MTH 250 Graded Assignment 9, part four
Proof of law of cosines
The Law of Cosines states that in a triangle
ABC
a 2  b 2  c 2  2bc cos A
b 2  a 2  c 2  2ac cos B
c 2  a 2  b 2  2ab cos C
There are three possible cases:
(1)
ABC is acute (all angles are acute)
(2)
ABC is obtuse (one obtuse angle)
(3)
ABC is right
Proof by cases.
Within cases (2) and (3), the choice of which angle is obtuse or right is arbitrary; it's not going to
make difference if A  90 , B  90 , or C  90 as the triangle can be oriented and labeled
any way.
Case 1:
ABC is acute ( A  90, B  90, C  90 )
1. Orient the triangle any way you like (I'll put vertex B at the top), and
construct an altitude, dividing side length b into b1 and b2 .
2. Use the Pythagorean Theorem to relate a, h, b2 .
[Fill in 1]
3. Use the Pythagorean Theorem to relate c, h, b1 , and solve for h 2 in terms of the other two
sides.
[Fill in 2]
2
4. Substitute, so you have an expression for a in terms of c, b1 , b2 .
[Fill in 3]
2
5. Add and subtract a term b on the right hand side.
[Fill in 4]
2
6. And observe that b  b1  b2 . Make a substitution into the subtracted b . You should now
have an expression:
[Fill in 5]
7. Which, if things have been going well up to this point, should simplify to
a 2  b 2  c 2  2b1 (b1  b2 )
[Fill in 6 - Show algebra that takes you from your expression in step 6 to this expression]
(If it doesn't, you need to go back and reconsider what you've got above.)
8. By definition, cos A  [Fill in 7], and so b1  [Fill in 8]. Also, b1  b2  b .
9. Substitute, and end up with the law of cosines.
[Do that.]
Since all angles acute, the proof for this case holds no matter how the triangle is oriented, and
we also get
b 2  a 2  c 2  2ac cos B
c 2  a 2  b 2  2ab cos C
Case 2:
ABC is obtuse (say A  90, B  90, C  90 )
Suppose ABC is obtuse, and we'll arbitrarily say that
is the obtuse angle.
B
As before, with the B vertex at the top, the proof of the
a 2  b 2  c 2  2bc cos A
variant is identical to Case 1, and the proof of the
c 2  a 2  b 2  2ab cos C
is the same (it's the mirror image on the other side) because of the acute angles. The one we
need to establish is the
b 2  a 2  c 2  2ac cos B
for B  90 ,
1. Re-orient the triangle so it sits on base AB , and construct an
altitude from vertex C . This requires you to extend AB as shown.
2
2. Use the Pythagorean Theorem to relate a, h, c1 . Solve for h .
[Fill in 1]
3. Use the Pythagorean Theorem to relate b, h, c, c1 .
[Fill in 2]
2
4. Substitute, so you have an expression for b in terms of a, c1 , c .
[Fill in 3]
5. Simplify. The c12 terms should cancel.
[Fill in 4]
6. Since B is obtuse, we define its cosine in terms of its supplement; by definition cos B  [Fill
in 5].
7. Since cos(180  B) = [Fill in 6], cos B  [Fill in 7], and c1  [Fill in 8].
8. Substitute, and end up with the law of cosines.
[Do that.]
The law of cosines is now established for this case.
Case 3:
ABC is right (say A  90, B  90, C  90 )
a 2  b 2  c 2  2bc cos A
c 2  a 2  b 2  2ab cos C
as in Case 1. We need to establish
b 2  a 2  c 2  2ac cos B
with B  90 .
1. With B  90 , the Pythagorean Theorem gives us (using lengths marked on triangle):
[Fill in 1]
2. Also, since B  90 , by definition, cos B  [Fill in 2]
3. And so since the term 2ac cos B is therefore equal to [Fill in 3] as well, we can just add it in
to the equation, and get the Law.
[Do that.]
And...done!