1. Use Pythagorean theorem to rewrite each trigonometric equation in terms of the numerator.
- [tex]$\sin (A)=\frac{h}{b} \Rightarrow h = b \sin(A)$[/tex]
- [tex]$\cos (A)=\frac{c+r}{b} \Rightarrow c + r = b \cos(A)$[/tex]
2. Then, Carson can write an expression for side [tex]\( r \)[/tex] in terms of [tex]\( c \)[/tex].
- From the second equation, [tex]\( r = b \cos(A) - c \)[/tex]
3. Next, he can use the Pythagorean theorem to relate [tex]\( a, b, c \)[/tex], and [tex]\( A \)[/tex].
- Using the right triangle with side lengths [tex]\( r, h \)[/tex], and [tex]\( a \)[/tex]:
[tex]\[
a^2 = r^2 + h^2
\][/tex]
Substituting [tex]\( r = b \cos(A) - c \)[/tex] and [tex]\( h = b \sin(A) \)[/tex]:
[tex]\[
a^2 = (b \cos(A) - c)^2 + (b \sin(A))^2
\][/tex]
[tex]\[
a^2 = (b \cos(A) - c)^2 + (b \sin(A))^2
\][/tex]
[tex]\[
a^2 = b^2 \cos^2(A) - 2bc \cos(A) + c^2 + b^2 \sin^2(A)
\][/tex]
Using the identity [tex]\(\cos^2(A) + \sin^2(A) = 1\)[/tex]:
[tex]\[
a^2 = b^2 (\cos^2(A) + \sin^2(A)) - 2bc \cos(A) + c^2
\][/tex]
[tex]\[
a^2 = b^2 \cdot 1 - 2bc \cos(A) + c^2
\][/tex]
[tex]\[
a^2 = b^2 + c^2 - 2bc \cos(A)
\][/tex]
And that's the proof of the law of cosines.
