PRE-CALCULUS LESSON 6.1
LAW OF SINES
Spring 2011
Question! How to measure the depth?
http://www.mcs.uvawise.edu/dbl5h/resources/latex_examples/files/triglawsines.pdf
At the end of this lesson you should be able to




Use the Law of Sines to solve oblique triangles (AAS
or ASA).
Use the Law of Sines to solve oblique triangles
(SSA).
Find areas of oblique triangles.
Use the Law of Sines to model and solve real-life
problems.
Introduction
1.
To solve an oblique triangle, we need to be given at least
one side and then any other two parts of the triangle.
3.
The sum of the interior angles is 180°.
Why can’t we use the Pythagorean Theorem?
4.
State the Law of Sines.
2.
C
a
b
5.
6.
Area = ½ base*height
Area= ½*a*b*sineC
A
c
B
Introduction (continued)
7.
Draw a diagram to represent the information. ( Do not
solve this problem.)
A triangular plot of land has interior angles A=95° and
C=68°. If the side between these angles is 115 yards long,
what are the lengths of the other two sides?
Playing with a the triangle
If we think of h as
being opposite to
both A and B, then
C
b
A
a
h
c
Let’s drop an altitude
and call it h.
sin A 
h
and sin B 
b
B
h
a
Let’s solve both for h.
h  b sin A and h  a sin B
This means
b sin A  a sin B and dividing by ab .
sinA
a

sin B
b
C
If I were to drop an altitude to
side a, I could come up with
a
b
A
c
sin B
B

sin C
b
c
Putting it all together gives us
the Law of Sines.
sin A
Taking reciprocals,
we have

sin B
a
b
a
b
sin A

sin B

sin C
c

c
sin C
What good is it?
The Law of Sines can be used to solve the following types of
oblique triangles
•Triangles with 2 known angles and 1 side (AAS or ASA)
•Triangles with 2 known sides and 1 angle opposite one of the
sides (SSA)
With these types of triangles, you will almost always have
enough information/data to fill out one of the fractions.
sin A
a

sin B
b

sin C
c
Example 1 (AAS)
C
Let A  45  , B  50  , a  30
85°
b
50°
45°
A
a =30
c
B
I’m given both pieces for
sinA/a and part of sinB/b,
so we start there.
sin 45 
30
Once I have 2 angles, I can
find the missing angle by
subtracting from 180.
C=180 – 45 – 50 = 85°

sin 50 
b
Cross multiply
and divide to get
b sin 45   30 sin 50 
b
30 sin 50 
sin 45 
C
85°
b  3 2 .5 
sin A
a =30
50°
45°
c  42.3 
A
b

sin C
a
c
sin 45 
B
30 sin 50 
sin 45 

sin 85 
30
c
c sin 4 5   3 0 sin 8 5 
c
3 0 sin 8 5 
sin 4 5 
U sing a calculator,
b  32.5
We’ll repeat the process to find
side c.
Remember to avoid rounded
values when computing.
U sing a calculator,
c  42.3
We’re done when we
know all 3 sides and all 3
angles.
Example 2 ASA
A triangular plot of land has interior angles A=95° and C=68°. If the side
between these angles is 115 yards long, what are the lengths of the other two
sides?
The Ambiguous Case (SSA)
Three possible situations
1.
No such triangle exists.
2.
Only one such triangle exists.
3.
Two distinct triangles can satisfy the conditions.
Example 3(SSA)
Use the Law of Sines to solve the triangle.
A = 26, a = 21 inches, b = 5 inches
C
a

sin A
21
sin 26 
sin B 
a = 21 in
b
sin B

5
26
B
b = 5 in
A
a > b : One Triangle
sin B
1
5  sin 26 
B  sin (
21
B  5.99 
5  sin 26 
21
)
Example 3(SSA)
Use the Law of Sines to solve the triangle.
A = 26, a = 21 inches, b = 5 inches
C
C  180  ( 26  5 . 99 )  148 . 01 
a = 21 in
B
21
sin 26 
c

5.99°
26
b = 5 in
A
c
sin 148 . 01 
21  sin 148 . 01 
sin 26 
 25 . 38
a > b : One Triangle
Example 4(SSA)
Use the Law of Sines to solve the triangle.
A = 76, a = 18 inches, b = 20 inches
a  b
sin A sin B
a = 18 in
18
 20
sin 7 6  sin B
sin B  1 .0 7 8 
There is no angle whose sine is 1.078.
B
b = 20 in
h
76
a < h:None
h  b sin A
h  20 sin 76 
h  19 . 4
There is no triangle satisfying the given conditions.
A
Example 5(SSA)
Let A = 40°, b = 10, and a = 9.
C
We have enough information for
a=9
b = 10
sin A

sin B
a
sin 40 
9
h=6.4
b

A
sin B
c
B
h < a < b: Two
10
Cross multiply and
divide
40°
9 sin B  10 sin 40 
sin B 
10 sin 40 
9
h  b sin A
h  10 sin 40 
h  6 .4
C
To get to angle B, you must
unlock sin using the inverse.
sin
1
 sin B   sin
B  sin
1
1
 10 sin 40  


9


 10 sin 40  

  45.6 
9


Once you know 2 angles, find the
third by subtracting from 180.
b = 10
A
a=9
45.6°
40°
B
c =14.0
sin A

sin C
a
c
sin 40 
C = 180 – (40 + 45.6) = 94.4°
We’re ready to look for
side c.
94.4°

sin 94.4 
9
c
c sin 4 0   9 sin 9 4 .4 
c 
9 sin 9 4 .4 
sin 4 0 
 1 4 .0
Example 5
Finding the Second Triangle
Let A = 40°, b = 10, and a = 9.
Start by finding B’ = 180 - B
Now solve this triangle.
b= 10
C’
40°
a=9
B’ B = 45.6°
A
b= 10
a=9
A
40° 134.4°
B’
c’
B’ = 180 – 45.6 = 134.4°
C’
5.6°
Next, find C’ = 180 – (40 + 134.4)
C’ = 5.6°
b= 10
a=9
134.4°
40°
A
c =1.4
B’
sin A

sin C
a
c
sin 40 

sin 5.6
9
c
c sin 40   9 sin 5.6 
c
9 sin 5.6 
sin 40 
 1.4
A New Way to Find Area
C
We all know that A = ½ bh.
And a few slides back we
found this.
h  b sin A and h  a sin B
b
A
a
h
c
Area  1 bc sin A  1 ab sin C  1 ac sin B
2
2
2
Area = ½*product of two given sides * sine of the included angle
B
Example 6
Finding the area of the triangle
Find the area of a triangle with side a = 10, side b
= 12, and angle C = 40°.
Example 7 Application
Two fire ranger towers lie on the east-west line and are 5
miles apart. There is a fire with a bearing of N27°E from tower
1 and N32°W from tower 2. How far is the fire from tower 1?
The angle at the fire is 180° - (63° + 58°) = 59°.
N
1
S
N
x
63°
58°
2
5 mi
S
22
Example 8 Application
http://www.mcs.uvawise.edu/dbl5h/resources/latex_examples/files/triglawsines.pdf
Practice
p.398-400 #s 2-38, even
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