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Keesha is proving the law of cosines for a triangle with three acute angles. What is the next step of the proof?

Prove: [tex]a^2 = b^2 + c^2 - 2bc \cos A[/tex] for a triangle with three acute angles.

1. Draw an altitude from [tex]B[/tex] to segment [tex]\overline{AC}[/tex]. Label the point where the altitude and segment [tex]\overline{AC}[/tex] intersect [tex]D[/tex]. The altitude forms two right triangles within [tex]\triangle ABC[/tex].

2. By the Pythagorean theorem, [tex]c^2 = h^2 + x^2[/tex] and [tex]a^2 = h^2 + (b - x)^2[/tex].

3. Expanding the binomial squared, [tex]a^2 = h^2 + b^2 - 2bx + x^2[/tex].

4. Subtract the equation for [tex]c^2[/tex] from [tex]a^2[/tex], resulting in [tex]a^2 - c^2 = b^2 - 2bx[/tex].

Subtract [tex]b^2[/tex] from both sides of the equation, resulting in [tex]a^2 - b^2 - c^2 = -2bx[/tex].

Add [tex]2bx[/tex] to both sides of the equation, resulting in [tex]a^2 - c^2 + 2bx = b^2[/tex].

Add [tex]c^2[/tex] to both sides of the equation, resulting in [tex]a^2 = b^2 + c^2 - 2bx[/tex].

Note: Ensure that [tex]x = c \cos A[/tex] for the proof to align with the law of cosines.



Answer :

GinnyAnswer
Let's go through the next steps in detail after the initial steps provided:

1. Draw an altitude from [tex]\(B\)[/tex] to segment [tex]\(\overline{AC}\)[/tex]. Label the point where the altitude and segment [tex]\(\overline{AC}\)[/tex] intersect as [tex]\(D\)[/tex]. The altitude forms two right triangles within [tex]\(\triangle ABC\)[/tex].

2. By the Pythagorean theorem:
- For [tex]\(\triangle ABD\)[/tex]: [tex]\(c^2 = h^2 + x^2\)[/tex]
- For [tex]\(\triangle BDC\)[/tex]: [tex]\(a^2 = h^2 + (b - x)^2\)[/tex]

3. Expanding the binomial squared:
[tex]\[ a^2 = h^2 + b^2 - 2bx + x^2 \][/tex]

4. Subtract the equation for [tex]\(c^2\)[/tex] from [tex]\(a^2\)[/tex]:
[tex]\[ a^2 - c^2 = (h^2 + b^2 - 2bx + x^2) - (h^2 + x^2) \][/tex]
Simplify:
[tex]\[ a^2 - c^2 = b^2 - 2bx \][/tex]

5. Subtract [tex]\(b^2\)[/tex] from both sides of the equation:
[tex]\[ a^2 - b^2 - c^2 = -2bx \][/tex]

6. Add [tex]\(2bx\)[/tex] to both sides of the equation:
[tex]\[ a^2 - b^2 - c^2 + 2bx = 0 \][/tex]
Simplify:
[tex]\[ a^2 - c^2 + 2bx = b^2 \][/tex]

7. Add [tex]\(c^2\)[/tex] to both sides of the equation:
[tex]\[ a^2 = b^2 + c^2 - 2bx \][/tex]

To continue the proof and simplify this expression further, consider the relationship between [tex]\(x\)[/tex] and [tex]\(\cos(A)\)[/tex]:

8. By construction, [tex]\(x = c \cos(A)\)[/tex]. Substitute this into the equation:
[tex]\[ a^2 = b^2 + c^2 - 2b \cdot (c \cos(A)) \][/tex]

9. Therefore, the proof of the law of cosines is completed:
[tex]\[ a^2 = b^2 + c^2 - 2bc \cos(A) \][/tex]

And thus, we have successfully proven the law of cosines for a triangle with any angles, including acute angles.

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