Transcript The Law of Sines

The Law
of
Sines
What if the triangle we want to solve is NOT a right
triangle? In the next two sections we’ll develop ways of
solving these triangles if we know at least one side and
two other pieces of info (sides or angles or one of each).

c
a


b
We’ll label these triangles with sides a, b and c and use
their Greek alphabet counterparts for the angles
opposite those sides as shown above.

c
a
h


b
Draw a perpendicular line
and call the length h. We do
this so that we have a right
triangle which we already
know how to work with.
Let’s write some trig functions we know from the right
triangles formed.
h
sin  
c
Solve these for h
Since these both = h
we can substitute
c sin   h
h
sin  
a
a sin   h
divide both sides by ac
c sin   a sin 
ac
ac
sin  sin 

a
c
This process can be repeated dropping a
perpendicular from a different vertex of the triangle.
What we get when we combine these is:
THE LAW OF SINES
THE LAW OF SINES
sin  sin  sin 


a
b
c
a
b
c


sin  sin  sin 
Use these to find
missing angles
Use these to find
missing sides
What this says is that you can set up the ratio of the
sine of any angle in a triangle and the side opposite
it and it will equal the ratio of the sine of any other
angle and the side opposite it. If you know three of
these pieces of information, you can then solve for
the fourth.
There are three possible configurations that will enable
us to use the Law of Sines. They are shown below.
You don’t have an angle and side opposite it here but can easily find
the angle opposite the side you know since the sum of the angles in a
triangle must be 180°.
ASA
You may have an angle, a
side and then another angle
SSA
You may have two sides
and then an angle not
between them.
SAA
You may have a side and
then an angle and then
another angle
What this means is that you
need to already know an
angle and a side opposite it
(and one other side or angle)
to use the Law of Sines.
Solve a triangle where  = 55°,  = 82° and c = 9
Draw a picture (just draw and
label a triangle. Don't worry
about having lengths and This is SAA
angles look right size)
Do we know an angle and side
opposite it? If so the Law of
9
Sines will help us determine

the other sides.
sin 82
How can you find ?
Hint: The sum of all the
angles in a triangle is 180°.
  180  55  82  43
a
9

sin 43 sin 82
9
c

55
43
a
6.20

82
7.44
b
b
b sin 82  9 sin 55
sin 55
9 sin 55  7.44
b
sin 82
How can you find a? (Remember it is
NOT a right triangle so Pythagorean
theorem will not work).
9 sin 43
a
 6.20
sin 82
You can use the
Law of Sines again.
Let's look at a triangle where you have SSA.
b
a

It could be that you can't get the sides to
join with the given info so these would be
"no solution".
It is easy to remember that the
It could tricky
be that there
one SSA
triangle that
onesis are
could be formed and you could solve the
(it's bassakwards).
triangle.
a
a

b
b

b
a

a, b and  are the same in both of these triangles.
It could be that since side c and  are not given that there are two ways
to draw the triangle and therefore 2 different solutions to the triangle
You can just check to see if there are two triangles whenever you have
the SSA case. The "no solution" case will be obvious when computing
as we will see.
Solve a triangle where  = 95°, b = 4 and c = 5

4
a
We have SSA. We know an angle and a side
opposite it so we'll use the Law of Sines.

95
5
sin   1.245
  sin 1 1.245
sin 95 sin 

4
5
5 sin 95
 sin 
4
We have the answer to sine and want to know the
angle so we can use inverse sine.
What happens when you put this in your calculator?
Remember the domain of the inverse sine function is numbers from
-1 to 1 since the sine values range from -1 to 1. What this means is
there is no solution. (You can't build a triangle like this).
Solve a triangle where  = 35°, b = 6 and c = 8

49.9
The smallest angle should
have the smallest side
opposite it and the largest
angle should have the largest
side opposite it.
6
a
10.42

95.1
35
8
sin 35 sin 

6
8
1
0.765


sin
sin   0.765
Knowing  and , can you find ?
Now how can you find a?
We have SSA again so we
know it could be the weird
one of no solution, one
solution or two solutions.
8 sin 35
 sin 
6
  49.9
180  35  49.9  95.1
sin 35 sin 95.1

6
a
6 sin 95.1
a
 10.42
sin 35
Since this was an SSA triangle we need to check to see if there are
two solutions. Remember your calculator only gives you one answer
on the unit circle that has the sine value of 0.765. You need to figure
out where the other one is and see if you can make a triangle with it.
Looking at the same problem: Solve a triangle where  = 35°,
b = 6 and c = 8
130.1

Let's check to see if there is another triangle possible.
We got 49.9° from the calculator. Draw a picture and
see if there is another angle whose sine is 0.765.
6
2.69
a
180 - 49.9 = 130.1
14.9

35
49.9
8
sin   0.765
Knowing  and , can you find ?
Now how can you find a?
So there IS another
triangle. (remember
our picture is not
drawn to scale).
180  35 130.1  14.9
6
a

sin 35 sin 14.9
6 sin 14.9
a
 2.69
sin 35
Solve a triangle where  = 42°, b = 22 and c = 12
21.4

sin 42 sin 

22
12
22
29.40
a
116.6

42
12
  sin 1 0.365  21.4
Since this is SSA we need to check the
other possible sine value for possibility
of a second triangle solution.
Knowing  and , can you find ?
sin 42
a

22
sin 116.6
sin   0.365
180  42  21.4  116.6
22 sin 116.6
a
 29.40
sin 42
Solve a triangle where  = 42°, b = 22 and c = 12
158.6

Not possible to build another triangle with these
stipulations.
22
a

42
180 - 21.4 = 158.6
21.4
12
Knowing  and , can you find ?
180  42 158.6  20.6
This negative number tells us that there is no second triangle so this is the
one triangle solution.
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au