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Prove: [tex]x^2 = z^2 + y^2 - 2yz \cos (X)[/tex]

Where is the first error made in the proof?

[tex]\[
\begin{array}{|l|l|}
\hline \text{Statements} & \text{Reasons} \\
\hline \Delta XYZ \text{ with altitude } h & \text{Given} \\
\hline \sin (X) = \frac{h}{z}, \cos (X) = \frac{r}{z} & \text{Definitions of sine and cosine} \\
\hline z \sin (X) = h, z \cos (X) = r & \text{Multiplication property of equality} \\
\hline x^2 = h^2 + (y - r)^2 & \text{Substitution property of equality} \\
\hline x^2 = (z \sin (X))^2 + (y - z \cos (X))^2 & \text{Substitution property of equality} \\
\hline
\end{array}
\][/tex]

Sagot :

GinnyAnsweridn
To prove the equation [tex]\(x^2 = z^2 + y^2 - 2 y z \cos(X)\)[/tex], we can break down the provided steps and verify their correctness:

1. Given: [tex]\(\Delta XYZ\)[/tex] with altitude [tex]\(h\)[/tex]
This is the basic setup of the problem, where a triangle [tex]\(XYZ\)[/tex] has an altitude [tex]\(h\)[/tex] from vertex [tex]\(X\)[/tex] to side [tex]\(YZ\)[/tex].

2. [tex]\(\sin(X) = \frac{h}{z}, \cos(X) = \frac{r}{z}\)[/tex]
These are definitions based on the given triangle. Specifically:
- [tex]\(\sin(X)\)[/tex] is the opposite side (altitude [tex]\(h\)[/tex]) over the hypotenuse [tex]\(z\)[/tex].
- [tex]\(\cos(X)\)[/tex] is the adjacent side ([tex]\(r\)[/tex]) over the hypotenuse [tex]\(z\)[/tex].

3. [tex]\(z \sin(X) = h, z \cos(X) = r\)[/tex]
By multiplying both sides of the trigonometric definitions by [tex]\(z\)[/tex], we solve for [tex]\(h\)[/tex] and [tex]\(r\)[/tex].

4. [tex]\(x^2 = h^2 + (y - r)^2\)[/tex]
This follows from the Pythagorean theorem in the right triangle formed by dropping the altitude [tex]\(h\)[/tex] from [tex]\(X\)[/tex] to side [tex]\(YZ\)[/tex]. The equation represents the sum of the squares of the legs of the right triangle [tex]\((h\)[/tex]) and the horizontal distance [tex]\((y - r)\)[/tex].

5. [tex]\(x^2 = (z \sin(X))^2 + (y - z \cos(X))^2\)[/tex]
Substituting the expressions [tex]\(h = z \sin(X)\)[/tex] and [tex]\(r = z \cos(X)\)[/tex] into the previous equation.

6. Expanding the squared terms:
[tex]\[ x^2 = (z \sin(X))^2 + (y - z \cos(X))^2 \][/tex]
[tex]\[ x^2 = z^2 \sin^2(X) + (y - z \cos(X))^2 \][/tex]
[tex]\[ x^2 = z^2 \sin^2(X) + (y^2 - 2yz \cos(X) + z^2 \cos^2(X)) \][/tex]
[tex]\[ x^2 = z^2 \sin^2(X) + y^2 - 2yz \cos(X) + z^2 \cos^2(X) \][/tex]

7. Using the trigonometric identity [tex]\(\sin^2(X) + \cos^2(X) = 1\)[/tex]:
[tex]\[ z^2 \sin^2(X) + z^2 \cos^2(X) = z^2 (\sin^2(X) + \cos^2(X)) = z^2 \cdot 1 = z^2 \][/tex]
Substituting this into the equation gives:
[tex]\[ x^2 = z^2 + y^2 - 2yz \cos(X) \][/tex]

Hence, all steps are valid, and no error is found in the proof. The final equation [tex]\(x^2 = z^2 + y^2 - 2 y z \cos(X)\)[/tex] is correct and consistent with trigonometric identities and the properties of the given triangle.
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