Download Module 6 Lesson 2 Solving Triangles using Law of Cosines Part 2

Module 6 Lesson 2
Solving Triangles using Law of Cosines Part 2
Let’s wrap up the Law of Cosines by considering the other situation for which it is used:
Example 1:
a  9 cm, b  5 cm, c  7 cm
Solve this triangle:
As always, you should begin with a picture:
A
5 cm
C
7 cm
9 cm
B
The situation here comes from our chart in Lesson 1 notes SSS
When you are given three sides of a triangle, the correct procedure to solve the triangle
begins with using the Law of Cosines. Recall that the Law of Cosines provides a
relationship between all three sides and one angle of the triangle. When we use Law of
Cosines first in this situation, we will find one of the missing angles.
Does it matter which angle we find first? Yes. An important thing to remember in this
situation:
Always find the largest angle first.
Why do we find the largest angle first? Without going into too much detail, think back
to our Lesson 1 Part 2 notes. We talked about how there is ambiguity when using the
Law of Sines because at times, multiple solutions are possible such as an equation as:
sin x  0.5
Let’s just say to avoid these types of problems, we find the largest angle first. How do
we know which is the largest angle? Recall from Geometry that the largest angle will be
opposite the largest side. In our triangle, “a” is the largest side. So we’ll begin by
finding Angle A.
a 2  b2  c 2  2bc cos A
Here is the Law of Cosines version we need:
Remember that when solving for an angle, it might be better to use the rearranged
a 2  b2  c 2
version of this Law of Cosines:
 cos A
2bc
Let’s plug in our values and solve:
a 2  b2  c 2
 cos A
2bc
92  52  72
 cos A
2(5)(7)
7
 cos A
70
0.1  cos A
cos 1 ( 0.1)  cos 1 (cos A)
cos 1 ( 0.1)  A
95.7  A
Angle A = 95.7
To be safe again at this point, we should use the Law of Cosines to find the next largest
angle, C. We use the Law of Cosines formula version for angle C:
c 2  a 2  b2
 cos C
2ab
Let’s plug in our values and solve:
Angle C = 50.7
c 2  a 2  b2
 cos C
2ab
7 2  9 2  52
 cos C
2(9)(5)
57
 cos C
90
57
cos 1 ( )  cos 1 (cos C )
90
57
cos 1 ( )  C
90
50.7  C
We now find Angle B by subtraction: 180  95.7  50.7  33.6
The solution to our triangle is:
Example 1:
A  95.7 , B  33.6 , C  50.7
Solve this triangle:
a  21 m, b  16.7 m, c  10.3 m
Solve for Angle A first, since “a” is the largest side.
212  16.7 2  10.32
 cos A
2(16.7)(10.3)
56.02
 cos A
344.02
56.02
cos 1 (
)  cos 1 (cos A)
344.02
56.02
cos 1 (
) A
344.02
99.4  A
Next find Angle B:
b2  a 2  c 2
 cos B
2ac
16.72  212  10.32
 cos B
2(21)(10.3)
268.2
 cos B
432.6
268.2
cos 1 (
)  cos 1 (cos B )
432.6
268.2
cos 1 (
)B
432.6
51.7  B
Finally, we find Angle C by subtraction:
180  99.4  51.7  28.9
The solution to our triangle is:
Angle B  33.6
Angle C  28.9
A  99.4 , B  51.7 , C  28.9
Earlier in this module, I presented you with this outline. It represented the different
possible combinations of three pieces of information you might know about a triangle:
a. AAS Two angles and the side NOT in between them
b. ASA Two angles and the side in between them
c. SSA Two sides and an angle NOT in between them
d. SAS Two sides and the angle in between them
e. SSS Three sides
f. AAA Three angles
We have covered five of these situations and named the appropriate Law to use with
each one…
We now know that the first step is to use the Law of Sines to solve the situations of
AAS, ASA, and SSA. We also know that the first step is the use the Law of Cosines to
solve the situations of SAS and SSS.
That leaves us with one situation about which we’ve had no discussion yet. That
situation is AAA.
This last situation is the easiest to discuss for this reason: It is impossible to solve.
Why? Knowing three angles of a triangle alone does NOT determine a unique triangle.
We do not have enough information to solve.
The easiest way to look at this situation is with equilateral triangles. Equilateral
triangles have all three angles identical. Each is 60 . (Look at next page.)
Example 2:
Solve:
A  60 , B  60 , C  60
Look at these pictures. Each has the angle situation described above.
So which is the solution? That’s the point. We can’t tell. Just because we have three
angles of a triangle known, that doesn’t mean we can solve the triangle. In fact, it’s just
the opposite. We can’t.
A  95 , B  37 , C  48
Example 3:
Solve:
Solution:
This problem has no solution.