Ensino Médio ⇒ Lados e Ângulos num Triângulo Tópico resolvido

[andreluiz]

Seja ABC um triângulo com lados [tex3]a[/tex3], [tex3]b[/tex3] e [tex3]c[/tex3] e seus respectivos ângulos correspondentes opostos [tex3]\alpha[/tex3], [tex3]\theta[/tex3] e [tex3]\gamma[/tex3].
Prove que a igualdade [tex3]\alpha=3\cdot\theta[/tex3] implica na relação: [tex3](a^2-b^2)\cdot (a - b) = b\cdot c^2[/tex3]

[petrasMOD]

andreluiz,
[tex3]\dfrac{a}{sen \alpha}=\dfrac{b}{\sen\theta}=\dfrac{c}{\sen\gamma}\Longleftrightarrow \dfrac{a}{sen 3\theta}=\dfrac{b}{\sen\theta}=\dfrac{c}{\sen 4\theta}\\
sen(3\theta)=sen(2\theta+\theta)=sen2\theta cos\theta+cos2\theta sen\theta=2sen\theta cos \theta.cos \theta+cos (2\theta) sen\theta=sen\theta(2cos^2\theta+cos (2\theta))\\
a= \frac{b.sen(3\theta)}{sen\theta}=b(2cos^2 \theta+cos (2\theta))\\
sen(4\theta) = 2sen(2\theta)cos(2\theta) \implies c =\frac{b.2sen(2\theta)cos(2\theta)}{sen\theta}=\frac{b.2.2sen(2\theta)(cos(\theta)cos(2\theta)}{sen\theta}=4bcos \theta cos2\theta


\\
\Longrightarrow \cdots \Longrightarrow \left\{\begin{array}{l}a=b[2\cos^2\theta+\cos(2\theta)]\\c=4b\cos\theta\cos(2\theta)\end{array}\right.

[/tex3]
[tex3](a^2-b^2)(a-b)=[b^2(2\cos^2\theta+\cos(2\theta))^2-b^2][b[2\cos^2\theta+\cos(2\theta)]-b]=\\
b^3[(2\cos^2\theta+\cos(2\theta))^2-1][2\cos^2\theta+\cos(2\theta)-1]=\\
b^3[2\cos^2\theta+\cos(2\theta)-1]^2[2\cos^2\theta+\cos(2\theta)+1]=\\
4b^3\cos^2(2\theta)\cdot 4\cos^2\theta=\\16b^3\cos^2(2\theta)\cos^2\theta=b\cdot (4b\cos\theta\cos(2\theta))^2=bc^2c.q.d.
[/tex3]
(Solução>anipascual-adaptada)