cosC
64  121 49 154 cosC
64  170 154 cosC
106  154 cosC
106
 cosC
C
154
.6883  cosC
C  cos1 (.6883)
C  46.5 o
8
11
7
practice
• Pg. 559
• Find the largest angle for # 13
• Set up diagrams for 19 and 20
Do Now:
• Find the area of the triangle that has a
base of 20 and an altitude of 5.
Area of a triangle
• K = ½ absin C
You must have two sides and the
INCLUDED angle.
Area of a triangle:
• Given b=5 and c = 7 and A = 650, find the
area.
• K = ½ absin C
5
650
7
Area of a triangle:
• Given b=5 and c = 7 and A = 650, find the
area.
• K = ½ absin C
• K = ½ (5)(7)(sin 65)
k 15.86
5
650
7
Apply to a parallelogram:
• Find the area of the parallelogram:
12
60
14
Apply to a parallelogram:
• Find the area of the parallelogram:
• (a parallelogram is 2 triangles)
12
60
14
k  .5(12)(14)sin 60
k  72.7
2(72.7)  145.4
Formula for a parallelogram: K=absinC
Missing information
• Find the area of a triangle if e=10,f=6 and
• Angle E = 80…draw it!
Missing information
• Find the area of a triangle if e=10,f=6 and
• Angle E = 80
• We are missing angle G!
G
6
E 80
10
F
Missing information
• Find the area : e=10,f=6,E=80
• We are missing angle G but we can find F
using the law of sines and then find G
G
6
E 80
10
F
10
6

sin 80 sin F
F  36
G  180  (80  36)  64
Missing information
• Find the area : e=10,f=6,E=80
• We are missing angle G but we can find F
using the law of sines and then find G
6
E 80
G
64 10
F
10
6

sin 80 sin F
F  36
G  180  (80  36)  64
K  .5(6)(10)(sin 64
K  26.96
Picture forces:
• 2 forces are 28 and 85 and the angle
between them is 55 degrees. Find the
resultant
25
x
55
85
Picture forces:
• Make a parallelogram
125
25
x
55
85
85
Picture forces:
• Make a parallelogram
125
85
25
x
55
85
x  25  85  2(25)(85)cos125
2
2
x  101
2
Trig Applications:
• Use law of sines when you have a
• complete ratio: SSA or AAS
• Use the law of cosines when you have
• SAS or SSS
AMBIGUOUS CASE
• ASS
• Set up the law of sines and draw 2
triangles
• Find the missing angle. Put it in the
first triangle and it’s supplement in the
second triangle
• Decide if both triangles work!
Trigonometric Identities
Half Angle Formulas
1  cos(A)
 A
sin    
2
2
1  cos(A)
 A
cos   
2
2
1  cos(A)
 A
tan   
1  cos(A)
2
A
The quadrant of 2
determines the sign.