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Given: [tex]$\triangle ABC$[/tex] with altitude [tex]$h$[/tex].

Two right triangles are formed: one with side lengths [tex]$c + r$[/tex], [tex]$h$[/tex], and [tex]$b$[/tex], and one with side lengths [tex]$r$[/tex], [tex]$h$[/tex], and [tex]$a$[/tex].

Carson starts the proof of the law of cosines with [tex]$\sin (A) = \frac{h}{b}$[/tex] by the definition of the sine ratio and [tex]$\cos (A) = \frac{c + r}{b}$[/tex] by the definition of the cosine ratio.

What are the next steps in the proof?

Use the [tex]$\square$[/tex] transitive property of equality to rewrite each trigonometric equation in terms of the numerator.

Then, Carson can write an expression for side [tex]$\square$[/tex] in terms of [tex]$\square$[/tex].

Next, he can use the [tex]$\square$[/tex] to relate [tex]$a, b, c$[/tex], and [tex]$A$[/tex].



Answer :

Carson starts the proof of the law of cosines using the definitions of sine and cosine. Here's a detailed step-by-step solution for completing the next parts of the proof:

1. Transitive Property of Equality:
- Transitive property states that if [tex]\(a = b\)[/tex] and [tex]\(b = c\)[/tex], then [tex]\(a = c\)[/tex]. This will be used to rewrite each trigonometric equation in terms of the numerator.

So, we start with:
[tex]\[ \sin(A) = \frac{h}{b} \][/tex]
and
[tex]\[ \cos(A) = \frac{c + r}{b} \][/tex]

Using the transitive property of equality to rewrite in terms of [tex]\(h\)[/tex] and [tex]\(c+r\)[/tex]:
[tex]\[ h = b \sin(A) \][/tex]
and
[tex]\[ c + r = b \cos(A) \][/tex]

2. Expression for Side [tex]\(r\)[/tex]:
- Next, we need to find the expression for side [tex]\( r \)[/tex]. We relate [tex]\( r \)[/tex] in terms of [tex]\( a, h, \)[/tex] and [tex]\(\cos(A)\)[/tex]:
[tex]\[ r = a \cos(A) \][/tex]

3. Relate [tex]\(a, b, c,\)[/tex] and [tex]\(A\)[/tex]:
- Use the Pythagorean theorem in both right triangles formed by the altitude [tex]\( h \)[/tex]:

For the triangle with side lengths [tex]\(r, h,\)[/tex] and [tex]\(a\)[/tex]:
[tex]\[ a^2 = r^2 + h^2 \][/tex]

Substitute [tex]\(r = a \cos(A)\)[/tex]:
[tex]\[ a^2 = (a \cos(A))^2 + h^2 \][/tex]
[tex]\[ a^2 = a^2 \cos^2(A) + h^2 \][/tex]

For the triangle with side lengths [tex]\(c + r, h,\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ b^2 = (c + r)^2 + h^2 \][/tex]

Substitute [tex]\(r = a \cos(A)\)[/tex]:
[tex]\[ b^2 = (c + a \cos(A))^2 + h^2 \][/tex]
[tex]\[ b^2 = c^2 + 2ac \cos(A) + a^2 \cos^2(A) + h^2 \][/tex]

4. Combining the Results:
- Finally, since [tex]\( h^2 = a^2 - a^2 \cos^2(A) \)[/tex], we substitute it into the equation:
[tex]\[ h^2 = a^2 \sin^2(A) \][/tex]
- Adding this to [tex]\( b^2 \)[/tex], we get:
[tex]\[ b^2 = c^2 + 2ac \cos(A) + a^2 \cos^2(A) + a^2 \sin^2(A) \][/tex]
- Simplifying using the trigonometric identity [tex]\( \cos^2(A) + \sin^2(A) = 1 \)[/tex]:
[tex]\[ b^2 = c^2 + 2ac \cos(A) + a^2 ( \cos^2(A) + \sin^2(A)) \][/tex]
[tex]\[ b^2 = c^2 + 2ac \cos(A) + a^2 \][/tex]
This is the law of cosines.

Thus, Carson successfully completes the proof using these steps.