Transcript Law of Sines - HCC Learning Web

Law of Sines
Section 7.1
Deriving the Law
of Sines
C

b
a
h
α
h
 sin A
b
A
h
 sin B
a
h  b sin A
h  a sin B
b sin A  a sin B
sin A sin B

a
b
β
c
B
Since we could draw another altitude and perform the
same operations, we can extend the last equality by
including the other ratio. This gives us the LAW OF SINES
sin A sin B sin C


a
b
c
Example 1
Find the missing parts of triangle ABC if A  60 , a  45 in, and b  20 in
sin A sin B

a
b
sin60
sin B

45
20
20 sin60
 sin B
45
.3849  sinB
B  22.6
C  180  60  22.6
C  97.4
c
a

sin C sin A
c
45

sin97.4
sin 60
c
45 sin97.4
sin60
c  51.5
If you use a chart or table it looks like this:
A  60
a  45 in.
B  22.6
b  20 in.
C  97.4
c  51.5 in.
Example 2
Solve the following triangle:
A  40 ,C  22.5 , b  12
Since we know two of the angles of the triangle, we can find the
third angle by subtracting the two known angles from 180.
180 - 40 - 22.5 = 117.5o
B = 117.5o
Now we can use the Law of Sines to solve for either a or c.
We will need to use the Law of Sines twice, once for each
remaining side.
12
a
=
sin117.5o sin 40o
12
c
=
sin117.5o sin22.5o
12 sin 40o
a=
sin117.5o
12 sin 22.5o
c=
sin117.5o
Use your calculator to solve for a and c.
a = 8.7
c = 5.2
Since the given side is an integer, round each value to the nearest
integer.
a9
c 5
Example 3
Solve the following triangle: b  100 ft., c  60 ft.,   28.0
Since we know both c and γ we can use the law of sines.
sin  sin28

100
60
100 sin28
sin  
60
sin   .7825
1
We find β by using sin .7825
The problem we have is that there is a first quadrant angle and a
second quadrant angle with this sine.
Find both angles that have the sine we found.
sin1 .7825   51.5
180  51.5  128.5
Is it possible for both values to work in this triangle?
We check by finding the sum of β and each of the values we found.
If the sum is less than 180, then the angle can be part of this
triangle.
28.0  51.5  79.5
28.0  128.5  156.5
Since each of these is less than 180, there is a third angle
that will make a triangle in both cases. Find each angle and
then use the Law of Sines to find the remaining side in
each case.
A  180  79.5  100.5
A  180  156.5  23.5
a
60

sin100.5 sin28.0
60 sin100.5
a
sin28.0
a  126
a
60

sin23.5 sin28.0
60 sin23.5
a
sin28.0
a  51
A  100.5 or 23.5
a  126 ft. or 51 ft.
B  51.5 or 128.5
b  100 ft.
C  28.0
c  60 ft.
Example 4
We have done two examples with one angle and two
sides given with different results. The first time there was
one triangle and the second time two triangles. Let’s look
at a third case.
a  21, b  14,   100
Since we know both b and β, we can use the Law of
Sines to find α.
sin  sin100

21
14
21sin100
sin  
14
At this point you can use the sin-1 function on your calculator or
you can find sinα.
sin α = 1.4772
Since the sine is never greater than 1, we know there is a
problem. This means there can be no triangle with the
given information.
If you use the sin-1 method, your calculator gives you an
error message. This also means there can be no triangle.
This case is called the ambiguous case because there are
three possible outcomes. When given two sides and an angle
opposite one of the sides, there could be none, one, or two
triangles. You must always check to see which case you have.
Example 5
A hot air balloon is sighted at the same time by two friends
who are 2 miles apart on the same side of the balloon. The
angles of elevation from the two friends to the balloon are
20.5° and 25.5° respectively. How high is the balloon?
h
20.5°
2 mi
25.5°
There are several triangles in
the diagram, including two right
triangles, but none of them have
enough given information that
can be used to solve for
anything. If you look at the small
triangle on the left, you should
be able to observe that we can
find the remaining two angles.
Then we can solve for a side.
180  25.5  154.5 This is the angle supplementary to 25.5
180  20.5  154.5  5 This is the angle at the top of the small triangle
With this information, we can solve for one of the
two unknown sides of the small triangle. The
problem can be solved using either side, so it
doesn’t matter which one we pick. Let’s choose
the one on the left of the diagram. We’ll call it x.
x
2

sin154.5 sin5
2 sin154.5
x
sin5
x  9.88
Using the right triangle with x the hypotenuse
and h the side opposite the 20.5° angle, we
can find h.
h
9.88
h  9.88 sin20.5
h  3.46
sin20.5 
The balloon is 3.5 mi high.
You have some problems assigned that you should now work. Practice will help
you understand the Law of Sines. Good Luck!