Question

where is the first error made in the proof?
statements
△xyz with altitude h
sin(x)=\frac{h}{z}, cos(x)=\frac{r}{z}
zsin(x)=h, zcos(x)=r
x^{2}=h^{2}+(y - r)^{2}
x^{2}=(zsin(x))^{2}+(y - zcos(x))^{2}
x^{2}=(z^{2}sin^{2}(x)+y^{2}-2yzcos(x)+z^{2}cos^{2}(x)
x^{2}=z^{2}sin^{2}(x)+cos^{2}(x)+y^{2}-2yzcos(x)
sin^{2}(x)+cos^{2}(x)=\frac{h^{2}}{z^{2}}+\frac{r^{2}}{z^{2}}
x^{2}=z^{2}(\frac{h^{2}}{z^{2}}+\frac{r^{2}}{z^{2}})+y^{2}-2yzcos(x)
x^{2}=z^{2}(\frac{h^{2}+r^{2}}{z^{2}})+y^{2}-2yzcos(x)
z^{2}=h^{2}+r^{2}
x^{2}=z^{2}(\frac{z^{2}}{z^{2}})+y^{2}-2yzcos(x)
x^{2}=z^{2}+y^{2}-2yzcos(x)
reasons
given
definitions of sine and cosine
multiplication property of equality
substitution property of equality
substitution property of equality
expanding the squared terms
factoring
square both sides of the equation
substitution property of equality
addition of fractions
pythagorean theorem
substitution property of equality
multiplicative inverse property

Explanation:

Step1: Analyze the Pythagorean theorem step

The Pythagorean theorem is used for right - angled triangles. In the proof of the law of cosines for \(\triangle XYZ\) with altitude \(h\), the step \(z^{2}=h^{2}+r^{2}\) is correct as it applies the Pythagorean theorem to the right - angled triangle formed by the altitude \(h\), segment \(r\), and side \(z\).

Step2: Analyze the substitution step \(x^{2}=z^{2}(\frac{h^{2}+r^{2}}{z^{2}})+y^{2}-2yz\cos(X)\)

We know that \(z\sin(X) = h\) and \(z\cos(X)=r\). If we substitute \(h = z\sin(X)\) and \(r = z\cos(X)\) into \(z^{2}=h^{2}+r^{2}\), we get \(z^{2}=z^{2}\sin^{2}(X)+z^{2}\cos^{2}(X)\). Then \(\frac{h^{2}+r^{2}}{z^{2}}=\sin^{2}(X)+\cos^{2}(X) = 1\). But the substitution \(x^{2}=z^{2}(\frac{h^{2}+r^{2}}{z^{2}})+y^{2}-2yz\cos(X)\) is wrong. The correct substitution should be based on expanding \((y - r)^{2}+h^{2}\). Expanding \((y - r)^{2}+h^{2}=y^{2}-2yr + r^{2}+h^{2}\). Since \(r = z\cos(X)\) and \(h = z\sin(X)\) and \(z^{2}=h^{2}+r^{2}\), we have \(x^{2}=y^{2}+z^{2}-2yz\cos(X)\)

Answer:

The first error is in the step \(x^{2}=z^{2}(\frac{h^{2}+r^{2}}{z^{2}})+y^{2}-2yz\cos(X)\). The Pythagorean theorem gives \(z^{2}=h^{2}+r^{2}\), so \(\frac{h^{2}+r^{2}}{z^{2}} = 1\), and this substitution is incorrect as it does not follow the proper expansion of \((y - r)^{2}+h^{2}\) (where \(r = z\cos(X)\) and \(h = z\sin(X)\)) to derive the law of cosines formula \(x^{2}=y^{2}+z^{2}-2yz\cos(X)\)