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Question

Explain how the Pythagorean Theorem is a special case of the Law of Cosines.

Solution

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Answered 2 years ago
Answered 2 years ago
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To find the formula for cosine on triangles ABC and ACD we apply the Pythagorean theorem.

ADC:b2=h2+x2h2=b2x2ABD:c2=h2+(ax)2c2=b2x2+(ax)2c2=b2x2+a22ax+x2 c2=b2+a22axcosC=xbx=bcosCc2=b2+a22abcosC a2=b2+c22bccosAb2=a2+c22accosB\begin{align*} \triangle ADC:\\ b^{2}&=h^{2}+x^{2}\\ h^{2}&=b^{2}-x^{2}\\ \triangle ABD:\\ c^{2}&=h^{2}+(a-x)^{2}\\ c^{2}&=b^{2}-x^{2}+(a-x)^{2}\tag{$h^{2}=b^{2}-x^{2}$}\\ c^{2}&=b^{2}-x^{2}+a^{2}-2ax+x^{2}~\tag{$(a-x)^{2}=a^{2}-2ax+x^{2}$}\\ c^{2}&=b^{2}+a^{2}-2ax\\ \cos C&=\dfrac{x}{b}\\ x&=b\cdot \cos C\\ c^{2}&=b^{2}+a^{2}-2ab \cos C~\tag{for $x=b\cdot \cos C$}\\ a^{2}&=b^{2}+c^{2}-2bc \cos A\\ b^{2}&=a^{2}+c^{2}-2ac \cos B\\ \end{align*}

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