Pré-Vestibular ⇒ (FUVEST - 2012) Trigonometria

[Lars]

No triângulo acutângulo ABC, ilustrado na figura, o comprimento do lado BC mede [tex3]\frac{\sqrt{15}}{5}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{\sqrt{15}}{5}[/tex3]

, o ângulo interno de vértice C mede α, e o ângulo interno de vértice B mede α/2. Sabe-se, também, que 2 cos(2α ) + 3 cos α + 1 = 0.

3201CDA5-CEB6-4D34-AEC8-E9F6CCE5DD29.jpeg (3.83 KiB) Exibido 1188 vezes

Nessas condições, calcule

a) o valor de sen α;

b) o comprimento do lado AC.

Resposta

---

[2.(raiz15)]/15

Não consegui fazer item b, se alguém puder ajudar agradeceria.

[rcompany]

1)

[tex3]
\begin{align}
2\cos(2\alpha)+3\cos(\alpha)+1=0&\implies 4\cos^2\!\!\alpha+3\cos\alpha-1=0\\
&\implies \left\{\begin{array}{ll}\cos\alpha=\dfrac{-3-\sqrt{9+16}}{8}=-1\quad\text{impossível já que }\alpha<\pi\\\text{ou}\\\cos\alpha=\dfrac{-3+5}{8}=-\dfrac{1}{4}\end{array}\right.\\
&\implies \left\{\begin{array}{ll}\sin\alpha=\sqrt{1-\cos^2\alpha}=\dfrac{\sqrt{15}}{4}\\\text{ou}\\\sin\alpha=-\sqrt{1-\cos^2\alpha}=-\dfrac{\sqrt{15}}{4}\quad\text{impossível já que }0<\alpha<\pi\end{array}\right.\\
&\implies \sin\alpha=\frac{\sqrt{15}}{4}
\end{align}\\[96pt]


[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]
\begin{align}
2\cos(2\alpha)+3\cos(\alpha)+1=0&\implies 4\cos^2\!\!\alpha+3\cos\alpha-1=0\\
&\implies \left\{\begin{array}{ll}\cos\alpha=\dfrac{-3-\sqrt{9+16}}{8}=-1\quad\text{impossível já que }\alpha<\pi\\\text{ou}\\\cos\alpha=\dfrac{-3+5}{8}=-\dfrac{1}{4}\end{array}\right.\\
&\implies \left\{\begin{array}{ll}\sin\alpha=\sqrt{1-\cos^2\alpha}=\dfrac{\sqrt{15}}{4}\\\text{ou}\\\sin\alpha=-\sqrt{1-\cos^2\alpha}=-\dfrac{\sqrt{15}}{4}\quad\text{impossível já que }0<\alpha<\pi\end{array}\right.\\
&\implies \sin\alpha=\frac{\sqrt{15}}{4}
\end{align}\\[96pt]


[/tex3]

2)

[tex3]
\begin{align}
\text{Temos }\left.\begin{array}[m]{rl}
AB\sin\widehat{B}=AC\sin\widehat{C}\\
AC\cos\widehat{C}+AB\cos\widehat{B}=BC
\end{array}\right\}
&\implies AC\cos\widehat{C}+AC\dfrac{\sin\widehat{C}}{\sin\widehat{B}}\cos\widehat{B}=BC\\
&\implies AC=\dfrac{BC}{\cos\widehat{C}+\cos\widehat{B}\cdot\dfrac{\sin\widehat{C}}{\sin\widehat{B}}}\\
&\implies AC=\dfrac{BC}{\cos\alpha+\cos(\dfrac{\alpha}{2})\cdot \dfrac{\sin\alpha}{\sin(\dfrac{\alpha}{2})}}\\
&\implies AC=\dfrac{BC}{2\cos^2(\dfrac{\alpha}{2})-1+\cos(\dfrac{\alpha}{2})\cdot \dfrac{2\sin(\dfrac{\alpha}{2})\cos(\dfrac{\alpha}{2})}{\sin(\dfrac{\alpha}{2})}}\\
&\implies AC=\dfrac{BC}{4\cos^2(\dfrac{\alpha}{2})-1}\\
&\implies AC=\dfrac{BC}{2(2\cos^2(\dfrac{\alpha}{2})-1)+1}=\dfrac{BC}{2\cos\alpha+1}\\
&\implies AC=\dfrac{\sqrt{15}}{5}\cdot \dfrac{2}{3}=2\,\dfrac{\sqrt{15}}{15}
\end{align}\\[48pt]
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]
\begin{align}
\text{Temos }\left.\begin{array}[m]{rl}
AB\sin\widehat{B}=AC\sin\widehat{C}\\
AC\cos\widehat{C}+AB\cos\widehat{B}=BC
\end{array}\right\}
&\implies AC\cos\widehat{C}+AC\dfrac{\sin\widehat{C}}{\sin\widehat{B}}\cos\widehat{B}=BC\\
&\implies AC=\dfrac{BC}{\cos\widehat{C}+\cos\widehat{B}\cdot\dfrac{\sin\widehat{C}}{\sin\widehat{B}}}\\
&\implies AC=\dfrac{BC}{\cos\alpha+\cos(\dfrac{\alpha}{2})\cdot \dfrac{\sin\alpha}{\sin(\dfrac{\alpha}{2})}}\\
&\implies AC=\dfrac{BC}{2\cos^2(\dfrac{\alpha}{2})-1+\cos(\dfrac{\alpha}{2})\cdot \dfrac{2\sin(\dfrac{\alpha}{2})\cos(\dfrac{\alpha}{2})}{\sin(\dfrac{\alpha}{2})}}\\
&\implies AC=\dfrac{BC}{4\cos^2(\dfrac{\alpha}{2})-1}\\
&\implies AC=\dfrac{BC}{2(2\cos^2(\dfrac{\alpha}{2})-1)+1}=\dfrac{BC}{2\cos\alpha+1}\\
&\implies AC=\dfrac{\sqrt{15}}{5}\cdot \dfrac{2}{3}=2\,\dfrac{\sqrt{15}}{15}
\end{align}\\[48pt]
[/tex3]

[tex3]
\boxed{\\\hspace{1cm}\\\hspace{1cm}AC=2\,\dfrac{\sqrt{15}}{15}\hspace{1cm}\\}
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]
\boxed{\\\hspace{1cm}\\\hspace{1cm}AC=2\,\dfrac{\sqrt{15}}{15}\hspace{1cm}\\}
[/tex3]

[Lars]

rcompany, na resposta da b, eu entendi a primeira equação, tirada da aplicação da lei dos senos.
A segunda equação eu não entendi, é uma outra relação ou foi algo tirado do raciocínio da imagem?