The first equation will be proven for . The remaining equations can be proven similarly.
a2=b2+c2−2bccosA
Consider
△ABC with side lengths of
a, b, and
c, respectively opposite the angles with measures
A, B, and
C, such that
m∠A is greater than
90∘.
The altitude of the is the segment from B to the extension of the base AC. Let D be the endpoint of this and x be the distance from D to A.
From the definition of an altitude, it follows that
△BDA and
△BDC are right triangles. Two equations can be obtained by applying the Pythagorean Theorem to both triangles.
Equation I: Equation II: a2=h2+(b+x)2 c2=h2+x2
Expand the binomial in Equation I.
a2=h2+(b+x)2
a2=h2+b2+2bx+x2
a2=h2+x2+b2+2bx
From Equation II,
h2+x2 is equal to
c2. Therefore, the expanded form of Equation I can be rewritten by using the .
a2=h2+x2+b2+2bx
a2=c2+b2+2bx
Note that
A and
180∘−A are . Using the cosine ratio of
180∘−A then gives an for the
x-term.
In
△BDA, the cosine of
180∘−A is the ratio of
x to
c. cos(180∘−A)=cx⇕x=ccos(180∘−A)
By the ,
cos(180∘−A) and
cosA have opposite values.
x=ccos(180∘−A)⇕x=-ccosA
Finally, by substituting
-ccosA for
x into
a2=b2+c2+2bx, the Law of Cosines is obtained.
a2=b2+c2+2bx
a2=b2+c2+2b(-ccosA)
a2=b2+c2−2bccosA