Sure, let's continue with the step-by-step proof for the law of cosines based on the given information.
1. Rewrite each trigonometric equation in terms of the numerator: Given
[tex]\[
\sin(A) = \frac{h}{b} \implies h = b \sin(A)
\][/tex]
and
[tex]\[
\cos(A) = \frac{c+r}{b} \implies c+r = b \cos(A).
\][/tex]
2. Expression for side length [tex]$r$[/tex] in terms of [tex]$b$[/tex] and [tex]$\cos(A)$[/tex]:
Since [tex]\(c+r = b \cos(A)\)[/tex],
we solve for [tex]\(r\)[/tex] as follows:
[tex]\[
r = b \cos(A) - c.
\][/tex]
3. Using the Pythagorean theorem for the two right triangles:
We know from the right triangle with side lengths [tex]\(r, h, a\)[/tex] that:
[tex]\[
a = \sqrt{r^2 + h^2}.
\][/tex]
Substitute [tex]\(r = b \cos(A) - c\)[/tex] and [tex]\(h = b \sin(A)\)[/tex]:
[tex]\[
a = \sqrt{(b \cos(A) - c)^2 + (b \sin(A))^2}.
\][/tex]
4. Relate [tex]\(a, b, c\)[/tex], and [tex]\(A\)[/tex]:
Expand and simplify:
[tex]\[
a^2 = (b \cos(A) - c)^2 + (b \sin(A))^2.
\][/tex]
[tex]\[
a^2 = (b \cos(A))^2 - 2b \cos(A) \cdot c + c^2 + (b \sin(A))^2.
\][/tex]
Since [tex]\((b \cos(A))^2 + (b \sin(A))^2 = b^2(\cos^2(A) + \sin^2(A)) = b^2\)[/tex]:
[tex]\[
a^2 = b^2 - 2bc \cos(A) + c^2.
\][/tex]
Hence, the final relationship is:
[tex]\[
a^2 = b^2 + c^2 - 2bc \cos(A).
\][/tex]
The missing terms for the drop-down menus are:
1. Use the [tex]\(\boxed{\sin}\)[/tex] to rewrite each trigonometric equation in terms of the numerator.
2. Then, Carson can write an expression for side [tex]\(\boxed{r}\)[/tex] in terms of [tex]\(\boxed{\b \cos(A) - c}\)[/tex].
3. Next, he can use the [tex]\(\boxed{Pythagorean~theorem}\)[/tex] to relate [tex]\(a, b, c,\)[/tex] and [tex]\(A\)[/tex].
